Rd Sharma 2021 for Class 9 Maths Chapter 5 - Factorization Of Algebraic Expressions

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Rd Sharma 2021 Solutions for Class 9 Maths Chapter 5 Factorization Of Algebraic Expressions are provided here with simple step-by-step explanations. These solutions for Factorization Of Algebraic Expressions are extremely popular among Class 9 students for Maths Factorization Of Algebraic Expressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 9 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 5.14:

Question 1:

Factorize each of the following expressions:

p³ + 27

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 2:

y³ + 125

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 3:

1 − 27a³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 4:

8x³y³ + 27a³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 5:

64a³ − b³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 6:

$\frac{x^{3}}{216} - 8 y^{3}$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 7:

10x⁴y − 10xy⁴

Answer:

The given expression to be factorized is

Take common from the two terms,. Then we have

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 8:

54x⁶y + 2x³y⁴

Answer:

The given expression to be factorized is

Take common from the two terms,. Then we have

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 9:

32a³ + 108b³

Answer:

The given expression to be factorized is

Take common from the two terms,. Then we have

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 10:

(a − 2b)³ − 512b³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 11:

8x²y³ − x⁵

Answer:

The given expression to be factorized is

Take common. Then we have

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 12:

1029 − 3x³

Answer:

The given expression to be factorized is

Take common 3. Then we have from the above expression,

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 13:

x³y³+ 1

Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 14:

x⁴y⁴ − xy

Answer:

The given expression to be factorized is

Take common. Then we have from the above expression,

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 15:

a³ + b³ + a + b

Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula, we have

Take common. Then we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 16:

Simplify:

(i) $\frac{173 \times 173 \times 173 + 127 \times 127 \times 127}{173 \times 173 - 173 \times 127 + 127 \times 127}$

(ii) $\frac{155 \times 155 \times 155 - 55 \times 55 \times 55}{155 \times 155 + 155 \times 55 + 55 \times 55}$

(iii) $\frac{1.2 \times 1.2 \times 1.2 - 0.2 \times 0.2 \times 0.2}{1.2 \times 1.2 + 1.2 \times 0.2 + 0.2 \times 0.2}$

Answer:

(i) The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for sum of two cubes

Using the above formula, the expression becomes

Note that both and b are positive. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

(ii) The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for difference of two cubes

Using the above formula, the expression becomes

Note that both, b is positive and unequal. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

(iii) The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for difference of two cubes

Using the above formula, the expression becomes

Note that both, b is positive and unequal. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

Page No 5.14:

Question 17:

(a + b)³ − 8(a − b)³

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 18:

(x + 2)³ + (x − 2)³

Answer:

The given expression to be factorized is

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 19:

x⁶ + y⁶

Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 20:

a¹²+ b¹²

Answer:

The given expression to be factorized is

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 21:

x³ + 6x² + 12x + 16

Answer:

The given expression to be factorized is

This can be written as

Take common x² from first two terms, 2x from the next two terms andfrom the last two terms. Then we have,

Finally, take common. Then we get,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 22:

$a^{3} - \frac{1}{a^{3}} - 2 a + \frac{2}{a}$

Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula and taking common from the last two terms, we get

Take common. Then we have,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 23:

a³ + 3a²b + 3ab² + b³ − 8

Answer:

The given expression to be factorized is

Recall the well known formula

The given expression can be written as

Recall the formula for difference of two cubes

Using the above formula and taking common –2 from the last two terms, we get

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.14:

Question 24:

8a³ − b³ − 4ax + 2bx

Answer:

The given expression to be factorized is

The given expression can be written as

Recall the formula for difference of two cubes

Using the above formula and taking common from the last two terms, we get

Take common. Then we have,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.17:

Question 1:

Factorize:

64a³ + 125b³ + 240a²b + 300ab²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.17:

Question 2:

125x³ − 27y³ − 225x²y + 135xy²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 5.17:

Question 3:

$\frac{8}{27} x^{3} + 1 + \frac{4}{3} x^{2} + 2 x$

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 5.17:

Question 4:

8x³ + 27y³ + 36x²y + 54xy²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 5.17:

Question 5:

a³ − 3a²b + 3ab² − b³ + 8

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the third and fourth terms. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

This can be written in the following form

Recall the formula for the sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 5.17:

Question 6:

x³ + 8y³ + 6x²y + 12xy²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 5.18:

Question 7:

8x³ + y³ + 12x²y + 6xy²

Answer:

The given expression to be factorized is

This can be written in the form

Take common 6xy from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 5.18:

Question 8:

8a³+ 27b³ + 36a²b + 54ab²

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 5.18:

Question 9:

8a³ − 27b³ − 36a²b + 54ab²

Answer:

The given expression to be factorized is

This can be written in the form

Take common – 18ab from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 5.18:

Question 10:

x³ − 12x(x − 4) − 64

Answer:

The given expression to be factorized is

This can be written in the form

Take common – 12x from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 5.18:

Question 11:

a³x³ − 3a²bx² + 3ab²x − b³

Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 5.23:

Question 1:

Factorize each of the following expressions:

a³ + 8b³ + 64c³ − 24abc

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.23:

Question 2:

x³− 8y³ + 27z³ + 18xyz

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 5.23:

Question 3:

27x³ − y³ − z³ − 9xyz

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is .

Page No 5.23:

Question 4:

$\frac{1}{27} x^{3} - y^{3} + 125 z^{3} + 5 x y z$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 5.24:

Question 5:

8x³ +27y³ − 216z³ + 108xyz

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 5.24:

Question 6:

125 + 8x³ − 27y³ + 90xy

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 5.24:

Question 7:

8x³ − 125y³ + 180xy + 216

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 5.24:

Question 8:

Multiply:
(i) x² + y² + z² − xy + xz + yz by x + y − z

(ii) x² + 4y² + z³ + 2xy + xz − 2yz by x − 2y − z

(iii) x²+ 4y² + 2xy − 3x + 6y + 9 by x − 2y + 3

(iv) 9x² + 25y² + 15xy + 12x − 20y + 16 by 3x − 5y + 4

(v) x²+ 4y² + z² + 2xy + xz – 2yz by (−z + x – 2y)

Answer:

(v) x²+ 4y² + z² + 2xy + xz – 2yz by (−z + x – 2y)

$(x^{2} + 4 y^{2} + z^{2} + 2 x y + x z - 2 y z) (- z + x - 2 y) = (- z) (x^{2} + 4 y^{2} + z^{2} + 2 x y + x z - 2 y z) + (x) (x^{2} + 4 y^{2} + z^{2} + 2 x y + x z - 2 y z) + (- 2 y) (x^{2} + 4 y^{2} + z^{2} + 2 x y + x z - 2 y z) = - x^{2} z - 4 y^{2} z - z^{3} - 2 x y z - x z^{2} + 2 y z^{2} + x^{3} + 4 x y^{2} + x z^{2} + 2 x^{2} y + x^{2} z - 2 x y z - 2 x^{2} y - 8 y^{3} - 2 y z^{2} - 4 x y^{2} - 2 x y z + 4 y^{2} z = x^{3} - 8 y^{3} - z^{3} - x^{2} z + 2 x^{2} y + x^{2} z - 2 x^{2} y - 4 y^{2} z + 4 x y^{2} - 4 x y^{2} + 4 y^{2} z - x z^{2} + 2 y z^{2} + x z^{2} - 2 y z^{2} - 2 x y z - 2 x y z - 2 x y z = x^{3} - 8 y^{3} - z^{3} - 6 x y z$

Hence, the required value is $x^{3} - 8 y^{3} - z^{3} - 6 x y z .$

Page No 5.24:

Question 9:

(3x − 2y)³ + (2y − 4z)³ + (4z − 3x)³

Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

when.

Using the above formula, the given expression can be written as

Put, and. Then we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 5.24:

Question 10:

(2x − 3y)³ + (4z − 2x)³ + (3y − 4z)³

Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the given expression can be written as

Put, and. Then we have

We cannot further factorize the expression.

So, the required factorization is ofis .

Page No 5.24:

Question 11:

${(\frac{x}{2} + y + \frac{z}{3})}^{3} + {(\frac{x}{3} - \frac{2 y}{3} + z)}^{3} + {(- \frac{5 x}{6} - \frac{y}{3} - \frac{4 z}{3})}^{3}$

Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the given expression can be written as

Put, and.

Then we have

We cannot further factorize the expression.

So, the required factorization is of is

Page No 5.24:

Question 12:

(a − 3b)³ + (3b − c)³ + (c − a)³

Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the given expression can be written as

Put, and. Then we have

We cannot further factorize the expression.

So, the required factorization is ofis.

Page No 5.24:

Question 13:

$2 {\sqrt{2 a}}^{3} + 3 {\sqrt{3 b}}^{3} + c^{3} - 3 \sqrt{6} a b c$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is .

Page No 5.24:

Question 14:

$3 \sqrt{3} a^{3} - b^{3} - 5 \sqrt{5} c^{3} - 3 \sqrt{15} a b c$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

Page No 5.24:

Question 15:

$2 \sqrt{2} a^{3} + 16 \sqrt{2} b^{3} + c^{3} - 12 a b c$

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is .

Page No 5.24:

Question 16:

(x – 2y)³ + (2y – 3z)³ + (3z – x)³

Answer:

$To solve : {(x - 2 y)}^{3} + {(2 y - 3 z)}^{3} + {(3 z - x)}^{3} Let a = (x - 2 y), b = (2 y - 3 z), c = (3 z - x) . a + b + c = (x - 2 y) + (2 y - 3 z) + (3 z - x) = x - 2 y + 2 y - 3 z + 3 z - x = 0 As we know that, if a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3 a b c Therefore, a^{3} + b^{3} + c^{3} = 3 a b c \Rightarrow {(x - 2 y)}^{3} + {(2 y - 3 z)}^{3} + {(3 z - x)}^{3} = 3 (x - 2 y) (2 y - 3 z) (3 z - x) Hence, {(x - 2 y)}^{3} + {(2 y - 3 z)}^{3} + {(3 z - x)}^{3} = 3 (x - 2 y) (2 y - 3 z) (3 z - x)$

Page No 5.24:

Question 17:

Find the value of x³ + y³ − 12xy + 64, when x + y =−4

Answer:

The given expression is

It is given that

The given expression can be written in the form

Recall the formula

Using the above formula, we have

Page No 5.24:

Question 18:

If a, b, c are all non-zero and a + b + c = 0, prove that $\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} = 3$ .

Answer:

$Given : a + b + c = 0 As we know that, if a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3 a b c Therefore, a^{3} + b^{3} + c^{3} = 3 a b c Dividing by (a b c) on both sides, we get \Rightarrow \frac{a^{3} + b^{3} + c^{3}}{a b c} = \frac{3 a b c}{a b c} \Rightarrow \frac{a^{3}}{a b c} + \frac{b^{3}}{a b c} + \frac{c^{3}}{a b c} = 3 \Rightarrow \frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{a b} = 3 Hence, \frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{a b} = 3 .$

Page No 5.25:

Question 1:

The factors of x³ −x²y − xy²+ y³ are

(a)(x + y) (x² − xy + y²)

(b) (x + y) (x² + xy + y²)

(c) (x + y)² (x − y)

(d) (x − y)² (x + y)

Answer:

The given expression to be factorized is

Take common from the first two terms and from the last two terms. That is

Finally, take commonfrom the two terms. That is

So, the correct choice is (d).

Page No 5.25:

Question 2:

The factors of x³ − 1 + y³ + 3xy are

(a) (x − 1 + y) (x² + 1 + y² + x + y − xy)

(b) (x + y + 1) (x² + y² + 1 −xy − x − y)

(c) (x − 1 + y) (x² − 1 − y² + x + y + xy)

(d) 3(x + y −1) (x² + y² − 1)

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

So, the correct choice is (a).

Page No 5.25:

Question 3:

The factors of 8a³ + b³ − 6ab + 1 are

(a) (2a + b − 1) (4a² + b² + 1 − 3ab − 2a)

(b) (2a − b + 1) (4a² + b² − 4ab + 1 − 2a + b)

(c) (2a + b + 1) (4a² + b² + 1 −2ab − b − 2a)

(d) (2a − 1 + b) (4a² + 1 − 4a − b − 2ab)

Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

So, the correct choice is (c).

Page No 5.25:

Question 4:

(x + y)³ − (x − y)³ can be factorized as

(a) 2y (3x² + y²)

(b) 2x (3x² + y²)

(c) 2y (3y² + x²)

(d) 2x (x²+ 3y²)

Answer:

The given expression to be factorized is

Recall the formula for difference of two cubes

Using the above formula, we have,

So, the correct choice is (a).

Page No 5.25:

Question 5:

The expression (a − b)³ + (b − c)³ + (c −a)³ can be factorized as

(a) (a − b) (b − c) (c −a)

(b) 3(a − b) (b − c) (c −a)

(c) −3(a − b) (b −c) (c − a)

(d) (a + b + c) (a² + b² + c² − ab − bc − ca)

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that:

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

So, the correct choice is (b).

Page No 5.26:

Question 6:

The value of $\frac{(2.3)^{3} - 0.027}{(2.3)^{2} + 0.69 + 0.09}$

(a) 2

(b) 3

(c) 2.327

(d) 2.273

Answer:

The given expression is

This can be written in the form

Assumeand. Then the given expression can be rewritten as

Recall the formula for difference of two cubes

Using the above formula, the expression becomes

Note that both a and b are positive, unequal. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

So, the correct choice is (a).

Page No 5.26:

Question 7:

The value of $\frac{(0.013)^{3} + (0.007)^{3}}{(0.013)^{2} - 0.013 \times 0.007 + (0.007)^{2}}$ is

(a) 0.006

(b) 0.02

(c) 0.0091

(d) 0.00185

Answer:

The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for sum of two cubes

Using the above formula, the expression becomes

Note that both and b are positive. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

So, the correct choice is (b).

Page No 5.26:

Question 8:

Mark the correct alternative in each of the following:

The factors of a² − 1 − 2x − x² are

(a) (a − x + 1) (a − x − 1)

(b) (a + x − 1) (a − x + 1)

(c) (a + x +1) (a − x + 1)

(d) none of these

Answer:

The given expression to be factorized is

Take commonfrom the last three terms and then we have

So, the correct choice is (c).

Page No 5.26:

Question 9:

The factors of x⁴ + x² + 25 are

(a) (x² + 3x + 5) (x² − 3x + 5)

(b) (x²+ 3x + 5) (x² + 3x − 5)

(c) (x² + x +5) (x² − x + 5)

(d) none of these

Answer:

The given expression to be factorized is

This can be written in the form

So, the correct choice is (a).

Page No 5.26:

Question 10:

The factors of x² + 4y² + 4y − 4xy − 2x − 8 are

(a) (x − 2y −4) (x − 2y + 2)

(b) (x − y + 2) (x − 4y − 4)

(c) (x + 2y − 4) (x + 2y + 2)

(d) none of these

Answer:

The given expression to be factorized is

This can be arrange in the form

Let. Then the above expression becomes

Put.

So, the correct choice is (a).

Page No 5.26:

Question 11:

The factors of x³ − 7x + 6 are

(a) x (x − 6) (x − 1)

(b) (x² − 6) (x − 1)

(c) (x + 1) (x + 2) (x + 3)

(d) (x − 1) (x + 3) (x − 2)

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms. Then we have

Finally, take commonfrom the above expression,

So, the correct choice is (d).

Page No 5.26:

Question 12:

The expression x⁴ + 4 can be factorized as

(a) (x² + 2x + 2) (x² − 2x + 2)

(b) (x² + 2x + 2) (x² + 2x − 2)

(c) (x² − 2x − 2) (x²− 2x + 2)

(d) (x² + 2) (x² − 2)

Answer:

The given expression to be factorized is

This can be written in the form

So, the correct choice is (a).

Page No 5.26:

Question 13:

If 3x = a + b + c, then the value of (x − a)³ + (x −b)³ + (x − c)³ − 3(x − a) (x − b) (x −c) is

(a) a + b + c

(b) (a − b) (b − c) (c − a)

(c) 0

(d) none of these

Answer:

The given expression is

Recall the formula

Using the above formula the given expression becomes

Given that

Therefore the value of the given expression is

So, the correct choice is (c).

Page No 5.26:

Question 14:

If (x + y)³ − (x − y)³ − 6y(x² − y²) = ky³, then k =

(a) 1

(b) 2

(c) 4

(d) 8

Answer:

The given equation is

Recall the formula

Using the above formula, we have

, provided.

So, the correct choice is (d).

Page No 5.26:

Question 15:

If x³ − 3x² + 3x − 7 = (x + 1) (ax² + bx + c), then a + b + c =

(a) 4

(b) 12

(c) −10

(d) 3

Answer:

The given equation is

x³ − 3x² + 3x − 7 = (x + 1) (ax² + bx + c)

This can be written as

$x^{3} - 3 x^{2} + 3 x - 7 = (x + 1) (a x^{2} + b x + c) \Rightarrow x^{3} - 3 x^{2} + 3 x - 7 = a x^{3} + b x^{2} + c x + a x^{2} + b x + c \Rightarrow x^{3} - 3 x^{2} + 3 x - 7 = a x^{3} + (a + b) x^{2} + (b + c) x + c$

Comparing the coefficients on both sides of the equation.

We get,

c = -7 .......(4)

Putting the value of a from (1) in (2)

We get,

So the value of a, b and c is 1, – 4 and -7 respectively.

Therefore,

a + b + c =1 - 4 - 7 = -10

So, the correct choice is (c).

Page No 5.27:

Question 1:

The factorized form of the expression y² + (x – 1)y – x is ____________.

Answer:

$y^{2} + (x - 1) y - x = y^{2} + x y - y - x = y (y + x) - 1 (y + x) = (y - 1) (y + x)$

Hence, the factorized form of the expression y² + (x – 1)y – x is (y – 1)(y + x).

Page No 5.27:

Question 2:

The factorized form of a³ + (b – a)³ – b³ is ____________.

Answer:

$a^{3} + {(b - a)}^{3} - b^{3} = a^{3} + (b^{3} - a^{3} - 3 b a (b - a)) - b^{3} (Using the identity : {(x - y)}^{3} = x^{3} - y^{3} - 3 x y (x - y)) = a^{3} + b^{3} - a^{3} - 3 b a (b - a) - b^{3} = - 3 b a (b - a) = 3 a b (a - b)$

Hence, the factorized form of a³ + (b – a)³ – b³ is 3ab(a – b).

Page No 5.27:

Question 3:

If $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 and a b c = 2, then a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c =____________.$

Answer:

$Given : \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 . . . (1) a b c = 2 . . . (2) Now, a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c = a^{2} b^{2} c^{2} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = {(a b c)}^{2} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = {(2)}^{2} (1) (From (1) and (2)) = 4$

Hence, if $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 and a b c = 2, then a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c = 4 .$

Page No 5.27:

Question 4:

Factorization of the polynomial $11 x^{2} - 10 \sqrt{3} x - 3$ gives ____________.

Answer:

$11 x^{2} - 10 \sqrt{3} x - 3 = 11 x^{2} - 11 \sqrt{3} x + \sqrt{3} x - 3 = 11 x (x - \sqrt{3}) + \sqrt{3} (x - \sqrt{3}) = (11 x + \sqrt{3}) (x - \sqrt{3})$

Hence, factorization of the polynomial $11 x^{2} - 10 \sqrt{3} x - 3$ gives $(11 x + \sqrt{3}) (x - \sqrt{3})$ .

Page No 5.27:

Question 5:

The polynomial x² + y² – z² – 2xy on factorization gives _____________.

Answer:

$x^{2} + y^{2} - z^{2} - 2 x y = (x^{2} + y^{2} - 2 x y) - z^{2} = {(x - y)}^{2} - {(z)}^{2} (Using the identity : {(a - b)}^{2} = a^{2} + b^{2} - 2 a b) = (x - y + z) (x - y - z) (Using the identity : a^{2} - b^{2} = (a + b) (a - b))$

Hence, the polynomial x² + y² – z² – 2xy on factorization gives $(x - y + z) (x - y - z)$ .

Page No 5.27:

Question 6:

The factors of the expression $a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a}$ are ____________.

Answer:

$a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a} = {(\sqrt{a})}^{2} + {(\sqrt{b})}^{2} + {(- \sqrt{c})}^{2} + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a} = {(\sqrt{a})}^{2} + {(\sqrt{b})}^{2} + {(- \sqrt{c})}^{2} + 2 \sqrt{a} \sqrt{b} + 2 \sqrt{b} (- \sqrt{c}) + 2 \sqrt{a} (- \sqrt{c}) = {(\sqrt{a} + \sqrt{b} - \sqrt{c})}^{2} (Using the identity : a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c = {(a + b + c)}^{2}) = (\sqrt{a} + \sqrt{b} - \sqrt{c}) (\sqrt{a} + \sqrt{b} - \sqrt{c})$

Hence, the factors of the expression $a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a}$ are $(\sqrt{a} + \sqrt{b} - \sqrt{c}) and (\sqrt{a} + \sqrt{b} - \sqrt{c}) .$ .

Page No 5.27:

Question 7:

The polynomial , x⁶ + 64y⁶ on factorization gives _____________.

Answer:

$x^{6} + 64 y^{6} = {(x^{2})}^{3} + {(4 y^{2})}^{3} = (x^{2} + 4 y^{2}) ({(x^{2})}^{2} + {(4 y^{2})}^{2} - (x^{2}) (4 y^{2})) (Using the identity : a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)) = (x^{2} + 4 y^{2}) (x^{4} + 16 y^{4} - 4 x^{2} y^{2})$

Hence, the polynomial x⁶ + 64y⁶ on factorization gives $(x^{2} + 4 y^{2}) (x^{4} + 16 y^{4} - 4 x^{2} y^{2})$ .

Page No 5.27:

Question 8:

The factorization form of a⁴ + b⁴– a²b² is _____________.

Answer:

$a^{4} + b^{4} - a^{2} b^{2} Adding and subtracting 2 a^{2} b^{2}, we get = {(a^{2})}^{2} + {(b^{2})}^{2} + 2 a^{2} b^{2} - 2 a^{2} b^{2} - a^{2} b^{2} = {(a^{2} + b^{2})}^{2} - 3 a^{2} b^{2} (Using the identity : a^{2} + b^{2} + 2 a b = {(a + b)}^{2}) = {(a^{2} + b^{2})}^{2} - {(\sqrt{3} a b)}^{2} = (a^{2} + b^{2} + \sqrt{3} a b) (a^{2} + b^{2} - \sqrt{3} a b) (Using the identity : a^{2} - b^{2} = (a + b) (a - b))$

Hence, the factorization form of a⁴ + b⁴– a²b² is $(a^{2} + b^{2} + \sqrt{3} a b) (a^{2} + b^{2} - \sqrt{3} a b)$ .

Page No 5.27:

Question 9:

If $3 x - \frac{y}{5} = 10 and x y = 5$ , then the value of $27 x^{3} - \frac{y^{3}}{125}$ is ____________.

Answer:

$Given : 3 x - \frac{y}{5} = 10 . . . (1) x y = 5 . . . (2) Now, 3 x - \frac{y}{5} = 10 Taking cube on both sides, we get \Rightarrow {(3 x - \frac{y}{5})}^{3} = 10^{3} \Rightarrow {(3 x)}^{3} - {(\frac{y}{5})}^{3} - 3 (3 x) (\frac{y}{5}) (3 x - \frac{y}{5}) = 1000 (Using the identity : {(a - b)}^{3} = a^{3} - b^{3} - 3 a b (a - b)) \Rightarrow 27 x^{3} - \frac{y^{3}}{125} - \frac{9}{5} x y (3 x - \frac{y}{5}) = 1000 \Rightarrow 27 x^{3} - \frac{y^{3}}{125} - \frac{9}{5} \times 5 \times 10 = 1000 (From (1) and (2)) \Rightarrow 27 x^{3} - \frac{y^{3}}{125} - 90 = 1000 \Rightarrow 27 x^{3} - \frac{y^{3}}{125} = 1000 + 90 \Rightarrow 27 x^{3} - \frac{y^{3}}{125} = 1090$

Hence, the value of $27 x^{3} - \frac{y^{3}}{125}$ is 1090.

Page No 5.27:

Question 10:

The factorized form of $\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is _____________.

Answer:

$\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = \frac{(x^{2} + y^{2} + z^{2})}{x y z} + 2 (\frac{y z + x z + x y}{x y x}) = \frac{x^{2} + y^{2} + z^{2} + 2 (y z + x z + x y)}{x y z} = \frac{{(x + y + z)}^{2}}{x y z} (Using the identity : {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c) = \frac{1}{x y z} {(x + y + z)}^{2}$

Hence, the factorized form of $\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is $\frac{1}{x y z} {(x + y + z)}^{2} .$

Page No 5.27:

Question 11:

The factorized form of a³ + b³ + 3ab – 1 is ____________.

Answer:

$a^{3} + b^{3} + 3 a b - 1 = a^{3} + b^{3} - 1 + 3 a b = a^{3} + b^{3} + {(- 1)}^{3} - 3 a b (- 1) = (a + b - 1) (a^{2} + b^{2} + {(- 1)}^{2} - a b - b (- 1) - a (- 1)) (Using the identity : a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)) = (a + b - 1) (a^{2} + b^{2} + 1 - a b + b + a)$

Hence, the factorized form of a³ + b³ + 3ab – 1 is $(a + b - 1) (a^{2} + b^{2} + 1 - a b + b + a) .$

Page No 5.27:

Question 1:

If a + b + c = 0, then write the value of a³ + b³ + c³.

Answer:

Recall the formula

When, we have

Page No 5.27:

Question 2:

If a² + b² + c² = 20 and a + b + c = 0, find ab + bc + ca.

Answer:

Recall the formula

Given that

Then we have

Page No 5.27:

Question 3:

If a + b + c = 9 and ab + bc + ca = 40, find a² + b² +c².

Answer:

Recall the formula

Given that

Then we have

Page No 5.27:

Question 4:

If a² + b² + c² = 250 and ab + bc + ca = 3, find a + b + c.

Answer:

Recall the formula

Given that

Then we have

Page No 5.27:

Question 5:

Write the value of 25³ − 75³+ 50³.

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

Page No 5.27:

Question 6:

Write the value of 48³ − 30³ − 18³.

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

Page No 5.28:

Question 7:

Write the value of ${(\frac{1}{2})}^{3} + {(\frac{1}{3})}^{3} - {(\frac{5}{6})}^{3} .$

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that:

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

Page No 5.28:

Question 8:

Write the value of 30³ + 20³ − 50³.

Answer:

The given expression is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

Page No 5.28:

Question 9:

Factorize: x⁴ + x² + 25.

Answer:

The given expression to be factorized is

This can be written in the form

We cannot further factorize the expression.

So, the required factorization is.

Page No 5.28:

Question 10:

Factorize : x² − 1 − 2a − a²

Answer:

The given expression to be factorized is

Take commonfrom the last three terms and then we have

We cannot further factorize the expression.

So, the required factorization is.

Page No 5.9:

Question 1:

Factorize:

1. $x^{3} + x - 3 x^{2} - 3$

Answer:

The given expression to be factorized is

Take common x from the first two terms and -3 from the last two terms. That is

Finally, take common x² + 1from the two terms. That is

We cannot further factorize the expression.

So, the required factorization is.

Page No 5.9:

Question 2:

Factorize:

2. a(a+b)³ − 3a²b (a + b)

Answer:

The given expression to be factorized is

Take common from the two terms. That is

Expand the term within the second braces.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 3:

$x (x^{3} - y^{3}) + 3 x y (x - y)$

Answer:

The given expression to be factorized is

We know that

The given expression then becomes

Take common from the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 4:

Factorize:

a²x² + (ax² + 1)x + a

Answer:

The given expression to be factorized is

Simplify the middle term. That is

Take common from the first two terms and 1 from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 5:

Factorize:
x² + y − xy − x

Answer:

The given expression to be factorized is

Rearrange the given expression as

Take common x from the first two terms and -1 from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 6:

Factorize:

x³ − 2x²y + 3xy² − 6y³

Answer:

The given expression to be factorized is

Take common from the first two terms and from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 7:

Factorize:

6ab − b² + 12ac − 2bc

Answer:

The given expression to be factorized is

Take common b from the first two terms and from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 8:

Factorize:

$x (x - 2) (x - 4) + 4 x - 8$

Answer:

The given expression to be factorized is

Take common 4 from the last two terms. That is

Again take commonfrom the two terms of the above expression.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 9:

Factorize:

(a − b + c)² + (b − c + a)² + 2(a − b + c) (b − c + a)

Answer:

The given expression to be factorized is

This can be written as

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 10:

Factorize:

a² + 2ab +b² − c²

Answer:

The given expression to be factorized is

This can be arrange in the form

Substitutingin the above expression, we get.

Put.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 11:

Factorize:

a² + 4b² − 4ab − 4c²

Answer:

The given expression to be factorized is

This can be arrange in the form

Substitute.

Put.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 12:

Factorize:

x² − y² − 4xz + 4z²

Answer:

The given expression to be factorized is

Rearrange the terms as

Substitutingin the avove expression,

Put.

We cannot further factorize the expression.

So, the required factorization ofis.

Page No 5.9:

Question 13:

Factorize:

$2 x^{2} - \frac{5}{6} x + \frac{1}{12}$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 14:

Factorize:

$x^{2} + \frac{12}{35} x + \frac{1}{35}$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 15:

Factorize:

$21 x^{2} - 2 x + \frac{1}{21}$

Answer:

The given expression to be factorized is

This can be written in the form

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 16:

Give possible expressions for the length and breadth of the rectangle having 35y² + 13y − 12 as its area.

Answer:

The area of the rectangle is

First we will factorize the above expression. This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

The area of a rectangle having length a and breadth bis ab.

Here we don’t know the bigger or the smaller factor. So, the two possibilities are

(i) Length isand breadth is

(ii) Length is and breadth is

Page No 5.9:

Question 17:

What are the possible expressions for the dimensions of the cuboid whose volume is 3x²− 12x.

Answer:

The volume of the cuboid is

First we will factorize the above expression.

Take commonfrom the two terms of the above expression,

The volume of a cuboid having length, breadth b and height is.

Here the word ‘dimensions’ stands for the length, breadth and height of the cuboid. So, the three possibilities are

(i) Length is, breadth is x and height is

(ii) Length is x , breadth isand height is

(iii) Length is, breadth isand height is x

There are many other possibilities also, because we can consider the product of two simple factors as a single factor.

Page No 5.9:

Question 18:

Factorize:

$(x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 6$

Answer:

The given expression to be factorized is

We have

Use the above result in the original expression to get

Substituting in the above , we get

Put.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 19:

Factorize:

(x+2) (x²+25) − 10x² − 20x

Answer:

The given expression to be factorized is

Take common from the last two terms. That is

Again take commonfrom the two terms of the above expression. Then

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 20:

Factorize:

$2 a^{2} + 2 \sqrt{6} a b + 3 b^{2}$

Answer:

The given expression to be factorized is

This can be written in the form

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 21:

Factorize:

a² + b² + 2(ab + bc + ca)

Answer:

The given expression to be factorized is

This can be written as

Take commonfrom the last two terms.

Finally, take common from the two terms of the above expression.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 22:

Factorize:

4(x − y)² − 12(x − y) (x + y) + 9(x + y)²

Answer:

The given expression to be factorized is

Substituting andin the above expression, we get

This can be arrange in the form

Putand.

Take common -1 from the expression within the braces.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 23:

Factorize:
a² + b² + 2bc − c²

Answer:

The given expression to be factorized is

This can be arrange in the form

Substitutingin the above expression, we get.

Put.

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 24:

Factorize:

xy⁹ − yx⁹

Answer:

The given expression to be factorized is

This can be written in the form

Take commonfrom the two terms of the above expression

We cannot further factorize the expression.

So, the required factorization of is

Page No 5.9:

Question 25:

Factorize:

x⁴+ x²y² + y⁴

Answer:

The given expression to be factorized is

Add and subtract the termin the given expression.

Substitutingin the above expression, we get

Putin the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 26:

Factorize:

$x^{2} + 6 \sqrt{2} x + 10$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms.

Finally, take commonfrom the above expression. Then we have

We cannot further factorize the expression.

So, the required factorization is.

Page No 5.9:

Question 27:

Factorize:

$x^{2} - 2 \sqrt{2} x - 30$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 28:

Factorize:

$x^{2} - \sqrt{3} x - 6$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms. Then we have

Finally take commonfrom the above expression. Then we have

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 29:

Factorize:

$x^{2} + 5 \sqrt{5} x + 30$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms. Then we have

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 30:

Factorize:

$x^{2} + 2 \sqrt{3} x - 24$

Answer:

The given expression to be factorized is

This can be written in the form

Take common x from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 31:

Factorize:

$5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$

Answer:

The given expression to be factorized is

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 32:

Factorize:

$2 x^{2} + 3 \sqrt{5} x + 5$

Answer:

The given expression to be factorized is

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 33:

Factorize:

9(2a − b)² − 4(2a − b) − 13

Answer:

The given expression to be factorized is

Substitutingin the above expression, we get

This can be written in the form

Take common x from the first two terms and 1 from the last two terms,

Finally take commonfrom the above expression,

Put,

We cannot further factorize the expression.

So, the required factorization ofis.

Page No 5.9:

Question 34:

Factorize:

7(x − 2y)² − 25(x − 2y) + 12

Answer:

The given expression to be factorized is

Substitutingin the above expression, we get

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

Put in the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

Page No 5.9:

Question 35:

Factorize:

2(x + y)² − 9(x + y) − 5

Answer:

The given expression to be factorized is

Substitutingin the above expression, we get

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

Put. Then we have

We cannot further factorize the expression.

So, the required factorization of is.

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