Rd Sharma 2021 for Class 9 Maths Chapter 6 - Factorization Of Polynomials

Question 1:

In each of the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the result by actual division: (1−8)

1. f(x) = x³ + 4x² − 3x + 10, g(x) = x + 4

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will show by actual division

So the remainder by actual division is 22

Question 2:

f(x) = 4x⁴ − 3x³ − 2x² + x − 7, g(x) = x − 1

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will show remainder by actual division

So the remainder by actual division is −7

Question 3:

f(x) = 2x⁴ − 6x³ + 2x² − x + 2, g(x) = x + 2

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will calculate the remainder by actual division

So the remainder by actual division is 92

Question 4:

f(x) = 4x³ − 12x² + 14x − 3, g(x) = 2x − 1

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will calculate the remainder by actual division

So the remainder by actual division is

Question 5:

f(x) = x³ − 6x² + 2x − 4, g(x) = 1 − 2x

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will calculate remainder by actual division

So the remainder is

Question 6:

f(x) = x⁴ − 3x² + 4, g(x) = x − 2

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

We will calculate remainder by actual division

So the remainder is 8

Question 7:

f(x) = 9x³ − 3x² + x − 5, g(x) = $x - \frac{2}{3}$

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Remainder by actual division

Remainder is −3

Question 8:

$f (x) = 3 x^{4} + 2 x^{3} - \frac{x^{2}}{3} - \frac{x}{9} + \frac{2}{27}, g (x) = x + \frac{2}{3}$

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Remainder by actual division

Remainder is 0

Question 9:

If the polynomials 2x³ + ax² + 3x − 5 and x³ + x² − 4x +a leave the same remainder when divided by x −2, find the value of a.

Answer:

Let us denote the given polynomials as

Now, we will find the remaindersandwhenandrespectively are divided by.

By the remainder theorem, whenis divided bythe remainder is

By the given condition, the two remainders are same. Then we have,

Question 10:

If the polynomials ax³ + 3x² − 13 and 2x³ − 5x + a, when divided by (x − 2) leave the same remainder, find the value of a.

Answer:

Let us denote the given polynomials as

Now, we will find the remaindersandwhenandrespectively are divided by.

By the remainder theorem, whenis divided bythe remainder is

By the given condition, the two remainders are same. Then we have, R₁ = R₂

Question 11:

Find the remainder when x³ + 3x² + 3x + 1 is divided by

(i) x + 1

(ii) $x - \frac{1}{2}$

(iii) x

(iv) $x + π$

(v) 5 + 2x

Answer:

Let us denote the given polynomials as

(i) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(ii) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(iii) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(iv) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(v) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Question 12:

The polynomials ax³ + 3x² − 3 and 2x³ − 5x + a when divided by (x − 4) leave the remainders R₁ and R₂ respectively. Find the values of a in each of the following cases, if

(a) R₁ = R₂

(b) R₁ + R₂ = 0

(c) 2R₁ − R₂ = 0

Answer:

Let us denote the given polynomials as

Now, we will find the remaindersandwhenandrespectively are divided by.

By the remainder theorem, whenis divided bythe remainder is

(i) By the given condition,

R₁ = R₂

(ii) By the given condition,

R₁ + R₂ = 0

(iii) By the given condition,

2R₁ − R₂ = 0

Question 1:

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:

(i) 3x² − 4x + 15

(ii) y² + 2 $\sqrt{3}$

(iii) $3 \sqrt{x} + \sqrt{2} x$

(iv) $x - \frac{4}{x}$

(v) x¹² + y³ + t⁵⁰

Answer:

(i) is a polynomial of degree 2.i.e Quadratic polynomial.

(ii) is a polynomial of degree 2 in y variable. i.e. Quadratic polynomial.

(iii)

It is not a polynomial because exponent of x is 1/2 which is not a positive integer.

(iv)

It is not a polynomial because is fractional part.

(v)

It is a polynomial in three variables x, y and t.

Question 2:

Write the coefficient of x² in each of the following:

(i) 17 − 2x + 7x²

(ii) 9 − 12x + x³

(iii) $\frac{π}{6} x^{2} - 3 x + 4$

(iv) $\sqrt{3} x - 7$

Answer:

(i)

Coefficient of

(ii)

Coefficient of

(iii)

Coefficient of

(iv)

Coefficient of

Question 1:

In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1−7)

f(x) = x³ − 6x² + 11x − 6; g(x) = x − 3

Answer:

Given that:

By the factor theorem,

If g(x) is a factor of f(x)

i.e.

Then

As is zero therefore g(x), is the factor of polynomial f(x).

Question 2:

f(x) = 3x⁴ + 17x³ + 9x² − 7x − 10; g(x) = x + 5

Answer:

It is given that and

By the factor theorem, g(x) is a factor of polynomial f(x)

i.e.

Therefore,

$f (- 5) = 3 {(- 5)}^{4} + 17 {(- 5)}^{3} + 9 {(- 5)}^{2} - 7 (- 5) - 10 = 3 \times 625 + 17 \times (- 125) + 225 + 35 - 10 = 1875 - 2125 + 250 = 0$

Hence, g(x) is the factor of polynomial f(x).

Question 3:

f(x) = x⁵ + 3x⁴ − x³ − 3x² + 5x + 15, g(x) = x + 3

Answer:

It is given that and

By the factor theorem, g(x) is the factor of polynomial f(x).

i.e.

f (−3) = 0

Hence, g(x) is the factor of polynomial f (x).

Question 4:

f(x) = x³ −6x² − 19x + 84, g(x) = x − 7

Answer:

It is given that and

By the factor theorem, g(x) is the factor of polynomial f(x), if f (7) = 0.

Therefore, in order to prove that (x − 7) is a factor of f(x).

It is sufficient to show that f(7) = 0

Now,

Hence, (x − 7) is a factor of polynomial f(x).

Question 5:

f(x) = 3x³ + x² − 20x +12, g(x) = 3x − 2

Answer:

It is given that and

By the factor theorem,

(3x − 2) is the factor of f(x), if

Therefore,

In order to prove that (3x − 2) is a factor of f(x).

It is sufficient to show that

Now,

Hence, (3x − 2) is the factor of polynomial f(x).

Question 6:

f(x) = 2x³ − 9x² + x + 12, g(x) = 3 − 2x

Answer:

It is given that and

By factor theorem, (3 − 2x) is the factor of f(x), if = 0

Therefore,

In order to prove that (3 − 2x) is a factor of f(x). It is sufficient to show that

Now,

Hence, (3 − 2x), is the factor of polynomial f(x).

Question 7:

f(x) = x³ − 6x² + 11x − 6, g(x) = x² − 3x + 2

Answer:

It is given that and

We have

$\Rightarrow (x - 2)$ and (x − 1) are factor of g(x) by the factor theorem.

To prove that (x − 2) and (x − 1) are the factor of f(x).

It is sufficient to show that f(2) and f(1) both are equal to zero.

And

Hence, g(x) is the factor of the polynomial f(x).

Question 8:

Show that (x − 2), (x + 3) and (x − 4) are factors of x³ − 3x² − 10x + 24.

Answer:

Let be the given polynomial.

By factor theorem,

and are the factor of f(x).

If and f(4) are all equal to zero.

Now,

also

And

Hence, and are the factor of polynomial f(x).

Question 9:

Show that (x + 4) , (x − 3) and (x − 7) are factors of x³ − 6x² − 19x + 84

Answer:

Let be the given polynomial.

By the factor theorem,

and are the factor of f(x).

If and f(7) are all equal to zero.

Therefore,

Also

And

Hence, and are the factor of the polynomial f(x).

Question 10:

For what value of a is (x − 5) a factor of x³− 3x² + ax − 10?

Answer:

Let be the given polynomial.

By factor theorem, is the factor of f(x), if f (5) = 0

Therefore,

Hence, a = − 8.

Question 11:

Find the value of a such that (x − 4) is a factors of 5x³ − 7x² − ax − 28.

Answer:

Let be the given polynomial.

By the factor theorem,

(x − 4) is a factor of f(x).

Therefore f(4) = 0

Hence ,

$\Rightarrow 320 - 112 - 4 a - 28 = 0 \Rightarrow 180 - 4 a = 0 \Rightarrow a = \frac{180}{4} = 45$

Hence,

Question 12:

Find the value of a, if x + 2 is a factor of 4x⁴ + 2x³ − 3x² + 8x + 5a.

Answer:

Let 4x⁴ + 2x³ − 3x² + 8x + 5a be the polynomial.

By the factor theorem,

is a factor of f(x) if f(−2) = 0.

Therefore,

Hence,

Question 13:

Find the value k if x − 3 is a factor of k²x³ − kx² + 3kx − k.

Answer:

Let be the given polynomial.

By the factor theorem,

(x − 3) is a factor of f(x) if f (3) = 0

Therefore,

$\Rightarrow 27 k^{2} - 9 k + 9 k - k = 0 \Rightarrow 27 k^{2} - k = 0 \Rightarrow k (27 k - 1) = 0 \Rightarrow k = 0 o r k = \frac{1}{27}$

Hence, the value of k is 0 or .

Question 14:

Find the values of a and b, if (x² − 4) is a factor of ax⁴ + 2x³ − 3x² + bx − 4

Answer:

Let and be the given polynomial.

We have,

are the factors of g(x).

By factor theorem, if and both are the factor of f(x)

Then f(2) and f(−2) are equal to zero.

Therefore,

and

Adding these two equations, we get

Putting the value of a in equation (i), we get

Hence, the value of a and b are 1, − 8 respectively.

Question 15:

Find α and β, if (x + 1) and (x + 2) are factors of x³ + 3x²− 2αx + β.

Answer:

Let be the given polynomial.

By the factor theorem, and are the factor of the polynomial f(x) if and both are equal to zero.

Therefore,

$f (- 1) = (- 1)^{3} + 3 (- 1)^{2} - 2 α (- 1) + β = 0 \Rightarrow f (- 1) = - 1 + 3 + 2 α + β = 0 \Rightarrow 2 α + β = - 2 . . . (i)$

and

Subtracting (i) from (ii)

We get,

$α = - 1$

Putting the value of $α$ in equation (i), we get

Hence, the value of α and β are −1, 0 respectively.

Question 16:

If x − 2 is a factor of each of the following two polynomials, find the values of a in each case

(i) x³ − 2ax² + ax − 1

(ii) x⁵ − 3x⁴ − ax³ + 3ax² + 2ax + 4

Answer:

(i) Let be the given polynomial.

By factor theorem, if (x − 2) is a factor of f(x), then f (2) = 0

Therefore,

Thus the value of a is 7/6.

(ii) Let f(x) = x⁵ − 3x⁴ − ax³ + 3ax² + 2ax + 4 be the given polynomial.

By the factor theorem, (x − 2) is a factor of f(x), if f (2) = 0

Therefore,

Thus, the value of a is .

Question 17:

In each of the following two polynomials, find the value of a, if (x − a) is factor:

(i) x⁶ − ax⁵ + x⁴ − ax³ + 3x − a + 2

(ii x⁵ − a²x³ + 2x + a + 1)

Answer:

(i) Let be the given polynomial.

By factor theorem, (x − a) is a factor of the polynomial if f(a) = 0

Therefore,

Thus, the value of a is − 1.

(ii) Let be the given polynomial.

By factor theorem, (x − a) is a factor of f(x), if f(a) = 0.

Therefore,

Thus, the value of a is − 1/3.

Question 18:

In each of the following two polynomials, find the value of a, if (x + a) is a factor.

(i) x³ + ax² − 2x +a + 4

(ii) x⁴ − a²x² + 3x −a

Answer:

(i) Let be the given polynomial.

By the factor theorem, (+ a) is the factor of f(x), if f(− a) = 0, i.e.,

Thus, the value of a is − 4/3.

(ii) Let be the polynomial. By factor theorem, (x + a) is a factor of the f(x), if f(− a) = 0, i.e.,

Thus, the value of a is 0.

Question 19:

Find the values of p and q so that x⁴ + px³ + 2x² − 3x + q is divisible by (x² − 1).

Answer:

Let and be the given polynomials.

We have,

Here, are the factor of g(x).

If f(x) is divisible by and , then and are factor of f(x).

Therefore, f(1) and f(−1) both must be equal to zero.

Therefore,

and

Adding both the equations, we get,

Putting this value in (i)

Hence, the value of p and q are 3, −3 respectively.

Question 20:

Find the values of a and b so that (x + 1) and (x − 1) are factors of x⁴ + ax³ − 3x² + 2x + b.

Answer:

Let be the given polynomial.

By factor theorem, and are the factors of f(x) if f(−1) and f(1) both are equal to zero.

Therefore,

and

Adding equation (i) and (ii), we get

Putting this value in equation (i), we get,

Hence, the value of a and b are – 2 and 2 respectively.

Question 21:

If x³ + ax² − bx+ 10 is divisible by x² − 3x + 2, find the values of a and b.

Answer:

Let and be the given polynomials.

We have,

Here, and are the factors of g(x),

Now,

By factor theorem,

and

Subtracting (ii) by (i), we get,

$(2 a - b) - (a - b) = - 9 - (- 11) a = 2$
Putting this value in equation (ii), we get,

Hence, the value of a and b are 2 and 13 respectively.

Question 22:

If both (x + 1) and (x − 1) are factors of ax³ + x² − 2x + b, find the values of a and b.

Answer:

Let be the given polynomial.

By factor theorem, if and both are factors of the polynomial f (x). if f(−1) and f(1) both are equal to zero.

Therefore,

And

Adding (i) and (ii), we get

And putting this value in equation (ii), we get,

a = 2

Hence, the value of a and b are 2 and −1 respectively.

Question 23:

What must be added to x³ − 3x² − 12x + 19 so that the result is exactly divisibly by x² + x - 6 ?

Answer:

Let and be the given polynomial.

When p(x) is divided by q(x), the reminder is a linear polynomial in x.

So, let r(x) = ax + b is added to p(x), so that p(x) + r(x) is divisible by q(x).

Let

Then,

We have,

Clearly, q(x) is divisible by and i.e., and are the factors of q(x).

Therefore, f (x) is divisible by q(x), if and are factors of f(x), i.e.,

and f(2) = 0

Now, f(-3) = 0

$\Rightarrow$ f(-3) = (-3)³ -3(-3)² + (a-12)(-3)+19+b = 0

$\Rightarrow$ -27 - 27 - 3a + 36 + 19 + b = 0

$\Rightarrow$ -27 - 27 - 3a + 36 + 19 + b = 0

$\Rightarrow$ -54 - 3a + b + 55 = 0

$\Rightarrow$ -3a + b + 1 = 0 ---- (i)

And

Subtracting (i) from (ii), we get,

Putting this value in equation (ii), we get,

Hence, p(x) is divisible by q(x) if added to it.

Question 24:

What must be subtracted from x³ − 6x² − 15x + 80 so that the result is exactly divisible by x² + x − 12?

Answer:

By divisible algorithm, when is divided by the reminder is a linear polynomial

Let be subtracted from p(x) so that the result is divisible by q(x).

Let

We have,

Clearly, and are factors of q(x), therefore, f(x) will be divisible by q(x) if and are factors of f(x), i.e. f (−4) and f (3) are equal to zero.

Therefore,

and

Adding (i) and (ii), we get,

Putting this value in equation (i), we get,

Hence, will be divisible by if 4 x − 4 is subtracted from it

Question 25:

What must be added to 3x³ + x² − 22x + 9 so that the result is exactly divisible by 3x² + 7x − 6?

Answer:

By division algorithm, when is divided by the reminder is a linear polynomial. So, let r(x) = ax + b be added to p(x) so that the result is divisible by q(x)

Let

We have,

$q (x) = 3 x^{2} + 7 x - 6 = 3 x^{2} + 9 x - 2 x - 6 = 3 x (x + 3) - 2 (x + 3) = (3 x - 2) (x + 3)$

Clearly, $(3 x - 2)$ and $(x + 3)$ are factors of q(x).

Therefore, f(x) will be divisible by q(x) if and are factors of f(x), i.e.,

and f(−3) are equal to zero.

Now,

$f (\frac{2}{3}) = 0 \Rightarrow 3 {(\frac{2}{3})}^{3} + {(\frac{2}{3})}^{2} + (a - 22) (\frac{2}{3}) + 9 + b = 0 \Rightarrow 3 \times \frac{8}{27} + \frac{4}{9} + \frac{2 a}{3} - \frac{44}{3} + 9 + b = 0 \Rightarrow \frac{8}{9} + \frac{4}{9} - \frac{44}{3} + 9 + \frac{2 a}{3} + b = 0 \Rightarrow \frac{8 + 4 - 132 + 81}{9} + \frac{2 a}{3} + b = 0 \Rightarrow - \frac{39}{9} + \frac{2 a}{3} + b = 0 \Rightarrow \frac{2 a}{3} + b = \frac{13}{3} \Rightarrow 2 a + 3 b = 13 . . . . . . . . (i)$

And

$f (- 3) = 0 \Rightarrow 3 {(- 3)}^{3} + {(- 3)}^{2} + (a - 22) (- 3) + 9 + b = 0 \Rightarrow - 81 + 9 - 3 a + 66 + 9 + b = 0 \Rightarrow - 3 a + b = - 3 \Rightarrow b = - 3 + 3 a . . . . . . . . . (ii)$

Substituting the value of b from (ii) in (i), we get,

2 a + 3 (3 a - 3) = 13 \Rightarrow 2 a + 9 a - 9 = 13 \Rightarrow 11 a = 13 + 9 \Rightarrow 11 a = 22 \Rightarrow a = 2

Now, from (ii), we get

b = - 3 + 3 (2) = - 3 + 6 = 3

So, we have a = 2 and b = 3

Hence, p(x) is divisible by q(x), if is added to it.

Question 3:

Write the degrees of each of the following polynomials:

(i) 7x³ + 4x² − 3x + 12

(ii) 12 − x + 2x³

(iii) $5 y - \sqrt{2}$

(iv) 7

(v) 0

Answer:

(i) 7x³ + 4x² − 3x + 12

Degree of the polynomial = 3

Because the highest power of x is 3.

(ii)

Degree of the polynomial = 3. Because the highest power of x is 3.

(iii)

Degree of the polynomial = 1. Because the highest power of y is 1.

(iv) 7

Degree of the polynomial = 0. Because there is no variable term in the expression

(v) 0

Degree of the polynomial is not defined. As there is no variable or constant term

Question 4:

Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials:

(i) x + x² + 4

(ii) 3x − 2

(iii) 2x + x²

(iv) 3y

(v) t² + 1

(vi) 7t⁴ + 4t³ + 3t − 2

Answer:

(i)

The degree of the polynomial is 2. It is quadratic in x.

So, it is quadratic polynomial.

(ii)

The degree of the polynomial is 1. It is a linear polynomial in x.

(iii)

The degree of the polynomial is 1.

It is linear a polynomial in x.

(iv) 3y

The degree of the polynomial is 1. It is linear in y.

(v)

The degree of the polynomial is 2. It is quadratic polynomial in t.

(vi)

The degree of the polynomial is 4. Therefore, it is bi-quadratic polynomial in t.

Question 5:

Classify the following polynomials as polynomials in one-variable, two variables etc.:

(i) x² − xy + 7y²

(ii) x² − 2tx + 7t² − x + t

(iii) t³ − 3t² + 4t − 5

(iv) xy + yz + zx

Answer:

(i)

Here, x and y are two variables.

So, it is polynomial in two variables.

(ii)

Here, x and t are two variables.

So, it is polynomial in two variables.

(iii)

Here, only t is variable.

So, it is polynomial in one variable.

(iv)

Here, x, y and z are three variables

So, it is polynomial in three variables.

Question 6:

Identify polynomials in the following:

(i) f(x) = 4x³ − x² − 3x + 7

(ii) g(x) = 2x³ − 3x²+ $\sqrt{x}$ − 1

(iii) p(x) = $\frac{2}{3} x^{2} - \frac{7}{4} x + 9$

(iv) q(x) = 2x² − 3x + $\frac{4}{x}$ + 2

(v) h(x) = $x^{4} - x^{\frac{3}{2}} + x - 1$

(vi) f(x) = $2 + \frac{3}{x} + 4 x$

Answer:

(i)

It is cubic in x, so, it is cubic polynomial in x variable.

(ii)

Here, exponent of x in is not a positive integer, so, it is not a polynomial.

(iii)

It is a quadratic polynomial.

(iv) q(x) = 2x² − 3x + $\frac{4}{x}$ + 2

Here, exponent of x in is not a positive integer. So it is not a polynomial.

(v)

Here, exponent of x in x³^/2 is not a positive integer. So, it is not a polynomial.

(vi)

Here, exponent of x in is not a positive integer, so, it not a polynomial.

Question 7:

Identify constant, linear, quadratic and cubic polynomials from the following polynomials:

(i) f(x) = 0

(ii) g(x) = 2x³ − 7x + 4

(iii) h(x) = - $3 x + \frac{1}{2}$

(iv) p(x) = 2x² − x + 4

(v) q(x) = 4x + 3

(vi) r(x) = 3x³ + 4x² + 5x − 7

Answer:

(i)

The given expression is a Constant polynomial as there is no variable term in it.

(ii)

The given expression is Cubic polynomial as the highest exponent of x is 3.

(iii)

The given expression is linear polynomial as the highest exponent of x is 1.

(iv)

The given expression is Quadratic polynomial as the highest exponent of x is 2.

(v)

The given polynomial is an linear polynomial as the highest exponent of x is 1.

(vi)

The given polynomial is Cubic polynomial as the highest exponent of x is 3.

Question 8:

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer:

An example of binomial of degree 35 is $f (t) = 4 t^{35} - \frac{1}{2}$ . It is a binomial as it has two terms and degree is 35 because highest exponent of t is 35.

An example of monomial of degree 100 is . It is a monomial as it has only one term and degree is 100 because highest exponent of x is 100

Question 1:

Using factor theorem, factorize each of the following polynomials:

x³ + 6x² + 11x + 6

Answer:

Let f(x) = x³ + 6x² + 11x + 6 be the given polynomial.

Now, put the we get

Therefore, is a factor of f(x).

Now,
$f (x) = x^{3} + 5 x^{2} + x^{2} + 5 x + 6 x + 6$

Hence, are the factors of f(x).

Question 2:

x³ + 2x² − x − 2

Answer:

Let be the given polynomial.

Now, put the x = -1, we get

$f (- 1) = {(- 1)}^{3} + 2 {(- 1)}^{2} - (- 1) - 2 = 0$

Therefore, is a factor of polynomial f(x).

Now, x³ + 2x² − x − 2 can be written as,
$f (x) = x^{3} + 3 x^{2} - x^{2} - 3 x + 2 x - 2$

Hence, are the factors of the polynomial f(x).

Question 3:

x³ − 6x² + 3x + 10

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,
$f (x) = x^{3} - 7 x^{2} + x^{2} + 10 x - 7 x + 10$

Hence, are the factors of the polynomial f(x).

Question 4:

x⁴ − 7x³+ 9x² + 7x − 10

Answer:

Let be the given polynomial.

Now, putting we get

$f (1) = {(1)}^{4} - 7 {(1)}^{3} + 9 {(1)}^{2} + 7 (1) - 10 = 1 - 7 + 9 + 7 - 10 = 0$

Therefore, is a factor of polynomial f(x).

Now,
$f (x) = x^{4} - x^{3} - 6 x^{3} + 6 x^{2} + 3 x^{2} - 3 x + 10 x - 10$

Where

Putting we get

Therefore, is a factor of g(x).

Now,
$g (x) = x^{3} - 7 x^{2} + x^{2} - 7 x + 10 x + 10$

From equation (i) and (ii), we get

Hence are the factors of polynomial f(x).

Question 5:

3x³ − x² − 3x + 1

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Question 6:

x³ − 23x² + 142x − 120

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Question 7:

y³ − 7y + 6

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(y).

Now,

Hence are the factors of polynomial f (y).

Question 8:

x³ − 10x² − 53x − 42

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Question 9:

y³ − 2y² − 29y − 42

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(y).

Now,

Hence are the factors of polynomial f(y).

Question 10:

2y³ − 5y² − 19y + 42

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(y).

Now,

Hence are the factors of polynomial f(y).

Question 11:

x³ + 13x² + 32x + 20

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Question 12:

x³ − 3x² − 9x − 5

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Question 13:

2y³ + y² − 2y − 1

Answer:

Let be the given polynomial.

Now, putting we get

Now,

Hence are the factors of polynomial f(y).

Question 14:

x³ − 2x² − x + 2

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Question 15:

Factorize each of the following polynomials:

(i) x³ + 13x² + 31x − 45 given that x + 9 is a factor

(ii) 4x³ + 20x² + 33x + 18 given that 2x + 3 is a factor.

Answer:

(i) Let be the given polynomial.

is a factor of the polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

(ii) Let be the given polynomial.

Therefore is a factor of the polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Question 16:

x⁴ − 2x³ − 7x² + 8x + 12

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,
$f (x) = x^{4} - 3 x^{3} + x^{3} - 3 x^{2} - 4 x^{2} + 12 x - 4 x + 12$

Where

Putting we get

Therefore, is the factor of g(x).

Now,
$g (x) = x^{3} - 2 x^{2} - x^{2} - 6 x + 2 x + 12$

From equation (i) and (ii), we get

Hence are the factors of polynomial f(x).

Question 17:

x⁴ + 10x³ + 35x² + 50x + 24

Answer:

Let $f (x) = x^{4} + 10 x^{3} + 35 x^{2} + 50 x + 24$ be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Where

Putting we get

Therefore, is the factor of g(x).

Now,

From equation (i) and (ii), we get

Hence are the factors of polynomial f(x).

Question 18:

2x⁴ − 7x³ − 13x² + 63x − 45

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Where

Putting we get

Therefore, is a factor of g(x).

Now,

From equation (i) and (ii), we get

Hence are the factors of polynomial f(x).

Question 1:

Which one of the following is a polynomial?

(a) $\frac{x^{2}}{2} - \frac{2}{x^{2}}$

(b) $\sqrt{2 x} - 1$

(c) $x^{2} + \frac{3 x^{3 / 2}}{\sqrt{x}}$

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

Answer:

(a) $\frac{x^{2}}{2} - \frac{2}{x^{2}} = \frac{x^{2}}{2} - 2 x^{- 2}$

Exponent of x cannot be negative.
Therefore, it is not a polynomial.

(b) $\sqrt{2 x} - 1$

Exponent of x must be a whole number.
Therefore, it is not a polynomial.

(c) $x^{2} + \frac{3 x^{3 / 2}}{\sqrt{x}} = x^{2} + 3 x^{\frac{3}{2} - \frac{1}{2}} = x^{2} + 3 x$

It is a polynomial.

(d) $\frac{x - 1}{x + 1}$

It is not a polynomial.

Hence, the correct option is (c).

Question 2:

Degree of the polynomial f(x) = 4x⁴ + 0x³+ 0x⁵ + 5x + 7 is
(a) 4
(b) 5
(c) 3
(d) 7

Answer:

Given: f(x) = 4x⁴ + 0x³+ 0x⁵ + 5x + 7 = 4x⁴ + 5x + 7

Degree is the highest power of x in the polynomial.

Here, the highest power of x is 4.

Thus, degree of the polynomial is 4.

Hence, the correct option is (a).

Question 3:

Degree of the zero polynomial is
(a) 0
(b) 1
(c) any natural number
(d) not defined

Answer:

The zero polynomial is a polynomial in which all the coefficients are zero.
We can't say about powers of x.

Thus, degree of the zero polynomial is not defined.

Hence, the correct option is (d).

Question 4:

$\sqrt{2}$ is a polynomial of degree
(a) 2
(b) 0
(c) 1
(d) $\frac{1}{2}$

Answer:

Given: f(x) = $\sqrt{2}$
It can also be written as $f (x) = \sqrt{2} x^{0}$ .

Degree is the highest power of x in the polynomial.

Here, the highest power of x is 0.

Thus, degree of the polynomial is 0.

Hence, the correct option is (b).

Question 5:

Zero of the zero polynomial is
(a) 0
(b) 1
(c) any real number
(d) not defined

Answer:

Given: zero polynomial i.e., f(x) = 0

Zero of a polynomial is the value of x at which the polynomial becomes zero.

Here, any value of x can be a zero of the zero polynomial.

Hence, the correct option is (c).

Question 6:

If f(x) = x + 3, then f(x) + f(–x) is equal to
(a) 3
(b) 2x
(c) 0
(d) 6

Answer:

Given: f(x) = x + 3

f(–x) = –x + 3

Thus,
f(x) + f(–x) = x + 3 + (–x + 3)
= x + 3 – x + 3
= 6

Hence, the correct option is (d).

Question 7:

Zero of the polynomial f(x) = 3x + 7 is
(a) $\frac{7}{3}$
(b) $\frac{- 3}{7}$
(c) $- \frac{7}{3}$
(d) –7

Answer:

Given: f(x) = 3x + 7

Zero of a polynomial is the value of x at which the polynomial becomes zero.

$f (x) = 0 \Rightarrow 3 x + 7 = 0 \Rightarrow 3 x = - 7 \Rightarrow x = - \frac{7}{3}$

Therefore, zero of the polynomial f(x) = 3x + 7 is $- \frac{7}{3}$ .

Hence, the correct option is (c).

Question 8:

On of the zeros of the polynomial f(x) = 2x² + 7x – 4 is
(a) 2
(b) $\frac{1}{2}$
(c) $- \frac{1}{2}$
(d) –2

Answer:

Given: f(x) = 2x² + 7x – 4

Zero of a polynomial is the value of x at which the polynomial becomes zero.

$f (x) = 0 \Rightarrow 2 x^{2} + 7 x - 4 = 0 \Rightarrow 2 x^{2} + 8 x - x - 4 = 0 \Rightarrow 2 x (x + 4) - 1 (x + 4) = 0 \Rightarrow (2 x - 1) (x + 4) = 0 \Rightarrow (2 x - 1) = 0 or (x + 4) = 0 \Rightarrow 2 x = 1 or x = - 4 \Rightarrow x = \frac{1}{2} or x = - 4$

Therefore, zeroes of the polynomial f(x) = 2x² + 7x – 4 are $\frac{1}{2} and - 4 .$

Hence, the correct option is (b).

Question 9:

If $f (x) = x^{2} - 2 \sqrt{2} x + 1, then f (2 \sqrt{2})$ is equal to
(a) 0
(b) 1
(c) $4 \sqrt{2}$
(d) $8 \sqrt{2} + 1$

Answer:

Given: f(x) = $x^{2} - 2 \sqrt{2} x + 1$

$f (2 \sqrt{2}) = {(2 \sqrt{2})}^{2} - 2 \sqrt{2} (2 \sqrt{2}) + 1 = {(2 \sqrt{2})}^{2} - {(2 \sqrt{2})}^{2} + 1 = 1$

Hence, the correct option is (b).

Question 10:

x + 1 is a factor of the polynomial
(a) x³ + x²– x + 1
(b) x³ + x² + x + 1
(c) x⁴ + x³ + x² + 1
(d) x⁴ + 3x³ + 3x² + x + 1

Answer:

The polynomial f(x) which becomes zero when x = −1 is the required polynomial whose factor is x + 1.

(a) Let f(x) = x³ + x²– x + 1

f(−1) = (−1)³ + (−1)² − (−1) + 1
= −1 + 1 + 1 + 1 = 2 ≠ 0

Therefore, x + 1 is not a factor of the polynomial,

(b) Let f(x) = x³ + x² + x + 1

f(−1) = (−1)³ + (−1)² + (−1) + 1
= −1 + 1 − 1 + 1 = 0

Therefore, x + 1 is a factor of the polynomial,

(c) Let f(x) = x⁴ + x³ + x² + 1

f(−1) = (−1)⁴ + (−1)³ + (−1)² + 1
= 1 − 1 + 1 + 1 = 2 ≠ 0

Therefore, x + 1 is not a factor of the polynomial,

(d) Let f(x) = x⁴ + 3x³ + 3x² + x + 1

f(−1) = (−1)⁴ + 3(−1)³ + 3(−1)² + (−1) + 1
= 1 − 3 + 3 − 1 + 1 = 1 ≠ 0

Therefore, x + 1 is not a factor of the polynomial,

Hence, the correct option is (b).

Question 11:

If x² + kx + 6 = (x + 2) (x + 3) for all x, then the value of k is
(a) 1
(b) –1
(c) 5
(d) 3

Answer:

Given: x² + kx + 6 = (x + 2) (x + 3)

$x^{2} + k x + 6 = (x + 2) (x + 3) \Rightarrow x^{2} + k x + 6 = x^{2} + 2 x + 3 x + 6 \Rightarrow x^{2} + k x + 6 = x^{2} + 5 x + 6 \Rightarrow k x = 5 x \Rightarrow k = 5$

Hence, the correct option is (c).

Question 12:

If x − 2 is a factor of x² + 3ax − 2a, then a =

(a) 2

(b) − 2

(c) 1

(d) −1

Answer:

As is a factor of

i.e.

Hence, correct option is (d).

Question 13:

If x³ + 6x² + 4x + k is exactly divisible by x + 2, then k =

(a) −6

(b) −7

(c) −8

(d) −10

Answer:

As is exactly divisible by

i.e., is a factor of f(x).

Therefore,

Hence, the correct option is (c).

Question 14:

If x − a is a factor of x³ −3x²a + 2a²x + b, then the value of b is

(a) 0

(b) 2

(c) 1

(d) 3

Answer:

As is a factor of

i.e.

Thus, b = 0

Thus, the correct option is (a).

Question 15:

If x¹⁴⁰ + 2x¹⁵¹ + k is divisible by x + 1, then the value of k is

(a) 1

(b) −3

(c) 2

(d) −2

Answer:

As is divisible by

i.e., is a factor of f(x).

Therefore,

Hence, the correct option is (a).

Question 16:

If x + 2 is a factor of x² + mx + 14, then m =

(a) 7

(b) 2

(c) 9

(d) 14

Answer:

As is a factor of

Therefore,

Hence, the correct option is (c).

Question 17:

If x − 3 is a factor of x² − ax − 15, then a =

(a) −2

(b) 5

(c) −5

(d) 3

Answer:

As is a factor of polynomial

i.e.

Therefore,

Hence, the correct option is (a)

Question 23:

Let f(x) be a polynomial such that $f (- \frac{1}{2})$ = 0, then a factor of f(x) is
(a) 2x − 1

(b) 2x + 1

(c) x − 1

(d) x + 1

Answer:

Let f(x) be a polynomial such that

i.e., is a factor.

On rearranging can be written as

Thus, is a factor of f(x).

Hence, the correct option is (b).

Question 24:

When x³ − 2x² + ax − b is divided by x² − 2x − 3, the remainder is x − 6. The values of a and b are respectively

(a) −2, −6

(b) 2 and −6

(c) −2 and 6

(d) 2 and 6

Answer:

If the reminder (x −6) is subtracted from the given polynomial then rest of part of this polynomial is exactly divisible by x² − 2x − 3.

Therefore,

Now,

Therefore, are factors of polynomial p(x).

Now,

And

and

Solving (i) and (ii) we get

Hence, the correct option is (c).

Question 25:

One factor of x⁴ + x² − 20 is x² + 5. The other factor is

(a) x² − 4

(b) x − 4

(c) x² − 5

(d) x + 2

Answer:

It is given that is a factor of the polynomial

Here, reminder is zero. Therefore, is a factor of polynomial.

Thus, the correct option is (a).

Question 26:

If (x − 1) is a factor of polynomial f(x) but not of g(x) , then it must be a factor of

(a) f(x) g(x)

(b) −f(x) + g(x)

(c) f(x) − g(x)

(d) $\{f (x) + g (x)\} g (x)$

Answer:

Asis a factor of polynomial f(x) but not of g(x)

Therefore

Now,

Let

Now

Therefore (x − 1) is also a factor of f(x).g(x).

Hence, the correct option is (a).

Question 27:

(x+1) is a factor of xⁿ + 1 only if

(i) n is an odd integer

(ii) n is an even integer

(iii) n is a negative integer

(iv) n is a positive integer

Answer:

The linear polynomial is a factor of only if

If n is odd integer, then

Hence, the correct option is (a).

Question 28:

If x² + x + 1 is a factor of the polynomial 3x³ + 8x²+ 8x + 3 + 5k, then the value of k is

(a) 0

(b) 2/5

(c) 5/2

(d) −1

Answer:

Let be the given polynomial,

Since is the factor of f(x). Therefore, re`maider will be zero.

Now,

Hence, the correct option is (b).

Question 29:

If (3x − 1)⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵ +...+ a₁x + a₀, then a₇ + a₅+ ...+a₁ + a₀=

(a) 0

(b) 1

(c) 128

(d) 64

Answer:

Given that,

Putting

We get

Hence, the correct option is (c).

Question 30:

If both x − 2 and $x - \frac{1}{2}$ are factors of px² + 5x + r, then

(a) p = r

(b) p + r = 0

(c) 2p + r = 0

(d) p + 2r = 0

Answer:

As and are the factors of the polynomial

i.e., and

Now,

And

From equation (i) and (ii), we get

Hence, the correct option is (a).

Question 18:

If x⁵¹ + 51 is divided by x + 1, the remainder is

(a) 0

(b) 1

(c) 49

(d) 50

Answer:

As is divided by

The remainder will be

Hence, the correct option is (d).

Question 19:

If x + 1 is a factor of the polynomial 2x² + kx, then k =

(a) −2

(b) −3

(c) 4

(d) 2

Answer:

As is a factor of polynomial Therefore,

So,

Hence, the correct option is (d).

Question 20:

If x + a is a factor of x⁴ − a²x² + 3x − 6a, then a =

(a) 0

(b) −1

(c) 1

(d) 2

Answer:

Asis a factor of polynomial

Therefore,

Hence, the correct option is (a).

Question 21:

The value of k for which x − 1 is a factor of 4x³ + 3x² − 4x + k, is

(a) 3

(b) 1

(c) −2

(d) −3

Answer:

As is a factor of polynomial f(x) = 4x³ + 3x² − 4x + k

Therefore,

Hence, the correct option is (d).

Question 22:

If x + 2 and x − 1 are the factors of x³ + 10x² + mx + n, then the values of m and n are respectively

(a) 5 and −3

(b) 17 and −8

(c) 7 and −18

(d) 23 and −19

Answer:

It is given and are the factors of the polynomial

i.e., and

Now

$- 8 + 40 - 2 m + n = 0 \Rightarrow - 2 m + n = - 32 \Rightarrow 2 m - n = 32 . . . (i)$

And

Solving equation (i) and (ii) we get

m = 7 and n = − 18

Hence, the correct option is (c)

Question 31:

If x² − 1 is a factor of ax⁴ + bx³ + cx² + dx + e, then

(a) a + c + e = b + d

(b) a + b +e = c + d

(c) a + b + c = d + e

(d) b + c + d = a + e

Answer:

Asis a factor of polynomial

Therefore,

And

f(1) = 0

$a {(1)}^{4} + b {(1)}^{3} + c {(1)}^{2} + d (1) + e = 0 \Rightarrow a + b + c + d + e = 0$

And

Hence,

The correct option is (a).

Question 1:

7 – 9x + 2x² is called a _________ polynomial.

Answer:

Polynomial with degree 2 is known as a quadratic polynomial.

Hence, 7 – 9x + 2x² is called a quadratic polynomial.

Question 2:

12x – 7x² + 4 – 2x³ is called a ___________ polynomial.

Answer:

Polynomial with degree 3 is known as a cubic polynomial.

Hence, 12x – 7x² + 4 – 2x³ is called a cubic polynomial.

Question 3:

13x² – 88x³ + 9x⁴ is called a _________ polynomial.

Answer:

Polynomial with degree 4 is known as a biquadratic polynomial.

Hence, 12x – 7x² + 4 – 2x³ is called a biquadratic polynomial.

Question 4:

If x + 1 is a factor of the polynomial ax³ + x² – 2x + 4a – 9, then a = __________.

Answer:

Let f(x) = ax³ + x² – 2x + 4a – 9

It is given that one factor of f(x) is (x + 1).

Therefore, $f (x) = 0 when x = - 1$ .

On putting x = –1 in f(x) = 0, we get

$a {(- 1)}^{3} + {(- 1)}^{2} - 2 (- 1) + 4 a - 9 = 0 \Rightarrow - a + 1 + 2 + 4 a - 9 = 0 \Rightarrow 3 a - 6 = 0 \Rightarrow 3 a = 6 \Rightarrow a = 2$

Hence, if x + 1 is a factor of the polynomial ax³ + x² – 2x + 4a – 9, then a = 2.

Question 5:

If 3x – 1 is a factor of the polynomial 81x³ – 45x² +3a – 6, then the value of a is ___________.

Answer:

Let f(x) = 81x³ – 45x² +3a – 6

It is given that one factor of f(x) is (3x – 1).

Therefore, $f (x) = 0 when x = \frac{1}{3}$ .

On putting x = $\frac{1}{3}$ in f(x) = 0, we get

$81 {(\frac{1}{3})}^{3} - 45 {(\frac{1}{3})}^{2} + 3 a - 6 = 0 \Rightarrow 81 (\frac{1}{27}) - 45 (\frac{1}{9}) + 3 a - 6 = 0 \Rightarrow 3 - 5 + 3 a - 6 = 0 \Rightarrow 3 a - 8 = 0 \Rightarrow 3 a = 8 \Rightarrow a = \frac{8}{3}$

Hence, if 3x – 1 is a factor of the polynomial 81x³ – 45x² +3a – 6, then the value of a is $\frac{8}{3} .$

Question 6:

The remainders obtained when x³ + x² – 9x – 9 is divided by x, x + 1 and x + 2 respectively are _________.

Answer:

Let f(x) = x³ + x² – 9x – 9

To find the remainder obtained when x³ + x² – 9x – 9 is divided by x,

we use remainder theorem, put x = 0.

f(0) is the remainder.

Now,

f(0) = (0)³ + (0)² – 9(0) – 9
= –9

Hence, the remainder obtained when x³ + x² – 9x – 9 is divided by x is –9.

To find the remainder obtained when x³ + x² – 9x – 9 is divided by x + 1,

put x +1 = 0.

f(–1) is the remainder.

Now,

f(–1) = (–1)³ + (–1)² – 9(–1) – 9
= –1 + 1 + 9 – 9
= 0

Hence, the remainder obtained when x³ + x² – 9x – 9 is divided by x + 1 is 0.

To find the remainder obtained when x³ + x² – 9x – 9 is divided by x + 2,

put x +2 = 0.

f(–2) is the remainder.

Now,

f(–2) = (–2)³ + (–2)² – 9(–2) – 9
= –8 + 4 + 18 – 9
= 5

Hence, the remainder obtained when x³ + x² – 9x – 9 is divided by x + 2 is 5.

Hence, the remainders obtained when x³ + x² – 9x – 9 is divided by x, x + 1 and x + 2 respectively are –9, 0 and 5.

Question 7:

The remainder when f(x) = 4x³– 3x² + 2x – 1 is divided by 2x + 1 is __________.

Answer:

Let f(x) = 4x³– 3x² + 2x – 1

To find the remainder obtained when 4x³– 3x² + 2x – 1 is divided by 2x + 1,

we use remainder theorem, put 2x + 1 = 0.

f( $- \frac{1}{2}$ ) is the remainder.

Now,

$f (- \frac{1}{2}) = 4 {(- \frac{1}{2})}^{3} - 3 {(- \frac{1}{2})}^{2} + 2 (- \frac{1}{2}) - 1 = 4 (- \frac{1}{8}) - 3 (\frac{1}{4}) - 1 - 1 = - \frac{1}{2} - \frac{3}{4} - 2 = \frac{- 2 - 3 - 8}{4} = - \frac{13}{4}$

Hence, the remainder when f(x) = 4x³– 3x² + 2x – 1 is divided by 2x + 1 is $- \frac{13}{4} .$

Question 8:

The degree of a polynomial f(x) is 7 and that of polynomial f(x) g(x) is 56, then degree of g(x) is ________.

Answer:

Given:
Degree of a polynomial f(x) = 7
Degree of polynomial f(x) g(x) = 56

Degree of polynomial f(x) g(x) = Degree of polynomial f(x) × Degree of polynomial g(x)
⇒ 56 = 7 × Degree of polynomial g(x)
⇒ Degree of polynomial g(x) = $\frac{56}{7}$
⇒ Degree of polynomial g(x) = 8

Hence, degree of g(x) is 8.

Question 9:

The remainder when x¹⁵is divided by x + 1 is __________.

Answer:

Let f(x) = x¹⁵

To find the remainder obtained when x¹⁵ is divided by x + 1,

we use remainder theorem, put x + 1 = 0.

f(−1) is the remainder.

Now,

$f (- 1) = {(- 1)}^{15} = - 1$

Hence, the remainder when x¹⁵is divided by x + 1 is −1.

Question 10:

If $p (x) = x^{2} - 4 x + 3, then p (2) - p (- 1) + p (\frac{1}{2}) =_____________.$

Answer:

Let p(x) = x² – 4x + 3

$p (2) = {(2)}^{2} - 4 (2) + 3 = 4 - 8 + 3 = - 1 p (- 1) = {(- 1)}^{2} - 4 (- 1) + 3 = 1 + 4 + 3 = 8 p (\frac{1}{2}) = {(\frac{1}{2})}^{2} - 4 (\frac{1}{2}) + 3 = \frac{1}{4} - 2 + 3 = \frac{1}{4} + 1 = \frac{1 + 4}{4} = \frac{5}{4} Now, p (2) - p (- 1) + p (\frac{1}{2}) = - 1 - 8 + \frac{5}{4} = - 9 + \frac{5}{4} = \frac{- 36 + 5}{4} = - \frac{31}{4}$

Hence, if $p (x) = x^{2} - 4 x + 3, then p (2) - p (- 1) + p (\frac{1}{2}) = - \frac{31}{4} .$

Question 11:

If the polynomial f(x) = 5x⁵ – 3x³ + 2x² – k gives remainder 1 when divided by x + 1, then k = __________.

Answer:

Let f(x) = 5x⁵ – 3x³ + 2x² – k

To find the remainder obtained when 5x⁵ – 3x³ + 2x² – k is divided by x + 1,

we use remainder theorem, put x + 1 = 0.

f(−1) is the remainder.

Now,

$f (- 1) = 5 {(- 1)}^{5} - 3 {(- 1)}^{3} + 2 {(- 1)}^{2} - k \Rightarrow 1 = 5 (- 1) - 3 (- 1) + 2 (1) - k \Rightarrow 1 = - 5 + 3 + 2 - k \Rightarrow 1 = - k \Rightarrow k = - 1$

Hence, k = –1.

Question 12:

The remainder when f(x) = x⁴⁵ is divided by x² – 1 is ____________.

Answer:

Let f(x) =  x⁴⁵

To find the remainder obtained when  x⁴⁵ is divided by  x² – 1,

Let the remainder (r) be ax + b.

Then,
f(x) =  (x² – 1) q + r
$\Rightarrow x^{45} = (x^{2} - 1) q + (a x + b) . . . (1) Putting x = 1 in equation (1), we get {(1)}^{45} = ({(1)}^{2} - 1) q + (a (1) + b) \Rightarrow 1 = 0 + a + b \Rightarrow a + b = 1 . . . (2) Putting x = - 1 in equation (1), we get {(- 1)}^{45} = ({(- 1)}^{2} - 1) q + (a (- 1) + b) \Rightarrow - 1 = 0 - a + b \Rightarrow - a + b = - 1 . . . (3) Solving (2) and (3), we get b = 0 and a = 1$

Therefore, the remainder (r) = 1(x) + 0 = x.

Hence, the remainder when f(x) = x⁴⁵ is divided by x² – 1 is x.

Question 1:

Define zero or root of a polynomial.

Answer:

A real number a is a zero (or root) of the polynomial f(x), if f (a) = 0

Question 2:

If $x = \frac{1}{2}$ is a zero of the polynomial f(x) = 8x³ + ax² − 4x + 2, find the value of a.

Answer:

Since is a zero of polynomial f(x).

Therefore

The value of a is .

Question 3:

Write the remainder when the polynomial f(x) = x³ + x² − 3x + 2 is divided by x + 1.

Answer:

When the polynomial f(x), divided by the remainder will be

Thus, the reminder = 5

Question 4:

Find the remainder when x³ + 4x² + 4x − 3 is divided by x.

Answer:

When the polynomial f(x) = x³ + 4x² + 4x − 3 is divided by x the remainder will be.

Thus, the reminder = −3

Question 5:

If x + 1 is a factor of x³ + a, then write the value of a.

Answer:

As is a factor of polynomial

i.e.

${(- 1)}^{3} + a = 0 \Rightarrow a = 1$

Thus, the value of a = 1

Question 6:

If f(x) = x⁴ − 2x³ + 3x² − ax − b when divided by x − 1, the remainder is 6, then find the value of a + b

Answer:

When polynomial divided by

The remainder is 6.

i.e.

Thus, the value of

Question 1:

If f(x) = 2x³ − 13x² + 17x + 12, find (i) f(2) (ii) f(−3) (iii) f(0)

Answer:

Let be the given polynomial

(i) The value of f (2) can be found by putting x = 2

= 10

(ii) The value of f (–3) can be found by putting x = –3

(iii) The value of f (0) can be found by putting x = 0

Question 2:

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:

(i) $f (x) = 3 x + 1, x = - \frac{1}{3}$

(ii) $f (x) = x^{2} - 1, x = 1, - 1$

(iii) $g (x) = 3 x^{2} - 2, x = \frac{2}{\sqrt{3}}, - \frac{2}{\sqrt{3}}$

(iv) $p (x) = x^{3} - 6 x^{2} + 11 x - 6, x = 1, 2, 3$

(v) $f (x) = 5 x - π, x = \frac{4}{5}$

(vi) $f (x) = x^{2}, x = 0$

(vii) $f (x) = l x + m, x = - \frac{m}{1}$

(viii) $f (x) = 2 x + 1, x = \frac{1}{2}$

Answer:

(i) To check whether the given number is the zero of the polynomial or not we have to find

Hence, is the zeros of the given polynomial.

(ii) To check whether the given number is the zero of the polynomial or not we have to find and

Now,

And ,

Hence, and x = 1 are the zeros of the polynomial.

(iii) To check whether the given number is the zero of the polynomial or not we have to find

Now,

And

Here, both the value of , are not satisfied the polynomial. Therefore, they are not the zeros of the polynomial.

(iv) To check whether the given number is the zero of the polynomial or not we have to find

And

Hence, x = 1, 2, 3 are the zeros of the polynomial p(x).

(v)To check whether the given number is the zero of the polynomial or not we have to find

Here, , does not satisfy therefore, is not a zero of the polynomial.

(vi)To check whether the given number is the zero of the polynomial or not we have to find

Hence, x = 0 is the zeros of the polynomial.

(vii) To check whether the given number is the zero of the polynomial or not we have to find

Hence, is zeros of the polynomial.

(viii)To check whether the given number is the zero of the polynomial or not we have to find

Hence, does not satisyf the polynomial, so, is not zero of the polynomial.

Question 3:

If x = 2 is a root of the polynomial f(x) = 2x² − 3x + 7a, find the value of a.

Answer:

The given polynomial is

If x = 2 is the root of the polynomial.

Then

Question 4:

If $x = - \frac{1}{2}$ is a zero of the polynomial p(x) = 8x³ − ax²−x + 2, find the value of a.

Answer:

The given polynomial is

If is a zeros of the polynomial p(x).

then

Therefore,

Hence the value of

Question 5:

If x = 0 and x = −1 are the roots of the polynomial f(x) =2x³ − 3x²+ ax + b, find the value of a and b.

Answer:

The given polynomial is

f(x) =2x³ − 3x²+ ax + b

If is zeros of the polynomial f(x), then f(0) = 0

Similarly, if x = − 1 is the zeros of the polynomial of,

Then,

Putting the value of b from equation (1)

Thus,

Question 6:

Find the integral roots of the polynomial f(x) = x³ + 6x² + 11x + 6.

Answer:

The given polynomial is

Here, f(x) is a polynomial with integer coefficient and the coefficient of highest degree term is 1. So, the integer roots of f(x) are factors of 6. Which are by observing.

= 0

Also,

And similarly,

f(−3) = 0

Therefore, the integer roots of the polynomial f(x) are −1, −2, − 3

Question 7:

Find rational roots of the polynomial f(x) = 2x³ + x² − 7x − 6.

Answer:

The given polynomial is

f(x) is a cubic polynomial with integer coefficients. If $\frac{b}{c}$ is rational root in lowest terms, then the values of b are limited
to the factors of 6 which are $\pm 1, \pm 2, \pm 3, \pm 6$ and the values of c are limited to the factor of 2 as $\pm 1, \pm 2$ . Hence, the possible
rational roots are $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$ .

Since,

So, 2 is a root of the polynomial

Now, the polynomial can be written as,

Also,

Therefore,

Hence, the rational roots of the polynomialare 2, – 3/2 and – 1.

Rd Sharma 2021 for Class 9 Maths Chapter 6 - Factorization Of Polynomials

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