Rd Sharma 2019 2020 for Class 8 Maths Chapter 8 - Division Of Algebraic Expressions

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$$\begin{gathered} \mathrm{Quotient} = 7x^2 + x + 5 \\ \mathrm{Remainder} = 4 \end{gathered}$$

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$$\begin{gathered} \mathrm{Quotient} = 3x^2+ 4x + 1 \\ \mathrm{Remainder} = 0 \end{gathered}$$

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$$\begin{gathered} \mathrm{Quotient} = 2x-5 \\ \mathrm{Remainder} =0 \end{gathered}$$

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$$\begin{gathered} \mathrm{Quotient} =10x^2-3x-12 \\ \mathrm{Remainder}= 0 \end{gathered}$$

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$\therefore$ Quotient = 3x² + 4x + 2 and remainder = 0.

Answer:

(i)

Quotient = 2x + 3
Remainder = $-$3
Divisor = 7x $-$ 4
Divisor $\times$ Quotient + Remainder = (7x $-$ 4) (2x + 3) $-$ ​3 
                                                = 14x² + 21x $-$ 8x $-$ 12 $-$ ​3 
                                                = 14x² + 13x $-$ 15
                                                = Dividend
Thus,
Divisor $\times$ Quotient + Remainder = Dividend
Hence verified.

(ii)

$$\begin{gathered} \mathrm{Quotient} = 5z^2+\frac{10}{3}z+11 \\ \mathrm{Remainder} = 54 \\ \mathrm{Divisor} = 3z-6 \\ \mathrm{Divisor} \times \mathrm{Quotient} +\mathrm{Remainder} = (3z-6)\left( 5z^2+\frac{10}{3}z+11\right)+54 \\ = 15z^3+10z^2+33z-30z^2-20z-66+54 \\ = 15z^3-20z^2+13z-12 \\ = \mathrm{Dividend} \\ \text{Thus}, \\ \mathrm{Divisor} \times \mathrm{Quotient} + \mathrm{Remainder} = \mathrm{Dividend} \end{gathered}$$
Hence verified.

(iii)


Quotient = $3y^3-5y+\frac{3}{2}$
Remainder = 0
Divisor = 2y² $-$ 6
Divisor $\times$ Quotient + Remainder =
$$\begin{gathered} (2y^2-6) \left(3y^3-5y+\frac{3}{2}\right)+0 \\ =6y^5-10y^3+3y^2-18y^3+30y-9 \\ =6y^5-28 y^3+3y^2+30y-9 \end{gathered}$$
= Dividend
 
Thus, Divisor $\times$ Quotient + Remainder = Dividend
Hence verified.

(iv)

Quotient  = $-$ 4x³ + 2x² $-$ 8x + 30
Remainder  = $-$ 285 
Divisor  = 3x + 7
Divisor $\times$ Quotient + Remainder =  (3x + 7) ($-$ 4x³ + 2x² $-$ 8x + 30) $-$ 285 
                                                 = $-$ 12x⁴ + 6x³ $-$ 24x² + 90x $-$ 28x³ + 14x² $-$ 56x + 210 $-$ ​285
                                                 = $-$ 12x ⁴ $-$ 22x³ $-$ 10x² + 34x $-$ 75
                                                 =  Dividend
Thus,
Divisor $\times$ Quotient + Remainder = Dividend
Hence verified.

(v)


Quotient =  $5y^3-2y^2+\frac{5}{3}y$
Remainder =  6
Divisor = 3y $-$ 2
Divisor $\times$ Quotient  + Remainder = (3y $-$ 2) (5y³ $-$ 2y² + $\frac{5}{3}y$) + 6
                                                = $15y^4-6y^3+5y^2-10y^3+4y^2-\frac{10}{3}y+6$
                                                = $15y^4-16y^3+9y^2-\frac{10}{3}y+6$
                                                =  Dividend
Thus,
Divisor $\times$ Quotient + Remainder = Dividend
Hence verified.

(vi)

Quotient =  2y + 5
Remainder =  11y + 2
Divisor =  2y² $-$ y + 1
Divisor $\times$ Quotient + Remainder =  (2y² $-$ y + 1) (2y + 5) + 11y + 2
                                                =  4y³ +10y² $-$ 2y² $-$ 5y + 2y + 5 + 11y + 2
                                                =  4y³ + 8y² + 8y + 7
                                                =  Dividend
Thus,
Divisor $\times$ Quotient + Remainder  = Dividend
Hence verified.

(vii)




Quotient = 3y² + 2y + 2
Remainder = 4y² + 25y + 4
Divisor = 2y³ + 1
Divisor $\times$ Quotient + Remainder = (2y³ + 1) (3y² + 2y + 2) + 4y² + 25y + 4
                                                = 6y⁵ + 4y⁴ + 4y³ + 3y² + 2y + 2 + 4y² + 25y + 4
                                                = 6y⁵ + 4y⁴ + 4y³ + 7y² + 27y + 6
                                                = Dividend
Thus,
Divisor $\times$ Quotient + Remainder = Dividend
Hence verified.

Answer:

$\therefore$ Quotient = 5y³ + (26/3)y² + (25/9)y + (80/27)
Remainder = ($-$ 2/27)
Coefficient of y³ = 5
Coefficient of y² = (26/3)
Coefficient of y = (25/9)
Constant = (80/27)

Answer:

(i)

Remainder is zero. Hence (x+6) is a factor of x² -x-42
(ii)

As the remainder is non zero . Hence ( 4x-1) is not a factor of 4x² -13x-12



(iii)




$\because$ The remainder is non zero,
 2y $-$ 5 is not a factor of $4y^4-10y^3-10y^2+30y-15$.

(iv)

Remainder is zero.  Therefore, 3y² + 5 is a factor of $6y^5+15y^4+16y^3+4y^2+10y-35$.


(v)

Remainder is zero; therefore, z² + 3 is a factor of $z^{5 }-9z$.

(vi)



Remainder is zero ; therefore, $2x^2-x+3$ is a factor of $6x^5-x^{4 }+4x^3-5x^2-x-15$.

Answer:

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{a} \mathrm{if} (\mathrm{x}+2) \mathrm{is} \mathrm{a} \mathrm{factor} \mathrm{of} (4{\mathrm{x}}^4+2{\mathrm{x}}^3-3{\mathrm{x}}^2+8\mathrm{x}+5\mathrm{a}). \\ S\mathrm{ubstituting} \mathrm{x}=-2 \mathrm{in} 4{\mathrm{x}}^4+2{\mathrm{x}}^3-3{\mathrm{x}}^2+8\mathrm{x}+5\mathrm{a}, \mathrm{we} \mathrm{get}: \\ 4(-2{)}^4+2(-2{)}^3-3(-2{)}^2+8(-2)+5\mathrm{a}=0 \\ \mathrm{or}, 64-16-12-16+5\mathrm{a}=0 \\ \mathrm{or}, 5\mathrm{a}=-20 \\ \mathrm{or}, \mathrm{a}=-4 \\ \therefore \mathrm{If} (\mathrm{x}+2) \mathrm{is} \mathrm{a} \mathrm{factor} \mathrm{of} (4{\mathrm{x}}^4+2{\mathrm{x}}^3-3{\mathrm{x}}^2+8\mathrm{x}+5\mathrm{a}), \mathrm{a}=-4. \end{gathered}$$

Answer:

Thus, (x $-$ 2) should be added to ($x^4+2x^3-2x^2+x-1$) to make the resulting polynomial exactly divisible by ($x^2+2x-3$).

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{3{\mathrm{x}}^2+4\mathrm{x}+5}{\mathrm{x}-2} \\ =\frac{3\mathrm{x}(\mathrm{x}-2)+10(\mathrm{x}-2)+25}{(\mathrm{x}-2)} \\ =\frac{(\mathrm{x}-2)(3\mathrm{x}+10)+25}{(\mathrm{x}-2)} \\ =(3\mathrm{x}+10)+\frac{25}{(\mathrm{x}-2)} \\ \mathrm{Therefore}, \\ \text{q}\mathrm{uot}\text{ie}\mathrm{nt}=3\mathrm{x}+10 \mathrm{and} \text{r}\mathrm{emainder}=25. \\ \left(\mathrm{ii}\right) \frac{10{\mathrm{x}}^2-7\mathrm{x}+8}{5\mathrm{x}-3} \\ =\frac{2\mathrm{x}(5\mathrm{x}-3)-{\displaystyle \frac{1}{5}}(5\mathrm{x}-3)+{\displaystyle \frac{47}{5}}}{(5\mathrm{x}-3)} \\ =\frac{(5\mathrm{x}-3)(2\mathrm{x}-{\displaystyle \frac{1}{5}})+{\displaystyle \frac{47}{5}}}{(5\mathrm{x}-3)} \\ =(2\mathrm{x}-\frac{1}{5})+\frac{{\displaystyle \frac{47}{5}}}{5\mathrm{x}-3} \\ \mathrm{Therefore}, \\ \text{q}\mathrm{uot}\text{ie}\mathrm{nt}=2\mathrm{x}-\frac{1}{5} \mathrm{and} \text{r}\mathrm{emainder}=\frac{47}{5}. \\ \left(\mathrm{iii}\right) \frac{5{\mathrm{y}}^3-6{\mathrm{y}}^2+6\mathrm{y}-1}{5\mathrm{y}-1} \\ =\frac{{\mathrm{y}}^2(5\mathrm{y}-1)-\mathrm{y}(5\mathrm{y}-1)+1(5\mathrm{y}-1)}{(5\mathrm{y}-1)} \\ =\frac{(5\mathrm{y}-1)({\mathrm{y}}^2-\mathrm{y}+1)}{(5\mathrm{y}-1)} \\ =({\mathrm{y}}^2-\mathrm{y}+1) \\ \mathrm{Therefore}, \\ \mathrm{Quotient} = {\mathrm{y}}^2-\mathrm{y}+1 \mathrm{and} \mathrm{remainder} = 0 \end{gathered}$$

$$\begin{gathered} (iv)\frac{ x^4{-x}^3+5x}{\mathrm{x-1}} \\ =\frac{x^3(x-1)+5(x-1)+5}{\mathrm{x-1}} \\ =\frac{{(x-1)(x}^3+5)+5}{\mathrm{x-1}} \\ {=(x}^3+5)+\frac{5}{x-1} \\ \mathrm{Therefore,} \text{q}\mathrm{uot}\text{ie}\mathrm{nt} = x^3+5 \mathrm{and}\text{ r}\mathrm{emainder} = 5. \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right)\frac{ {\mathrm{y}}^4+{\mathrm{y}}^2}{{\mathrm{y}}^2-2} \\ =\frac{{\mathrm{y}}^2({\mathrm{y}}^2-2)+3({\mathrm{y}}^2-2)+6}{{\mathrm{y}}^2-2} \\ =\frac{({\mathrm{y}}^2-2)({\mathrm{y}}^2+3)+6}{{\mathrm{y}}^2-2} \\ =({\mathrm{y}}^2+3)+\frac{6}{{\mathrm{y}}^2-2} \\ \mathrm{Therefore}, \text{q}\mathrm{uotient} = {\mathrm{y}}^2+3 \mathrm{and} \text{r}\mathrm{emainder} = 6. \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{2{\mathrm{x}}^2+5\mathrm{x}+4}{\mathrm{x}+1} \\ =\frac{2\mathrm{x}(\mathrm{x}+1)+3(\mathrm{x}+1)+1}{\mathrm{x}+1} \\ =\frac{(\mathrm{x}+1)(2\mathrm{x}+3)+1}{(\mathrm{x}+1)} \\ =(2\mathrm{x}+3)+\frac{1}{\mathrm{x}+1} \\ \because \mathrm{Remainder}=1 \\ \mathrm{Therefore}, (\mathrm{x}+1) \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{factor} \mathrm{of} 2{\mathrm{x}}^2+5\mathrm{x}+4 \end{gathered}$$

$$\begin{gathered} (\mathrm{ii)} \frac{{3y}^3{+5y}^2+5y+2}{\mathrm{y-2}} \\ =\frac{{\mathrm{3y}}^2(y-2)+11y(y-2)+27(y-2)+56}{\mathrm{y-2}} \\ =\frac{{(y-2)(3y}^2+11y+27)+56}{\mathrm{y-2}} \\ {=(3y}^2+11y+27)+\frac{56}{\mathrm{y-2}} \\ \because \mathrm{Remainder} = 56 \\ \therefore (y-2) is \mathrm{not} a \mathrm{factor} of {\mathrm{3y}}^3{+5y}^2+5y+2. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right) \frac{4{\mathrm{x}}^4{+}^2+15}{{\mathrm{4x}}^2-5} \\ = \frac{{\mathrm{x}}^2{(4x}^2{-5)+3(4x}^2-5)+30}{{\mathrm{4x}}^2-5} \\ = \frac{({\mathrm{4x}}^2{-5)(x}^2+3)+30}{{\mathrm{4x}}^2-5} \\ {=(x}^2+3)+\frac{30}{{\mathrm{4x}}^2-5} \\ \because \mathrm{Remainder} = 30 \\ \mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}, (4{\mathrm{x}}^2-5) \mathrm{i}\mathrm{s} \mathrm{n}\mathrm{o}\mathrm{t} \mathrm{a} \mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r} \mathrm{o}\mathrm{f} 4{\mathrm{x}}^4+7{\mathrm{x}}^2+15 \end{gathered}$$

$$\begin{gathered} (iv) \frac{{\mathrm{3z}}^2-13z+4}{\mathrm{4-z}} \\ =\frac{{\mathrm{3z}}^2-12z-z+4}{\mathrm{4-z}} \\ =\frac{\mathrm{3z(z-4)-1(z-4)}}{\mathrm{4-z}} \\ =\frac{(z-4)(3z-1)}{\mathrm{4-z}} \\ =\frac{(\mathrm{4-z)(1-3z)}}{\mathrm{4-z}} \\ =1-3z \\ \because \mathrm{Remainder} = 0 \\ \therefore (4-z) \mathrm{is} a \mathrm{factor} of {\mathrm{3z}}^2-13z+4. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{V}\right) \frac{{\mathrm{10a}}^2-9a-5}{\mathrm{2a-3}} \\ =\frac{\mathrm{5a(2a-3)+3(2a-3)+4}}{\mathrm{2a-3}} \\ =\frac{(2a-3)(5a+3)+4}{\mathrm{2a-3}} \\ =(5a+3)+\frac{4}{\mathrm{2a-3}} \\ \because \mathrm{Remainder} = 4 \\ \therefore (\; 2a-3) \mathrm{is} \mathrm{not} a \mathrm{factor} \mathrm{of} {\mathrm{10a}}^2-9a-5. \end{gathered}$$

$$\begin{gathered} (vi) \frac{{\mathrm{8y}}^2-2y+1}{\mathrm{4y+1}} \\ =\frac{\mathrm{2y}(4y+1)-1(4y+1)+2}{\mathrm{4y+1}} \\ =\frac{(4y+1)(2y-1)+2}{\mathrm{4y+1}} \\ =(2y-1)+\frac{2}{\mathrm{4y+1}} \\ \because \mathrm{Remainder} = 2 \\ \therefore (4y+1) \mathrm{is} \mathrm{not} a \mathrm{factor} \mathrm{of} {\mathrm{8y}}^2-2y+1. \end{gathered}$$

Answer:

$$\begin{gathered} \frac{x^{\mathit{2}}\mathit{-}\mathit{5}x\mathit{+}\mathit{6}}{x\mathit{-}\mathit{3}} \\ \mathit{=}\frac{x^{\mathit{2}}\mathit{-}\mathit{3}x\mathit{-}\mathit{2}x\mathit{+}\mathit{6}}{x\mathit{-}\mathit{3}} \\ \mathit{=}\frac{x\mathit{(}x\mathit{-}\mathit{3}\mathit{)}\mathit{-}\mathit{2}\mathit{(}x\mathit{-}\mathit{3}\mathit{)}}{\mathit{(}x\mathit{-}\mathit{3}\mathit{)}} \\ \mathit{=}\frac{\mathit{(}x\mathit{-}\mathit{3}\mathit{)}\mathit{(}x\mathit{-}\mathit{2}\mathit{)}}{\mathit{(}x\mathit{-}\mathit{3}\mathit{)}} \\ \mathit{=}\mathit{ }x\mathit{-}\mathit{2} \end{gathered}$$

Answer:

$$\begin{gathered} \frac{a{x}^{\mathit{2}}\mathit{-}a{y}^{\mathit{2}}}{ax\mathit{+}ay} \\ \mathit{=}\frac{a\mathit{(}{x}^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}\mathit{)}}{a\mathit{(}x\mathit{+}y\mathit{)}} \\ \mathit{=}\frac{a\mathit{(}x\mathit{+}y\mathit{)}\mathit{(}x\mathit{-}y\mathit{)}}{a\mathit{(}x\mathit{+}y\mathit{)}} \\ \mathit{=}\mathit{ }x\mathit{-}y \end{gathered}$$

Answer:

$$\begin{gathered} \frac{x^{\mathit{4}}\mathit{-}{y}^{\mathit{4}}}{x^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}} \\ \mathit{=}\frac{\mathit{(}{x}^{\mathit{2}}{\mathit{)}}^{\mathit{2}}\mathit{-}\mathit{(}{y}^{\mathit{2}}{\mathit{)}}^{\mathit{2}}}{\mathit{(}{x}^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}\mathit{)}} \\ \mathit{=}\frac{\mathit{(}{x}^{\mathit{2}}\mathit{+}{y}^{\mathit{2}}\mathit{)}\mathit{(}{x}^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}\mathit{)}}{\mathit{(}{x}^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}\mathit{)}} \\ \mathit{=}\mathit{ }{x}^{\mathit{2}}\mathit{+}{y}^{\mathit{2}} \end{gathered}$$

Answer:

$$\begin{gathered} \frac{ac{x}^{\mathit{2}}\mathit{+}\mathit{(}bc\mathit{+}ad\mathit{)}x\mathit{+}bd}{\mathit{(}ax\mathit{+}b\mathit{)}} \\ \mathit{=}\frac{ac{x}^{\mathit{2}}\mathit{+}bcx\mathit{+}adx\mathit{+}bd}{\mathit{(}ax\mathit{+}b\mathit{)}} \\ \mathit{=}\frac{cx\mathit{(}ax\mathit{+}b\mathit{)}\mathit{+}d\mathit{(}ax\mathit{+}b\mathit{)}}{\mathit{(}ax\mathit{+}b\mathit{)}} \\ \mathit{=}\frac{\mathit{(}ax\mathit{+}b\mathit{)}\mathit{(}cx\mathit{+}d\mathit{)}}{\mathit{(}ax\mathit{+}b\mathit{)}} \\ \mathit{=}\mathit{ }cx\mathit{+}d \end{gathered}$$

Answer:

$$\begin{gathered} \frac{\mathit{(}{a}^{\mathit{2}}\mathit{+}\mathit{2}ab\mathit{+}{b}^{\mathit{2}}\mathit{)}\mathit{-}\mathit{(}{a}^{\mathit{2}}\mathit{+}\mathit{2}ac\mathit{+}{c}^{\mathit{2}}\mathit{)}}{\mathit{(}\mathit{2}a\mathit{+}b\mathit{+}c\mathit{)}} \\ \mathit{=}\frac{\mathit{(}a\mathit{+}b{\mathit{)}}^{\mathit{2}}\mathit{-}\mathit{(}a\mathit{+}c{\mathit{)}}^{\mathit{2}}}{\mathit{(}\mathit{2}a\mathit{+}b\mathit{+}c\mathit{)}} \\ \mathit{=}\frac{\mathit{(}a\mathit{+}b\mathit{+}a\mathit{+}c\mathit{)}\mathit{(}a\mathit{+}b\mathit{-}a\mathit{-}c\mathit{)}}{\mathit{(}\mathit{2}a\mathit{+}b\mathit{+}c\mathit{)}} \\ \mathit{=}\frac{\mathit{(}\mathit{2}a\mathit{+}b\mathit{+}c\mathit{)}\mathit{(}b\mathit{-}c\mathit{)}}{\mathit{(}\mathit{2}a\mathit{+}b\mathit{+}c\mathit{)}} \\ \mathit{=}b\mathit{-}c \end{gathered}$$

Answer:

$$\begin{gathered} \frac{{\displaystyle \frac{\mathit{1}}{\mathit{4}}}{x}^{\mathit{2}}\mathit{-}{\displaystyle \frac{\mathit{1}}{\mathit{2}}}x\mathit{-}\mathit{12}}{{\displaystyle \frac{\mathit{1}}{\mathit{2}}}x\mathit{-}\mathit{4}} \\ \mathit{=}\frac{{\displaystyle \frac{\mathit{1}}{\mathit{2}}}x\mathit{(}{\displaystyle \frac{\mathit{1}}{\mathit{2}}}x\mathit{-}\mathit{4}\mathit{)}\mathit{+}\mathit{3}\mathit{(}{\displaystyle \frac{\mathit{1}}{\mathit{2}}}x\mathit{-}\mathit{4}\mathit{)}}{{\displaystyle \frac{\mathit{1}}{\mathit{2}}}x\mathit{-}\mathit{4}} \\ \mathit{=}\frac{\mathit{(}{\displaystyle \frac{\mathit{1}}{\mathit{2}}}x\mathit{-}\mathit{4}\mathit{)}\mathit{(}{\displaystyle \frac{\mathit{1}}{\mathit{2}}}x\mathit{+}\mathit{3}\mathit{)}}{\mathit{(}{\displaystyle \frac{\mathit{1}}{\mathit{2}}}x\mathit{-}\mathit{4}\mathit{)}} \\ \mathit{=}\frac{\mathit{1}}{\mathit{2}}x\mathit{+}\mathit{3} \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \mathit{C}\mathit{o}\mathit{r}\mathit{r}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ } : It is 2{\mathrm{x}}^3+5{\mathrm{x}}^2-7 instead of 2{\mathrm{x}}^2+5{\mathrm{x}}^2-7. \\ \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polymonial} 2{\mathrm{x}}^3+5{\mathrm{x}}^2-7 \mathrm{is} 3. \\ \left(\mathrm{ii}\right) \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polymonial} 5{\mathrm{x}}^2-35\mathrm{x}+2 \mathrm{is} 2. \\ \left(\mathrm{iii}\right) \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polymonial} 2\mathrm{x}+{\mathrm{x}}^2-8 \mathrm{is} 2. \\ \left(\mathrm{iv}\right) \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polymonial} \frac{1}{2}{\mathrm{y}}^7-12{\mathrm{y}}^6+48{\mathrm{y}}^5-10 \mathrm{is} 7. \\ \left(\mathrm{v}\right) \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polymonial} 3{\mathrm{x}}^3+1 \mathrm{is} 3. \\ \left(\mathrm{vi}\right) 5 \mathrm{is} \mathrm{a} \mathrm{constant} \mathrm{polynomial} \mathrm{and} \mathrm{its} \mathrm{degree} \mathrm{is} 0. \\ \left(\mathrm{vii}\right) \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polymonial} 20{\mathrm{x}}^3+12{\mathrm{x}}^2{\mathrm{y}}^2-10{\mathrm{y}}^2+20 \mathrm{is} 4. \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) {\mathrm{x}}^2+2{\mathrm{x}}^{-2} \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{polynomial} \mathrm{because} -2 \mathrm{is} \mathrm{the} \mathrm{power} \mathrm{of} \mathrm{variable} \mathrm{x} \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{non} \mathrm{negative} \mathrm{integer}. \\ \left(\mathrm{ii}\right) \sqrt{\mathrm{ax}}+{\mathrm{x}}^2-{\mathrm{x}}^3 \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{polynomial} \mathrm{because} \frac{1}{2} \mathrm{is} \mathrm{the} \mathrm{power} \mathrm{of} \mathrm{variable} \mathrm{x} \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{non} \mathrm{negative} \mathrm{integer}. \\ \left(\mathrm{iii}\right) 3{\mathrm{y}}^3-\sqrt{5}\mathrm{y}+9 \mathrm{is} \mathrm{a} \mathrm{polynomial} \mathrm{because} \mathrm{the} \mathrm{powers} \mathrm{of} \mathrm{variable} \mathrm{y} \mathrm{are} \mathrm{non} \mathrm{negative} \mathrm{integer}\text{s}. \\ \left(\mathrm{iv}\right) {\mathrm{ax}}^{\frac{1}{2}}+\mathrm{ax}+9{\mathrm{x}}^2+4 \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{polynomial} \mathrm{because} \frac{1}{2} \mathrm{is} \mathrm{the} \mathrm{power} \mathrm{of} \mathrm{variable} \mathrm{x} \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{non} \mathrm{negative} \mathrm{integer}. \\ \left(\mathrm{v}\right) 3{\mathrm{x}}^{-2}+2{\mathrm{x}}^{-1}+4\mathrm{x}+5 \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{polynomial} \mathrm{because} -2 \mathrm{and} -1 \mathrm{are} \mathrm{the} \mathrm{powers} \mathrm{of} \mathrm{variable} \mathrm{x} \mathrm{are} \mathrm{not} \mathrm{non} \mathrm{negative} \mathrm{integer}s. \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Standard} \mathrm{form} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{polynomial} \mathrm{can} \mathrm{be} \mathrm{expressed} \mathrm{as}: \\ (5{\mathrm{x}}^4+{\mathrm{x}}^2+6\mathrm{x}+3) \mathrm{or} (3+6\mathrm{x}+{\mathrm{x}}^2+5{\mathrm{x}}^4) \\ \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polynomial} \mathrm{is} 4. \\ \left(\mathrm{ii}\right) \mathrm{Standard} \mathrm{form} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{polynomial} \mathrm{can} \mathrm{be} \mathrm{expressed} \mathrm{as}: \\ (5{\mathrm{a}}^6+{\mathrm{a}}^2+4) \mathrm{or} (4+{\mathrm{a}}^2+5{\mathrm{a}}^6) \\ \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polynomial} \mathrm{is} 6. \\ \left(\mathrm{iii}\right) ({\mathrm{x}}^3-1)({\mathrm{x}}^3-4)={\mathrm{x}}^6-5{\mathrm{x}}^3+4 \\ \mathrm{Standard} \mathrm{form} \mathrm{of} \mathrm{the} \text{given} \mathrm{polynomial} \mathrm{can} \mathrm{be} \mathrm{expressed} \mathrm{as}: \\ ({\mathrm{x}}^6-5{\mathrm{x}}^3+4) \mathrm{or} (4-5{\mathrm{x}}^3+{\mathrm{x}}^6) \\ \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polynomial} \mathrm{is} 6. \\ \left(\mathrm{iv}\right) ({\mathrm{y}}^3-2)({\mathrm{y}}^3+11)={\mathrm{y}}^6+9{\mathrm{y}}^3-22 \\ \mathrm{Standard} \mathrm{form} \mathrm{of} \mathrm{the}\text{ given} \mathrm{polynomial} \mathrm{can} \mathrm{be} \mathrm{expressed} \mathrm{as}: \\ ({\mathrm{y}}^6+9{\mathrm{y}}^3-22) \mathrm{or} (-22+9{\mathrm{y}}^3+{\mathrm{y}}^6) \\ \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polynomial} \mathrm{is} 6. \\ \left(\mathrm{v}\right) ({\mathrm{a}}^3-\frac{3}{8})({\mathrm{a}}^3+\frac{16}{17})={\mathrm{a}}^6+\frac{77}{136}{\mathrm{a}}^3-\frac{6}{17} \\ \mathrm{Standard} \mathrm{form} \mathrm{of} \mathrm{the} \text{given} \mathrm{polynomial} \mathrm{can} \mathrm{be} \mathrm{expressed} \mathrm{as}: \\ ({\mathrm{a}}^6+\frac{77}{136}{\mathrm{a}}^3-\frac{6}{17}) \mathrm{or} (-\frac{6}{17}+\frac{77}{136}{\mathrm{a}}^3+{\mathrm{a}}^6) \\ \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polynomial} \mathrm{is} 6. \\ \left(\mathrm{vi}\right) (\mathrm{a}+\frac{3}{4})(\mathrm{a}+\frac{4}{3})={\mathrm{a}}^2+\frac{25}{12}\mathrm{a}+1 \\ \mathrm{Standard} \mathrm{form} \mathrm{of} \mathrm{the} \text{given} \mathrm{polynomial} \mathrm{can} \mathrm{be} \mathrm{expressed} \mathrm{as}: \\ ({\mathrm{a}}^2+\frac{25}{12}\mathrm{a}+1) \mathrm{or} (1+\frac{25}{12}\mathrm{a}+{\mathrm{a}}^2) \\ \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{the} \mathrm{polynomial} \mathrm{is} 2. \end{gathered}$$

Answer:

$\frac{6x^3{y}^2{z}^2}{3x^{2}yz}=\frac{6\times x\times x\times x\times y\times y\times z\times z}{3\times x\times x\times y\times z} = 2x^{(3-2)}{y}^{(2-1)}{z}^{(2-1)}=2xyz$

Answer:

$$\begin{gathered} \frac{15m^2{n}^3}{5m^2{n}^2} \\ =\frac{15\times m\times m\times n\times n\times n}{5\times m\times m\times n\times n} \\ =3m^{(2-2)}{n}^{(3-2)} \\ =3m^0{n}^1 \\ =3n \end{gathered}$$

Answer:

$$\begin{gathered} \frac{24a^3{b}^3}{-8ab} \\ = \frac{24\times a\times a\times a\times b\times b\times b}{-8\times a\times b} \\ =-3a^{(3-1)}{b}^{(3-1)} \\ =-3a^2{b}^2 \end{gathered}$$

Answer:

$$\begin{gathered} \frac{-21ab{c}^2}{7abc} \\ = \frac{-21\times a\times b\times c\times c}{7\times a\times b\times c} \\ =-3a^{(1-1)}{b}^{(1-1)}{c}^{(2-1)} \\ =-3c \end{gathered}$$

Answer:

​$$\begin{gathered} \frac{72xy{z}^2}{-9xz} \\ =\frac{72\times x\times y\times z\times z}{-9\times x\times z} \\ =-8x^{(1-1)}y{z}^{(2-1)} \\ =-8yz \end{gathered}$$

Answer:

$$\begin{gathered} \frac{-72a^4{b}^5{c}^8}{-9a^2{b}^2{c}^3} \\ =\frac{-72\times a\times a\times a\times a\times b\times b\times b\times b\times b\times c\times c\times c\times c\times c\times c\times c\times c}{-9\times a\times a\times b\times b\times c\times c\times c} \\ =8a^{(4-2)}{b}^{(5-2)}{c}^{(8-3)} \\ =8a^2{b}^3{c}^5 \end{gathered}$$

Answer:

$$\begin{gathered} \frac{16m^3{y}^2}{4m^{2}y} \\ =\frac{16\times m\times m\times m\times y\times y}{4\times m\times m\times y} \\ =4m^{(3-2)}{y}^{(2-1)} \\ =4my \end{gathered}$$

Answer:

$$\begin{gathered} \frac{32m^2{n}^3{p}^2}{4mnp} \\ =\frac{32\times m\times m\times n\times n\times n\times p\times p}{4\times m\times n\times p} \\ =8m^{(2-1)}{n}^{(3-1)}{p}^{(2-1)} \\ =8m{n}^{2}p \end{gathered}$$

Answer:

$$\begin{gathered} \frac{x+2x^2+3x^4-x^5}{2x} \\ =\frac{x}{2x}+\frac{2x^2}{2x}+\frac{3x^4}{2x}-\frac{x^5}{2x} \\ =\frac{1}{2}+x+\frac{3}{2}{x}^3-\frac{1}{2}{x}^{\mathit{4}} \end{gathered}$$                                                                                   

Answer:

$$\begin{gathered} \frac{y^{\mathit{4}}\mathit{-}\mathit{3}{y}^{\mathit{3}}\mathit{+}\frac{\mathit{1}}{\mathit{2}}{y}^{\mathit{2}}}{\mathit{3}y} \\ \mathit{=}\frac{y^{\mathit{4}}}{\mathit{3}y}\mathit{-}\frac{\mathit{3}{y}^{\mathit{3}}}{\mathit{3}y}\mathit{+}\frac{\frac{\mathit{1}}{\mathit{2}}{y}^{\mathit{2}}}{\mathit{3}y} \\ \mathit{=}\frac{\mathit{1}}{\mathit{3}}{y}^{\mathit{(}\mathit{4}\mathit{-}\mathit{1}\mathit{)}}\mathit{-}{y}^{\mathit{(}\mathit{3}\mathit{-}\mathit{1}\mathit{)}}\mathit{+}\frac{\mathit{1}}{\mathit{6}}{y}^{\mathit{(}\mathit{2}\mathit{-}\mathit{1}\mathit{)}} \\ \mathit{=}\frac{\mathit{1}}{\mathit{3}}{y}^{\mathit{3}}\mathit{-}{y}^{\mathit{2}}\mathit{+}\frac{\mathit{1}}{\mathit{6}}y \end{gathered}$$

Answer:

$$\begin{gathered} \frac{\mathit{-}\mathit{4}{a}^{\mathit{3}}\mathit{+}\mathit{4}{a}^{\mathit{2}}\mathit{+}a}{\mathit{2}a} \\ \mathit{=}\frac{\mathit{-}\mathit{4}{a}^{\mathit{3}}}{\mathit{2}a}\mathit{+}\frac{\mathit{4}{a}^{\mathit{2}}}{\mathit{2}a}\mathit{+}\frac{a}{\mathit{2}a} \\ \mathit{=}\mathit{-}\mathit{2}{a}^{\mathit{(}\mathit{3}\mathit{-}\mathit{1}\mathit{)}}\mathit{+}\mathit{2}{a}^{\mathit{(}\mathit{2}\mathit{-}\mathit{1}\mathit{)}}\mathit{+}\frac{\mathit{1}}{\mathit{2}} \\ \mathit{=}\mathit{-}\mathit{2}{a}^{\mathit{2}}\mathit{+}\mathit{2}a\mathit{+}\frac{\mathit{1}}{\mathit{2}} \end{gathered}$$

                                                                       

Answer:

$$\begin{gathered} \frac{\mathit{-}{x}^{\mathit{6}}\mathit{+}\mathit{2}{x}^{\mathit{4}}\mathit{+}\mathit{4}{x}^{\mathit{3}}\mathit{+}\mathit{2}{x}^{\mathit{2}}}{\sqrt{\mathit{2}}{x}^{\mathit{2}}} \\ \mathit{=}\frac{\mathit{-}{x}^{\mathit{6}}}{\sqrt{\mathit{2}}{x}^{\mathit{2}}}\mathit{+}\frac{\mathit{2}{x}^{\mathit{4}}}{\sqrt{\mathit{2}}{x}^{\mathit{2}}}\mathit{+}\frac{\mathit{4}{x}^{\mathit{3}}}{\sqrt{\mathit{2}}{x}^{\mathit{2}}}\mathit{+}\frac{\mathit{2}{x}^{\mathit{2}}}{\sqrt{\mathit{2}}{x}^{\mathit{2}}} \\ \mathit{=}\frac{\mathit{-}\mathit{1}}{\sqrt{\mathit{2}}}{x}^{\mathit{(}\mathit{6}\mathit{-}\mathit{2}\mathit{)}}\mathit{+}\sqrt{\mathit{2}}{x}^{\mathit{(}\mathit{4}\mathit{-}\mathit{2}\mathit{)}}\mathit{+}\mathit{2}\sqrt{\mathit{2}}{x}^{\mathit{(}\mathit{3}\mathit{-}\mathit{2}\mathit{)}}\mathit{+}\sqrt{\mathit{2}}{x}^{\mathit{(}\mathit{2}\mathit{-}\mathit{2}\mathit{)}} \\ \mathit{=}\frac{\mathit{-}\mathit{1}}{\sqrt{\mathit{2}}}{x}^{\mathit{4}}\mathit{+}\sqrt{\mathit{2}}{x}^{\mathit{2}}\mathit{+}\mathit{2}\sqrt{\mathit{2}}x\mathit{+}\sqrt{\mathit{2}} \end{gathered}$$                                            
                                            

Answer:

$$\begin{gathered} \frac{\mathit{5}{z}^{\mathit{3}}\mathit{-}\mathit{6}{z}^{\mathit{2}}\mathit{+}\mathit{7}z}{\mathit{2}z} \\ \mathit{=}\frac{\mathit{5}{z}^{\mathit{3}}}{\mathit{2}z}\mathit{-}\frac{\mathit{6}{z}^{\mathit{2}}}{\mathit{2}z}\mathit{+}\frac{\mathit{7}z}{\mathit{2}z} \\ \mathit{=}\frac{\mathit{5}}{\mathit{2}}{z}^{\mathit{(}\mathit{3}\mathit{-}\mathit{1}\mathit{)}}\mathit{-}\mathit{3}{z}^{\mathit{(}\mathit{2}\mathit{-}\mathit{1}\mathit{)}}\mathit{+}\frac{\mathit{7}}{\mathit{2}} \\ \mathit{=}\frac{\mathit{5}}{\mathit{2}}{z}^{\mathit{2}}\mathit{-}\mathit{3}z\mathit{+}\frac{\mathit{7}}{\mathit{2}} \end{gathered}$$
                                                                

Answer:

$$\begin{gathered} \frac{\sqrt{\mathit{3}}{a}^{\mathit{4}}\mathit{+}\mathit{2}\sqrt{\mathit{3}}{a}^{\mathit{3}}\mathit{+}\mathit{3}{a}^{\mathit{2}}\mathit{-}\mathit{6}a}{\mathit{3}a} \\ \mathit{=}\frac{\sqrt{\mathit{3}}{a}^{\mathit{4}}}{\mathit{3}a}\mathit{+}\frac{\mathit{2}\sqrt{\mathit{3}}{a}^{\mathit{3}}}{\mathit{3}a}\mathit{+}\frac{\mathit{3}{a}^{\mathit{2}}}{\mathit{3}a}\mathit{-}\frac{\mathit{6}a}{\mathit{3}a} \\ \mathit{=}\frac{\mathit{1}}{\sqrt{\mathit{3}}}{a}^{\mathit{(}\mathit{4}\mathit{-}\mathit{1}\mathit{)}}\mathit{+}\frac{\mathit{2}}{\sqrt{\mathit{3}}}{a}^{\mathit{(}\mathit{3}\mathit{-}\mathit{1}\mathit{)}}\mathit{+}{a}^{\mathit{(}\mathit{2}\mathit{-}\mathit{1}\mathit{)}}\mathit{-}\mathit{2} \\ \mathit{=}\frac{\mathit{1}}{\sqrt{\mathit{3}}}{a}^{\mathit{3}}\mathit{+}\frac{\mathit{2}}{\sqrt{\mathit{3}}}{a}^{\mathit{2}}\mathit{+}a\mathit{-}\mathit{2} \end{gathered}$$