RD Sharma 2020 solution class 9 chapter 5 Factorization of Algebraic Expressions FBQS

FBQS

Page-5.27

Question 1:

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

Answer 1:

$$\begin{gathered} y^2+\left(x-1\right)y-x \\ =y^2+xy-y-x \\ =y\left(y+x\right)-1\left(y+x\right) \\ =\left(y-1\right)\left(y+x\right) \end{gathered}$$

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

Question 2:

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

Answer 2:

$$\begin{gathered} a^3+{\left(b-a\right)}^3-b^3 \\ =a^3+\left(b^3-a^3-3ba\left(b-a\right)\right)-b^3 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}:\hspace{0.17em}{\left(x-y\right)}^3=x^3-y^3-3xy\left(x-y\right)\right) \\ =a^3+b^3-a^3-3ba\left(b-a\right)-b^3 \\ =-3ba\left(b-a\right) \\ =3ab\left(a-b\right) \end{gathered}$$

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

Question 3:

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

Answer 3:

$$\begin{gathered} \mathrm{Given}: \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 ...\left(1\right) \\ abc=2 ...\left(2\right) \\ \mathrm{Now}, \\ a{b}^2{c}^2+a^{2}b{c}^2+a^2{b}^{2}c \\ =a^2{b}^2{c}^2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\ ={\left(abc\right)}^2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\ ={\left(2\right)}^2\left(1\right) \left(\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right) \\ =4 \end{gathered}$$

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

Question 4:

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

Answer 4:

$$\begin{gathered} 11x^2-10\sqrt{3}x-3 \\ =11x^2-11\sqrt{3}x+\sqrt{3}x-3 \\ =11x\left(x-\sqrt{3}\right)+\sqrt{3}\left(x-\sqrt{3}\right) \\ =\left(11x+\sqrt{3}\right)\left(x-\sqrt{3}\right) \end{gathered}$$

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

Question 5:

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

Answer 5:

$$\begin{gathered} x^2+y^2-z^2-2xy \\ =\left(x^2+y^2-2xy\right)-z^2 \\ ={\left(x-y\right)}^2-{\left(z\right)}^2 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a-b\right)}^2=a^2+b^2-2ab\right) \\ =\left(x-y+z\right)\left(x-y-z\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \end{gathered}$$

Hence, the polynomial x² + y² – z² – 2xy on factorization gives $\underline{\left(x-y+z\right)\left(x-y-z\right)}$.

Question 6:

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

Answer 6:

$$\begin{gathered} a+b+c+2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca} \\ ={\left(\sqrt{a}\right)}^2+{\left(\sqrt{b}\right)}^2+{\left(-\sqrt{c}\right)}^2+2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca} \\ ={\left(\sqrt{a}\right)}^2+{\left(\sqrt{b}\right)}^2+{\left(-\sqrt{c}\right)}^2+2\sqrt{a}\sqrt{b}+2\sqrt{b}\left(-\sqrt{c}\right)+2\sqrt{a}\left(-\sqrt{c}\right) \\ ={\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)}^2 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2+b^2+c^2+2ab+2bc+2ac={\left(a+b+c\right)}^2\right) \\ =\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right) \end{gathered}$$

Hence, the factors of the expression $a+b+c+2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}$ are $\underline{\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right) \mathrm{and} \left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)}.$.

Question 7:

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

Answer 7:

$$\begin{gathered} x^6+64y^6 \\ ={\left(x^2\right)}^3+{\left(4y^2\right)}^3 \\ =\left(x^2+4y^2\right)\left({\left(x^2\right)}^2+{\left(4y^2\right)}^2-\left(x^2\right)\left(4y^2\right)\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\right) \\ =\left(x^2+4y^2\right)\left(x^4+16y^4-4x^2{y}^2\right) \end{gathered}$$

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

Question 8:

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

Answer 8:

$$\begin{gathered} a^4+b^4-a^2{b}^2 \\ \mathrm{Adding} \mathrm{and} \mathrm{subtracting} 2a^2{b}^2, \mathrm{we} \mathrm{get} \\ ={\left(a^2\right)}^2+{\left(b^2\right)}^2+2a^2{b}^2-2a^2{b}^2-a^2{b}^2 \\ ={\left(a^2+b^2\right)}^2-3a^2{b}^2 \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2+b^2+2ab={\left(a+b\right)}^2\right) \\ ={\left(a^2+b^2\right)}^2-{\left(\sqrt{3}ab\right)}^2 \\ =\left(a^2+b^2+\sqrt{3}ab\right)\left(a^2+b^2-\sqrt{3}ab\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \end{gathered}$$

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

Question 9:

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

Answer 9:

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


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

Question 10:

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

Answer 10:

$$\begin{gathered} \frac{1}{xyz}\left(x^2+y^2+z^2\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \\ =\frac{\left(x^2+y^2+z^2\right)}{xyz}+2\left(\frac{yz+xz+xy}{xyx}\right) \\ =\frac{x^2+y^2+z^2+2\left(yz+xz+xy\right)}{xyz} \\ =\frac{{\left(x+y+z\right)}^2}{xyz} \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: {\left(a+b+c\right)}^2=a^2+b^2+c^2+2ab+2bc+2ac\right) \\ =\frac{1}{xyz}{\left(x+y+z\right)}^2 \end{gathered}$$

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

Question 11:

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

Answer 11:

$$\begin{gathered} a^3+b^3+3ab-1 \\ =a^3+b^3-1+3ab \\ =a^3+b^3+{\left(-1\right)}^3-3ab\left(-1\right) \\ =\left(a+b-1\right)\left(a^2+b^2+{\left(-1\right)}^2-ab-b\left(-1\right)-a\left(-1\right)\right) \left(\mathrm{Using} \mathrm{the} \mathrm{identity}: a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\right) \\ =\left(a+b-1\right)\left(a^2+b^2+1-ab+b+a\right) \end{gathered}$$

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