Factor the following algebraic expressions: Class 9

Factor the following algebraic expressions: Class 9

Question 3.

Factor the following algebraic expressions: Class 9

i. $4 y^2+1+\frac{1}{16 y^2}$

ii. $9 m^2-\frac{1}{25 n^2}$

iii. $\quad 27 b^3-\frac{1}{64 b^3}$

iv. $\quad x^2+\frac{5 x}{6}+\frac{1}{6}$

v. $27 u^3-\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25}$

vi. $64 y^3+\frac{1}{125} z^3$

vii. $p^3+27 q^3+r^3-9 p q r$

viii. $9 m^2-12 m+4$

ix. $\quad 9 x^3-\frac{8}{3} y^3+\frac{z^3}{3}+6 x y z$

x. 4x² + 9y² + 36z² + 12xy + 36yz + 24xz

xi. $27 u^3-\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4}$

Solution:

$\begin{array}{ll}\text { i. } & 4 y^2+1+\frac{1}{16 y^2} \\ & =(2 y)^2+2(2 y)\left(\frac{1}{4 y}\right)+\left(\frac{1}{4 y}\right)^2 \\ & \text { Here, } a=2 y, b=\frac{1}{4 y} \\ \therefore \quad & 4 y^2+1+\frac{1}{16 y^2} \\ & =\left(2 y+\frac{1}{4 y}\right)^2 \\ & \quad \ldots\left[\because(a+b)^2=a^2+2 a b+b^2\right]\end{array}$

$=\left(2 y+\frac{1}{4 y}\right)\left(2 y+\frac{1}{4 y}\right)$

ii. $9 m^2-\frac{1}{25 n^2}=(3 m)^2-\left(\frac{1}{5 n}\right)^2$

Here, a = 3m, b = $\frac{1}{5 n}$

$\begin{aligned} \therefore \quad 9 m^2-\frac{1}{25 n^2}=(3 m+ & \left.\frac{1}{5 n}\right)\left(3 m-\frac{1}{5 n}\right) \\ & \ldots\left[\because a^2-b^2=(a+b)(a-b)\right]\end{aligned}$

iii. $27 b^3-\frac{1}{64 b^3}=(3 b)^3-\left(\frac{1}{4 b}\right)^3$

Here, a = 3m, b = $\frac{1}{4 b}$

$\begin{aligned} \therefore \quad & 27 b^3-\frac{1}{64 b^3} \\ & =\left(3 b-\frac{1}{4 b}\right)\left((3 b)^2+(3 b)\left(\frac{1}{4 b}\right)+\left(\frac{1}{4 b}\right)^2\right) \\ & \quad \ldots\left[\because \mathrm{A}^3-\mathrm{B}^3=(\mathrm{A}-\mathrm{B})\left(\mathrm{A}^2+\mathrm{AB}+\mathrm{B}^2\right)\right] \\ & \quad=\left(3 b-\frac{1}{4 b}\right)\left(9 b^2+\frac{3}{4}+\frac{1}{16 b^2}\right)\end{aligned}$

iv. $x^2+\frac{5 x}{6}+\frac{1}{6}$

Since $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ and $\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$, the number are $\frac{1}{2}$ and $\frac{1}{3}$.

$\therefore \quad x^2+\frac{5 x}{6}+\frac{1}{6}=\left(x+\frac{1}{2}\right)\left(x+\frac{1}{3}\right)$

v. $\begin{aligned} & 27 u^3-\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25} \\ & =27 u^3-\frac{27 u^2}{5}+\frac{9 u}{25}-\frac{1}{125}\end{aligned}$

$=(3 u)^3-3(3 u)^2\left(\frac{1}{5}\right)+3(3 u)\left(\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^3$

Here, a = 3u, b = $\frac{1}{5}$$\begin{aligned} \therefore \quad 27 u^3-\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25} & =\left(3 u-\frac{1}{5}\right)^3 \\ \ldots\left[\because(a-b)^3\right. & \left.=a^3-3 a^2 b+3 a b^2-b^3\right]\end{aligned}$

vi. $64 y^3+\frac{1}{125} z^3=(4 y)^3+\left(\frac{z}{5}\right)^3$

Here, a = 4y, b = $\frac{z}{5}$

$\begin{aligned} \therefore \quad 64 y^3+\frac{1}{125} z^3 & \\ =\left(4 y+\frac{z}{5}\right)\left((4 y)^2-(4 y)\left(\frac{z}{5}\right)+\left(\frac{z}{5}\right)^2\right) & \\ & \ldots\left[\because a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]\end{aligned}$

$=\left(4 y+\frac{z}{5}\right)\left(16 y^2-\frac{4 y z}{5}+\frac{z^2}{25}\right)$

vii. p³ + 27q³ + r³ - 9pyr

= p³ + (3q)³ + r³ - 3(p)(3q)(r)

Here, a = p, b = 3q, c = r

∴ p³ + 27q³ + r³ - 9pqr

= (p + 3q + r)(p² + (3q)² + r² - p(3q) - (3q)r - pr)

...[∵ a³ + b³ + c³ + 3abc]

= (a + b + c) (a² + b² + c² - ab - bc - ac)]

= (p + 3q + r)( p² + 9q² + r² - 3pq - 3qr - pr)

viii. 9m² - 12m + 4 = (3m)² - 2(3m)(2)² + 2²

Here, a = 3m, b = 2

∴ 9m² - 12m + 4 = (3m - 2)²

...[∵ (a - b)² = a² - 2ab + b²]

ix. $9 x^3-\frac{8}{3} y^3+\frac{z^3}{3}+6 x y z$

= $\frac{1}{3}$[27x³ - 8y³ + z³ + 18xyz]

= $\frac{1}{3}$[(3x)³ + (-2y)³ + z³ - 3(3x)(-2y)(z)]

Here, a = 3x, b = -2y, c = z

$\therefore \quad 9 x^3-\frac{8}{3} y^3+\frac{z^3}{3}+6 x y z$

= $\frac{1}{3}$[(3x - 2y + z) ((3x)² + (-2y)² + z² -(3x) (-2y) - (-2y)z - (3x)z)]

....[∵ a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ac)]

= $\frac{1}{3}$[(3x - 2y + z) (9x² + 4y² + z² + 6xy + 2yz - 3xz)]

x. 4x² + 9y² + 36z² + 12xy + 36yz + 24xz

= (2x)² + (3y)² + (6z)² + 2(2x)(3y) + 2(3y)(6z) + 2(6z)(2x)

Here, a = 2x, b = 3y, c = 6z

∴ 4x² + 9y² + 36z² + 42xz + 36yz + 24xy

= (2x + 3y + 6z)²

....[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

[Note: The question has been modified]

xi. $\begin{aligned} & 27 u^3-\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4} \\ & =27 u^3-\frac{9 u^2}{2}+\frac{u}{4}-\frac{1}{216} \\ & =(3 u)^3-3(3 u)^2\left(\frac{1}{6}\right)+3(3 u)\left(\frac{1}{6}\right)^2-\left(\frac{1}{6}\right)^3\end{aligned}$

Here, a = 3u, b = $\frac{1}{6}$

$\begin{aligned} \therefore \quad 27 u^3-\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4} & =\left(3 u-\frac{1}{6}\right)^3 \\ \ldots\left[\because(a-b)^3\right. & \left.=a^3-3 a^2 b+3 a b^2-b^3\right]\end{aligned}$