Factor using suitable identities: i. 16y² - 24y + 9 Class 9

Factor using suitable identities: i. 16y² - 24y + 9 Class 9

Question 1.

Factor using suitable identities: Class 9

i. 16y² - 24y + 9

ii. $\frac{9}{4} s^2+6 s t+4 t^2$

iii. $\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2$

iv. $\frac{p^2}{16}-2+\frac{16}{p^2}$

v. 9a² + 4b² + c² - 12ab + 6ac - 4bc

Solution:

i. 16y² - 24y + 9

Writing in the form a² - 2ab + b², we get

16y² - 24y + 9 = (4y)² - 2(4y)(3) + (3)²,

where a = 4y, b = 3

∴ 16y² - 24y + 9 = (4y - 3)² ....[∵ (a - b)² = a² - 2ab + b²]

= (4y - 3)(4y - 3)

ii. $\frac{9}{4} s^2+6 s t+4 t^2$

Writing in the form a² + 2ab + b², we get

$\frac{9}{4} s^2+6 s t+4 t^2=\left(\frac{3 s}{2}\right)^2+2\left(\frac{3 s}{2}\right)(2 t)+(2 t)^2$

where a =$\frac{3 s}{2}$, b = 2t

$\begin{aligned} \therefore \quad \frac{9}{4} s^2+6 s t+4 t^2= & \left(\frac{3 s}{2}+2 t\right)^2 \\ & \ldots\left[\because(a+b)^2=a^2+2 a b+b^2\right] \\ = & \left(\frac{3 s}{2}+2 t\right)\left(\frac{3 s}{2}+2 t\right)\end{aligned}$

iii. $\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2$

Writing in the form

a² + b² + c² + 2ab + 2bc + 2cea, we get

$\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2$

$\begin{aligned}=\left(\frac{m}{3}\right)^2+\left(\frac{k}{2}\right)^2 & +(3 n)^2+2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right) \\ & +2\left(\frac{k}{2}\right)(3 n)+2\left(\frac{m}{3}\right)(3 n)\end{aligned}$

where a = $\frac{m}{3}$, b = $\frac{k}{2}$, c = 3n

$\begin{aligned} \therefore \quad & \frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2 \\ & =\left(\frac{m}{3}+\frac{k}{2}+3 n\right)^2 \\ & \ldots\left[\because(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a\right] \\ & =\left(\frac{m}{3}+\frac{k}{2}+3 n\right)\left(\frac{m}{3}+\frac{k}{2}+3 n\right)\end{aligned}$

iv. $\frac{p^2}{16}-2+\frac{16}{p^2}$

Writing in the form a² - 2ab + b², we get

$\begin{aligned} \frac{p^2}{16}-2+\frac{16}{p^2} & =\left(\frac{p}{4}\right)^2-2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right)+\left(\frac{4}{p}\right)^2, \text { where } \\ a=\frac{p}{4^{\prime}}, b=\frac{4}{p} & \\ \therefore \quad \frac{p^2}{16}-2+\frac{16}{p^2} & =\left(\frac{p}{4}-\frac{4}{p}\right)^2 \\ & \ldots\left[\because(a-b)^2=a^2-2 a b+b^2\right] \\ & =\left(\frac{p}{4}-\frac{4}{p}\right)\left(\frac{p}{4}-\frac{4}{p}\right)\end{aligned}$

v. 9a² + 4b² + c² - 12ab + 6ac - 4bc

Writing in the form

A² + B² + C² + 2AB + 2BC + 2CA, we get

9a² + 4b² + c² - 12ab + 6ac - 4bc

= (3a)² + (-2b)² + c² + 2(3a)(-2b) + 2(-2b)(c) + 2(3a)(c),

where A = 3a, B = -2b, C = c

9a² + 4b² + c² - 12ab + 6ac - 4bc

= (3a - 2b + c)² ...[∵ (A + B + C)²

= A² + B² + C² + 2AB + 2BC + 2CA = (3a - 2b + c)(3a - 2b + c)