Factor completely: i. 9x² + 24xy + 16y² ii. 4s² + 20st + 25t² iii. 49x² + 28xy + 4y² Class 9

Factor completely: i. 9x² + 24xy + 16y² ii. 4s² + 20st + 25t² iii. 49x² + 28xy + 4y² Class 9

Question 1.

Factor completely: Class 9

i. 9x² + 24xy + 16y²

ii. 4s² + 20st + 25t²

iii. 49x² + 28xy + 4y²

$\begin{aligned} & \text { iv. } 64 p^2+\frac{32}{3} p q+\frac{4}{9} q^2 \\ & { } \text { v. } 3 a^2+4 a b+\frac{4}{3} b^2 \\ & { } \text { vi. } \frac{9}{5} s^2+6 s v+5 v^2\end{aligned}$

(Hint: 2 was taken out as a common factor in Example 7 .Is it possible to do something similar in Exercises (v) and (vi) above?)

Solution:

i. 9x² + 24xy + 16y²

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

9x² + 24xy + 16y² = (3x)² + 2(3x)(4y) + (4y)²,

where a = 3x, b = 4y

∴ 9x² + 24xy + 16y² = (3x + 4y)² ....[∵ (a + b)² = a² + 2ab + b²]

= (3x + 4y)(3x + 4y)

ii. 4s² + 20st + 25t²

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

4s² + 10st + 25t² = (2s)² + 2(2s)(5t) + (5t)²,

where a = 2s, b = 5t

∴ 4s² + 20st + 25t² = (2s + 5t)² ...[∵ (a + b)² = a² + 2ab + b²]

= (2s + 5t)(2s + 5t)

iii. 49x² + 28xy + 4y²

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

49x² + 28xy + 4y² = (7x)² + 2(7x)(2y) + (2y)²,

where a = 7x, b = 2y

∴ 49x² + 28xy + 4y² = (7x + 2y)² ...[∵ (a + b)² = a² + 2ab + b²]

= (7x + 2y)(7x + 2y)

iv. 64p² + $\frac{32}{3}$ pq + $\frac{p}{q}$q²

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

64y² + $\frac{32}{3}$pq + $\frac{4}{93}$q² = (8p)² + 2(8p)$\left(\frac{2 q}{3}\right)+\left(\frac{2 q}{3}\right)^2$,

where a = 8p, b = $\frac{2q}{3}$

∴ 64y² + $\frac{32}{3}$pq + $\frac{4}{9}$q² = (8p + $\frac{2q}{3}$)² ......[∵ (a + b)² = a² + 2ab + b²]

= (8p + $\frac{2q}{3}$)(8p + $\frac{2q}{3}$)

v. 3a² + 4ab + $\frac{4}{3}$b² = 3(a² + $\frac{4}{3}$ab + $\frac{4}{9}$b²)

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

3a² + 4ab + $\frac{4}{3}$b² = 3[a² + 2(a)$\left(\frac{2 b}{3}\right)+\left(\frac{2 b}{3}\right)^2$],

where A = a, B = $\frac{2b}{3}$

∴ 3a² + 4ab + $\frac{4}{3}$b² = 3(a + $\frac{2 b}{3}$)²

...........[∵ (A + B)² = A² + 2AB + B²]

= 3(a+$\frac{2b}{3}$)(a+$\frac{2b}{3}$)

vi. $\frac{9}{5}$s² + 6sv + 5v² = $\frac{1}{5}$(9s² + 30sy + 15v²)

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

$\frac{9}{5}$s² + 6sv + 5v² = $\frac{1}{5}$[(3s)² + 2(3s)(5v) + (5v)²],

where a = 3s, b = 5v

∴ $\frac{9}{5}$s² + 6sv + 5v² = $\frac{1}{5}$(3s + 5v)²

....[∵ (a + b)² = a² + 2ab + b²]

= $\frac{1}{5}$(3s + 5v) (3s + 5v)