Question

Factories :

$\left(i\right)(a-3b{)}^2-36b^2(ii\left)25\right(a-5b{)}^2-4(a-3b{)}^2(iii)a^2-0.36b^2(iv)a^4-625(v)x^4-5x^2-36(vi\left)15\right(2x-y{)}^2-16(2x-y)-15$

Answer 1

(i) $(a-3b{)}^2-36b^2$

Using the identity $x^2-y^2=(x-y)(x+y)$, we get,

$\Rightarrow (a-3b{)}^2-36b^2=(a-3b{)}^2-(6b{)}^2$

$=[a-3b-(6b\left)\right][a-3b+(6b\left)\right]$

$=[a-9b][a+3b]$

$\therefore (a-3b{)}^2-36b^2=(a-9b\left)\right(a+3b)$

(ii)$25(a-5b{)}^2-4(a-3b{)}^2$

Using the identity $x^2-y^2=(x-y)(x+y)$, we get,

$\Rightarrow 25(a-5b{)}^2-4(a-3b{)}^2=\left[5\right(a-5b){]}^2-[2(a-3b){]}^2$

$=(5a-25b{)}^2-(2a-6b{)}^2$

$=[5a-25b-(2a-6b\left)\right][5a-25b+(2a-6b\left)\right]$

$=[5a-25b-2a+6b][5a-25b+2a-6b]$

$=[3a-19b][7a-31b]$

$\therefore 25(a-5b{)}^2-4(a-3b{)}^2=(3a-19b)(7a-31b)$

(iii)$a^2-0.36b^2$

Using the identity $x^2-y^2=(x-y)(x+y)$, we get,

$\Rightarrow a^2-0.36b^2=a^2-(0.6b{)}^2$

$=(a-0.6b)(a+0.6b)$

$\therefore a^2-0.36b^2=(a-0.6b)(a+0.6b)$

(iv) $a^4-625$

Using the identity $x^2-y^2=(x-y)(x+y)$, we get,

$\Rightarrow a^4-625=(a^2{)}^2-(25{)}^2$

$=(a^2-25)(a^2+25)$

$=(a^2-(5{)}^2\left)\right(a^2+25)$

Again using the same identity, we get,

$=(a-5)(a+5)(a^2+25)$

$\therefore a^4-625=(a-5)(a+5)(a^2+25)$

(v) $x^4-5x^2-36$

We have to find two numbers whose product is $-36x^4$ and whose sum is $-5x^2$

Thus, using Middle Term Splitting, we get,

$\Rightarrow x^4-5x^2-36=x^4-9x^2+4x^2-36$

$=x^2(x^2-9)+4(x^2-9)$

$=(x^2-9)(x^2+4)$

Using the identity $a^2-b^2=(a-b)(a+b)$, we get,

$=(x-3)(x+3)(x^2+4)$

$\therefore x^4-5x^2-36=(x-3)(x+3)(x^2+4)$

(vi) $15(2x-y{)}^2-16(2x-y)-15$

Let $t=2x-y$

$\Rightarrow 15(2x-y{)}^2-16(2x-y)-15=15t^2-16t-15$

Now we have to find find two numbers whose product is $225t^2$ and sum is $-16$

Thus, using Middle Term Splitting, we get,

$=15t^2-25t+9t-15$

$=5t(3t-5)+3(3t-5)$

$=(3+5t)(3-5t)$

$=[3+5(2x-y\left)\right][3-5(2x-y\left)\right]$

$=[3+10x-5y][3-10x+5y]$

$\therefore 15(2x-y{)}^2-16(2x-y)-15=[3+10x-5y\left]\right[3-10x+5y]$