Question

factorise (x^2+x)^2+4(x^2+x)-12

Answer 1

Solution :

∴ (x² + x)² + 4 (x² + x) - 12

= (x² + x)² + (6 - 2) (x² + x) - 12

= (x² + x)² + 6 (x² + x) - 2 (x² + x) - 12

= (x² + x) (x² + x + 6) - 2 (x² + x + 6)

= (x² + x + 6) (x² + x - 2)

= (x² + x + 6) {x² + (2 - 1) x - 2}

= (x² + x + 6) {x² + 2x - x - 2}

= (x² + x + 6) {x (x + 2) - 1 (x + 2)}

= (x² + x + 6) (x + 2) (x - 1) ,

which is the required factorization.

Method :

Here, 12, the product of the first term and the 3rd term can be factorised as the product of 6 and 2. We are using Middle-term factorization method where the middle term's coefficient is 4 and we see that 6 - 2 = 4 keeps the solution going.

In the 2nd factorization, the product of 1 and 2, i.e., 2 can be written as 2 = 2 - 1 while applying Middle-term factorization method.

Answer 2

Taking $x^2+x=k,$ we get,

$\begin{aligned}&k^2+4k-12\\ \\ \Longrightarrow\ \ &k^2-2k+6k-12\\ \\ \Longrightarrow\ \ &k(k-2)+6(k-2)\\ \\ \Longrightarrow\ \ &(k+6)(k-2)\\ \\ \Longrightarrow\ \ &(x^2+x+6)(x^2+x-2)\end{aligned}$

Factorising $x^2+x+6,$

$\begin{aligned}&x^2+x+6\\ \\ \Longrightarrow\ \ &x^2+\frac{1+\iota\sqrt{23}}{2}x+\frac{1-\iota\sqrt{23}}{2}x+6\\ \\ \Longrightarrow\ \ &x\left(x+\frac{1+\iota\sqrt{23}}{2}\right)+\frac{1-\iota\sqrt{23}}{2}\left(x+\frac{1+\iota\sqrt{23}}{2}\right)\\ \\ \Longrightarrow\ \ &\left(x+\frac{1+\iota\sqrt{23}}{2}\right)\left(x+\frac{1-\iota\sqrt{23}}{2}\right)\end{aligned}$

Factorising $x^2+x-2,$

$\begin{aligned}&x^2+x-2\\ \\ \Longrightarrow\ \ &x^2+2x-x-2\\ \\ \Longrightarrow\ \ &x(x+2)-(x+2)\\ \\ \Longrightarrow\ \ &(x-1)(x+2)\end{aligned}$

Thus,

$\footnotesize \boxed{(x^2+x)^2+4(x^2+x)-12=\left(x+\dfrac{1+\iota\sqrt{23}}{2}\right)\left(x+\frac{1-\iota\sqrt{23}}{2}\right)\bigg(x-1\bigg)\bigg(x+2\bigg)}$

Hence Factorised!