Question

Solve:$(x^3+6x^2+12x+16)$

Answer 1

Given,

$x^3+{6x}^2+12x+16=x^3+{6x}^2+12x+8+8=x^3+3\cdot 2\cdot x^2+3\cdot 2^2\cdot x+2^3+8={(x+2)}^3+2^3$

Using another identity we get,

$=(x+2+2)[{(x+2)}^2-2(x+2)+2^2]=(x+4)[x^2+4x+4-2x-4+4]=(x+4)(x^2+2x+4)$

This is the factorisation.

Now, to find solutions we have to make the factorisation equal to zero.

$\therefore (x+4)(x^2+2x+4)=0$

For,

$x+4=0\Rightarrow x=-4$

For,

$x^2+2x+4=0\Rightarrow x=\frac{(-2)\pm \sqrt{{(-2)}^2-4\cdot 1\cdot 4}}{2\cdot 1}=\frac{(-2)\pm \sqrt{-12}}{2},buttheserootsareimaginary\therefore x=-4$

Hence, $x=-4$ is the required solution.