Question

Solve the following equations:
$x^4-3x^2-42x-40=0$.

Answer 1

Given equation, $x^4-3x^2-42x-40=0$

Consider $f\left(x\right)=x^4-3x^2-42x-40$

$f(-1)=(-1{)}^4-3(-1{)}^2-42(-1)-40$

$=1-3+42-40=0$

By inspection we can easily see that $f(-1)=0$. Therefore, $(x+1)$ is a factor of the given equation

$\therefore f\left(x\right)=(x+1)\cdot g\left(x\right)$

$⟹g\left(x\right)=\frac{f\left(x\right)}{(x+1)}=x^3-x^2-2x-40$

We need to find the root of $g\left(x\right)=0$

$⟹x^3-x^2-2x-40=0$

$⟹x^3+3x^2-4x^2+10x-12x-40=0$

$⟹x^3-4x^2+3x^2-12x+10x-40=0$

$⟹x^2(x-4)+3x(x-4)+10(x-4)=0$

$⟹(x-4)(x^2+3x+10)=0$

on Solving the above quadratic equation, we have

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-3\pm \sqrt{(-3{)}^2-(4\times 1\times 10)}}{2\times 1}$

$=\frac{-3\pm \sqrt{9-40}}{2}$

$=\frac{-3\pm \sqrt{31}i}{2}$ [$\because \sqrt{-1}=i$]

$\therefore$ the roots of the given equation are $x=-1,4,\frac{-3\pm \sqrt{31}i}{2}$