Question

Solve the equation $x^6-18x^4+16x^3+28x^2-32x+8=0$, one of whose roots is $\sqrt{6}-2$.

Answer 1

Given equation, $x^6-18x^4+16x^3+28x^2-32x+8=0$

Consider $f\left(x\right)=x^6-18x^4+16x^3+28x^2-32x+8$

One root of the given equation is $\sqrt{6}–2$. Since irrational roots of an equation occurs in pairs, another root of the given equation is $-\sqrt{6}-2$

$\therefore f\left(x\right)=(x-\sqrt{6}+2)(x+\sqrt{6}+2)\cdot g\left(x\right)⟹g\left(x\right)=x^4-4x^3+8x-4$

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

$⟹x^4-4x^3+8x-4=0⟹x^4-4x^3+2x^2-2x^2+8x-4=0⟹(x^2-2)(x^2-4x+2)=0$

Solving the 2 quadratic equations, we have $x=\pm \sqrt{2},2\pm \sqrt{2}$

$\therefore$ the roots of the given equation are $x=\pm \sqrt{2},2\pm \sqrt{2},-2\pm \sqrt{6}$