Question

If the two zeroes of the Polynomial $x^4+3x^3-20x^2-6x+36$ are $\pm \sqrt{2},$ find the other zeroes.

Answer 1

$x^4+{3x}^3-20x^2-6x+36$

If two zeroes are $\pm \sqrt{2}$

Dividing given polynomial by $(x+\sqrt{2})(x-\sqrt{2})$ i.e, $x^2-2$

$x^2-2)x^4+{3x}^3-{20x}^2-6x+36(x^2+3x-18$

$x^4-{0x}^2-2x^2$

$\underset{–}{--+}$

${3x}^3-{18x}^2-6x$

${3x}^3-{0x}^2-6x$

$\underset{–}{-+}$

$-{18x}^2+36$

$\underset{–}{-{18x}^2+36}$

$\left(0\right)$

$(x^2-2)(x^2+3x-18)=0$

$x^2+3x-18=0$

$x^2+6x-3x-18=0$

$x(x+6)-3(x+6)=0$

$\therefore$ $x=3,-6$ are other zeroes.