Question

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Answer 1

First find Divisor

• Two zeros are -3,-2

$\\ \sf\longmapsto x^2-(-3-2)+(-3)(-2)$

$\\ \sf\longmapsto x^2+5x+6$

Dividend given by

$\\ \sf\longmapsto x^4-x^3-16x^2+4x+48$

We know

$\boxed{\sf Dividend=Divisor\times Quotient +Remainder}$

• Remainder is 0

$\\ \sf\longmapsto Dividend=Divisor\times Quotient$

$\\ \sf\longmapsto Quotient=\dfrac{Dividend}{Divisor}$

$\\ \sf\longmapsto \dfrac{x^4-x^3-16x^2+4x+48}{x^2+5x+6}$

$\\ \sf\longmapsto x^2-6x+8$

Now factor

$\\ \sf\longmapsto x^2-6x+8=0$

$\\ \sf\longmapsto x^2-4x-2x+8=0$

$\\ \sf\longmapsto x(x-4)-2(x-4)=0$

$\\ \sf\longmapsto (x-4)(x-2)=0$

$\\ \sf\longmapsto x-4=0\:or\:x-2=0$

$\\ \sf\longmapsto x=4\:or\:x=2$

Hence all zeros are

• 2,-2,-3,4