Question

Find the zeroes of the polynomials.
$x^4-2x^3-7x^2+8x+12$

Answer 1

Let $p\left(x\right)=x^4-2x^3-7x^2+8x+12$
Integral zeroes of p(x) are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$
$\because$ Positive term have more values. So we go for negative values first
$p(-1)=(-1{)}^4-2(-1{)}^3-7(-1{)}^2+8(-1)+12$
$=1+2-7-8+12=15-15=0$
$\Rightarrow -1$ is a zero of p(x).
Also
$p(-2)=(-2{)}^4-2(-2{)}^3-7(-2{)}^2+8(-2)+12$
$=16+16-28-16+12$
$=44-44=0$
$\Rightarrow -2$ is a zero of p(x).
Also
$p\left(2\right)=(2{)}^4-2(2{)}^3-7(2{)}^2+8(2)+12$
$=16-16-28+16+12=0$
$\Rightarrow 2$ is a zero of p(x).
Similarly
Also
$p\left(3\right)=(3{)}^4-2(3{)}^3-7(3{)}^2+8(3)+12$
$=81-54-63+24+12$
$=117-117=0$
$\Rightarrow -1,-2,2$ and $3$ are the zeroes of $p\left(x\right).$