Question

Find the zeroes of the polynomials.
$x^3-23x^2-142x-120$

Answer 1

Let $p\left(x\right)=x^3-23x^2+142x-120$
Integral zeroes of $120$ are
$\pm 1,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 8,\pm 10,\pm 12,\pm 15$ and $\pm 20.$
Here we go for only positive value, because negative value will be give only negative result.
$\therefore p\left(1\right)=(1{)}^3-23(1{)}^2+142\left(1\right)-120$
$=1-23+142-120$
$=143-143=0$
$\Rightarrow 1$ is a zero of $p\left(x\right).$
$p\left(10\right)=(10{)}^3-23(10{)}^2+142\left(10\right)-120$
$=1000-23000+1420-120$
$=2420-2420=0$
$\Rightarrow 10$ is a zero of $p\left(x\right).$
$p\left(12\right)=(12{)}^3-23(12{)}^2+142\left(12\right)-120$
$=1728-3312+1704-120$
$=3432-3432=0$
$\Rightarrow -12$ is a zero of $p\left(x\right).$
Hence, $1,10$ and $12$ are zeroes of p(x).