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Obtain all zeroes of $f(x)\ =\ x^3\ +\ 13x^2\ +\ 32x\ +\ 20$, if one of its zeros is $-2$.
Given:
$f(x)\ =\ x^3\ +\ 13x^2\ +\ 32x\ +\ 20$ and one of its zeros is $-2$.
To do:
We have to find all the zeros of $f(x)$.
Solution:
If $a$ is a zero of $f(x)$ then $(x-a)$ is a factor of $f(x)$.
Therefore,
$x-(-2)=x+2$ is a factor of $f(x)$.
On applying the division algorithm,
Dividend$f(x)\ =\ x^3\ +\ 13x^2\ +\ 32x\ +\ 20$
Divisor$=x+2$
$x+2$)$x^3+13x^2+32x+20$($x^2+11x+10$
$x^3+2x^2$
-----------------------------
$11x^2+32x+20$
$11x^2+22x$
-----------------------
$10x+20$
$10x+20$
--------------
$0$
Therefore,
Quotient$=x^2+11x+10$
$f(x)=(x+2)(x^2+11x+10)$
To get the other zeros, put $x^2+11x+10=0$.
$x^2+x+10x+10=0$
$x(x+1)+10(x+1)=0$
$(x+1)(x+10)=0$
$x+1=0$ and $x+10=0$
$x=-1$ and $x=-10$
All the zeros of the given polynomial $f(x)$ are $-2$, $-10$ and $-1$.
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