Question

If $1$ and $-2$ are the zeros of $f\left(x\right)=x^3+10x^2+ax+b$, then find the values of $a$.

Answer 1

$f\left(x\right)=x^3+10x^2+ax+b$
$1$ and $-2$ are the zeros of $f\left(x\right)$
Thus,

$f\left(1\right)=0$ and $f(-2)=0$
$f\left(1\right)=0$
$\Rightarrow 1^3+10(1{)}^2+a(1)+b=0$
$\Rightarrow 1+10+a+b=0$
$\Rightarrow a+b+11=0$-------------(i)

$f(-2)=0$
$\Rightarrow (-2{)}^3+10(-2{)}^2+a(-2)+b=0$
$\Rightarrow -8+40-2a+b=0$
$\Rightarrow -2a+b+32=0$ ---------------(ii)

Substracting (ii) and (i), we get
$\Rightarrow a+b+11-(-2a+b+32)=0$
$\Rightarrow a+b+11+2a-b-32=0$
$\Rightarrow 3a-21=0$
$\Rightarrow a=7$

Putting this value in (i), we get
$\Rightarrow 7+b+11=0$
$\Rightarrow b+18=0$
$\Rightarrow b=-18$

Thus,

$a=7$ and $b=-18$