Question

Find the values of ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ so that $(x+1)$ and $(x-1)$ are factors of $x^4+a{x}^3-3x^2+2x+b$

Answer 1

Let $f\left(x\right)=x^4+a{x}^3-3x^2+2x+b$

When $x+1$ is a factor of f(x), then $f(-1)=0$

$\therefore f(-1)=0$

$(-1{)}^4+a(-1{)}^3-3(-1{)}^2+2(-1)+b=0$

$1-a-3-2+b=0$

$\therefore$ $-4-a+b=0$

$\Rightarrow$ $-a+b=4..........\left(i\right)$

Again, When $x-1$ is a factor of f(x), then $f\left(1\right)=0$

$\therefore f\left(1\right)=0$

$\Rightarrow (1{)}^4+a(1{)}^3-3(1{)}^2+2\times 1+b=0$

$\Rightarrow 1+a-3+2+b=0$

$\Rightarrow a+b=\mathrm{0..........}\left(ii\right)$

Adding equation $\left(i\right)$ and $\left(ii\right),$ we have

$2b=4$

$\Rightarrow b=\frac{4}{2}=2$

From equation $\left(ii\right),$ $a+2=0$

$\Rightarrow a=-2$

$\therefore$ $a=-2,b=2$