Question

$x=0$ and $x=-1$ $P\left(x\right)=2x^3-3x^2+ax+b$ find a and b

Answer 1

It is given that $x=0$ and $x=-1$ are the roots of the polynomial $P\left(x\right)=2x^3-3x^2+ax+b$, therefore, $P\left(0\right)=0$ and $P(-1)=0$.

Let us first substitute $x=0$ in $P\left(x\right)=2x^3-3x^2+ax+b$ as follows:

$P\left(0\right)=2(0{)}^3-3(0{)}^2+a\left(0\right)+b\Rightarrow 0=0-0+0+b\Rightarrow b=0$

Now, substitute $x=-1$ in $P\left(x\right)=2x^3-3x^2+ax+b$ as follows:

$P(-1)=2(-1{)}^3-3(-1{)}^2+a(-1)+b\Rightarrow 0=(2\times -1)-(3\times 1)-a+b\Rightarrow 0=-2-3-a+b\Rightarrow 0=-5-a+b\Rightarrow a=b-5\Rightarrow a=0-5(\because b=0)\Rightarrow a=-5$

Hence, $a=-5,b=0$.