Question

If $1$ and $-3$ are zeroes of the polynomial $p\left(x\right)=x^3-a{x}^2-13x+b$, find the values of $a$ and $b$.

Answer 1

Step 1: Framing the relation between $a$ and $b$.

A real number $c$ is called a zero of the polynomial $p\left(x\right)$ if $p\left(c\right)=0$.

$$\begin{gathered} p\left(1\right)=0 \\ \Rightarrow 1^3-a{\left(1\right)}^2-13\left(1\right)+b=0 \\ \Rightarrow 1-a-13+b=0 \\ \Rightarrow b-a=12\dots \left(1\right) \end{gathered}$$

and

$$\begin{gathered} p\left(-3\right)=0 \\ \Rightarrow {\left(-3\right)}^3-a{\left(-3\right)}^2-13\left(-3\right)+b=0 \\ \Rightarrow -27-9a+39+b=0 \\ \Rightarrow b-9a=-12\dots \left(2\right) \end{gathered}$$

Step 2: Determine the value of $a$ and $b$.

Subtract equation $\left(2\right)$ from $\left(1\right)$.

$\begin{array}{rcl}b-a-\left(b-9a\right)& =& 12-\left(-12\right)\\ -a+9a& =& 24\\ 8a& =& 24\\ & \Rightarrow & a=3\end{array}$

Substitute $3$ for $a$ in equation $\left(1\right)$.

$\begin{array}{rcl}b-3& =& 12\\ & \Rightarrow & b=15\end{array}$

Hence, the value of $a$ and $b$ are $3$ and $15$ respectively.