Question

Find the values of $\mathrm{a}$ and $\mathrm{b}$ so that the polynomial ${\mathrm{x}}^3-{\mathrm{ax}}^2-13\mathrm{x}+\mathrm{b}$ has $(\mathrm{x}-1)$ and $(\mathrm{x}+3)$ as factors.

Answer 1

Factor Theorem:

Let $\mathrm{p}\left(\mathrm{x}\right)$ be any polynomial of degree greater than or equal to 1 and $\mathrm{a}$ be any real number.

If $\mathrm{p}\left(\mathrm{a}\right)=0$, then $(\mathrm{x}-\mathrm{a})$ is a factor of $\mathrm{p}\left(\mathrm{x}\right)$.

Solution:

Step 1. Apply factor theorem in the given polynomial using the given two factors:

Let $\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^3-{\mathrm{ax}}^2-13\mathrm{x}+\mathrm{b}$.

Since, $(\mathrm{x}-1)$ and $(\mathrm{x}+3)$ are the factors of $\mathrm{p}\left(\mathrm{x}\right)$, $\mathrm{p}\left(1\right)=0$ and $\mathrm{p}(-3)=0$.

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

and

$$\begin{gathered} \mathrm{p}(-3)={(-3)}^3-\mathrm{a}{(-3)}^2-13(-3)+\mathrm{b} \\ \mathrm{p}(-3)=-27-3\mathrm{a}+39+\mathrm{b} \\ \mathrm{p}(-3)=-3\mathrm{a}+\mathrm{b}+12 \end{gathered}$$

Step 2. Finding the value of a:

We are given that $\mathrm{p}\left(1\right)=\mathrm{p}(-3)$.

Therefore,

$$\begin{gathered} -\mathrm{a}+\mathrm{b}-12=-3\mathrm{a}+\mathrm{b}+12 \\ 2\mathrm{a}=24 \\ \mathrm{a}=12 \end{gathered}$$

Step 3. Finding the value of b:

On putting the value of $\mathrm{a}$ in $\mathrm{p}\left(1\right)$ we get,

$$\begin{gathered} \mathrm{p}\left(1\right)=-12+\mathrm{b}-12 \\ 0=\mathrm{b}-24 \\ \mathrm{b}=24 \end{gathered}$$

Hence, $\mathrm{a}=12,\mathrm{b}=24$.