Question

The polynomial $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4-2{\mathrm{x}}^3+3{\mathrm{x}}^2-\mathrm{ax}+\mathrm{b}$ when divided by $(\mathrm{x}-1)$ and $(\mathrm{x}+1)$ leaves remainder $5$and $19$. Find the values of $\mathrm{a}$ and $\mathrm{b}$. Hence, find the remainder when $\mathrm{f}\left(\mathrm{x}\right)$ is divided by $(\mathrm{x}-2)$.

Answer 1

Step 1. Divide the given polynomial by $\mathrm{x}-1$:

Remainder theorem:

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

If p(x) is divided by $(\mathrm{x}-\mathrm{a})$, then $\mathrm{p}\left(\mathrm{a}\right)$ is the remainder.

The remainder here will be $\mathrm{f}\left(1\right)$.

$$\begin{gathered} \mathrm{f}\left(1\right)={\left(1\right)}^4-2{\left(1\right)}^3+3{\left(1\right)}^2-\mathrm{a}+\mathrm{b} \\ \mathrm{f}\left(1\right)=1-2+3-\mathrm{a}+\mathrm{b} \\ \mathrm{f}\left(1\right)=2-\mathrm{a}+\mathrm{b} \end{gathered}$$

We are given that remainder is 5. Therefore,

$$\begin{gathered} 5=2-\mathrm{a}+\mathrm{b} \\ \mathrm{a}-\mathrm{b}=-3...................\left(1\right) \end{gathered}$$

Step 2: Divide the given polynomial by $\mathrm{x}+1$.

The remainder here will be $\mathrm{f}(-1)$.

$$\begin{gathered} \mathrm{f}(-1)={(-1)}^4-2{(-1)}^3+3{(-1)}^2-\mathrm{a}(-1)+\mathrm{b} \\ \mathrm{f}(-1)=1+2+3+\mathrm{a}+\mathrm{b} \\ \mathrm{f}(-1)=6+\mathrm{a}+\mathrm{b} \end{gathered}$$

We are given that remainder is 19. Therefore,

$$\begin{gathered} 19=6+\mathrm{a}+\mathrm{b} \\ 13=\mathrm{a}+\mathrm{b}.................\left(2\right) \end{gathered}$$

Step 3. Find values of a and b:

Adding (1) and (2) we get,

$$\begin{gathered} 2a=10 \\ a=5 \end{gathered}$$

Putting $\mathrm{a}=5$in (1) we get,

$$\begin{gathered} 13=5+\mathrm{b} \\ \mathrm{b}=8 \end{gathered}$$

Step 4: Put the values of a and b in the parent equation and find the remainder.

Parent equation is $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4-2{\mathrm{x}}^3+3{\mathrm{x}}^2-5\mathrm{x}+8$. For finding the remainder when the equation is divided by $(\mathrm{x}-2)$, we find $\mathrm{f}\left(2\right)$.

$$\begin{gathered} \mathrm{f}\left(2\right)={\left(2\right)}^4-2{\left(2\right)}^3+3{\left(2\right)}^2-5\left(2\right)+8 \\ \mathrm{f}\left(2\right)=16-16+12-10+8 \\ \mathrm{f}\left(2\right)=10 \end{gathered}$$

Hence, $\mathrm{a}=5$ and $\mathrm{b}=8$. Also, when $\mathrm{f}\left(\mathrm{x}\right)$ is divided by $(\mathrm{x}-2)$, the remainder is$10$.