Question

The polynomial p(x) = x⁴ − 2x³ + 3x² − ax + b when divided by (x − 1) and (x + 1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when p(x) is divided by (x − 2).

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ p\left(x\right)=x^4-2x^3+3x^2-ax+b \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \mathrm{When}p\left(x\right)\mathrm{is}\mathrm{divided}\mathrm{by}\left(x-1\right),\mathrm{the}\mathrm{remainder}\mathrm{is}p\left(1\right). \\ \mathrm{When}p\left(x\right)\mathrm{is}\mathrm{divided}\mathrm{by}\left(x+1\right),\mathrm{the}\mathrm{remainder}\mathrm{is}p\left(-1\right). \end{gathered}$$
Thus, we have:
$$\begin{gathered} p\left(1\right)=\left(1^4-2\times 1^3+3\times 1^2-a\times 1+b\right) \\ =\left(1-2+3-a+b\right) \\ =2-a+b \end{gathered}$$
And,
$$\begin{gathered} p\left(-1\right)=\left[{\left(-1\right)}^4-2\times {\left(-1\right)}^3+3\times {\left(-1\right)}^2-a\times \left(-1\right)+b\right] \\ =\left(1+2+3+a+b\right) \\ =6+a+b \end{gathered}$$

Now,

$$\begin{gathered} 2-a+b=5...\left(1\right) \\ 6+a+b=19...\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ 8+2b=24 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2b=16 \\ \Rightarrow b=8 \end{gathered}$$
By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,
$f\left(x\right)=x^4-2x^3+3x^2-5x+8$
Also,
$\mathrm{When}p\left(x\right)\mathrm{is}\mathrm{divided}\mathrm{by}\left(x-2\right),\mathrm{the}\mathrm{remainder}\mathrm{is}p\left(2\right).$
Thus, we have:
$$\begin{gathered} p\left(2\right)=\left(2^4-2\times 2^3+3\times 2^2-5\times 2+8\right)\left[a=5\mathrm{and}b=8\right] \\ =\left(16-16+12-10+8\right) \\ =10 \end{gathered}$$