Question

The polynomial $p\left(x\right)=x^4-2x^3+3x^2-ax+b$ 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

Let:$p\left(x\right)=x^4-2x^3+3x^2-ax+b$

Now,When p(x) is divided by x-1, the remainder is p(1).When p(x) is divided by x+1, the remainder is p(-1).

Thus, we have:

$p\left(1\right)=1^4-2\times 1^3+3\times 1^2-a\times 1+b=1-2+3-a+b=2-a+b$

And,

$p(-1)=-1^4-2\times -1^3+3\times -1^2-a\times -1+b=1+2+3+a+b=6+a+b$

Now,

$2-a+b=5----\left(1\right)$

$6+a+b=19---\left(2\right)$

Adding (1) and (2), we get:

$8+2b=24\Rightarrow 2b=16\Rightarrow b=8$

By putting the value of b, we get the value of a, i.e., 5.

∴ a = 5 and b = 8

Now,

$p\left(x\right)=x^4-2x^3+3x^2-5x+8$

Also,

When p(x) is divided by (x-2), the remainder is p(2).

Thus we have:

$p\left(2\right)=2^4-2\times 2^3+3\times 2^2-5\times 2+8=16-16+12-10+8=10$