Question

The polynomial $x^4$ - 2$x^3$ + 3$x^2$ - ax + b when divided by (x+1) and (x-1) gives remainders 19 and 5 respectively. Find the remainder when the polynomial is divided by (x-3).

Answer 1

Let p(x) = $x^4$ - 2$x^3$ + 3$x^2$ - ax + b

Writing the divisors in (x -a) form i.e. (x-(-1)) and (x-1).

According to remainder theorem, if (x-a) divides the polynomial p(x) and c is the remainder, then p(a) =c.

∴ p(-1) = 19 and p(1) = 5

$\Rightarrow$ p(-1) = ${(-1)}^4$ - 2${(-1)}^3$ + 3${(-1)}^2$ - a(-1) + b = 19

$\Rightarrow$ 1 - 2(-1) + 3 + a + b = 19

$\Rightarrow$ 1 + 2 + 3 + a + b = 19

$\Rightarrow$ a + b = 13 ...(I)

and p(1) = ${\left(1\right)}^4$- 2${\left(1\right)}^3$+ 3${\left(1\right)}^2$ - a(1) + b =5

$\Rightarrow$ 1 - 2 + 3 -a + b = 5

$\Rightarrow$ -a + b = 3 ...(II)

Adding equation I and II, gives

2b = 16 ⇒ b = 8

and a = 5

∴ p(x) = $x^4$- 2$x^3$ + 3$x^2$ - 5x + 8

The divisor is (x-3). Applying remainder theorem, remainder = p(3).

i.e. p(3) = ${\left(3\right)}^4$- 2${\left(3\right)}^3$ + 3${\left(3\right)}^2$ - 5(3) + 8 = 81 - 54 + 27 - 15 + 8 = 47