Question

The polynomial $x^4-{2x}^3+{3x}^2-ax+b$ when divided by $x+1$ and $x-1$ gives remainders 19 and 5 respectively. Find the remainder when the polynomial is divided by $x-3$. [3 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark
Calculation: 1 Mark

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

$p(-1)=19;p\left(1\right)=5$ [ Using Remainder theorem]

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

$a+b=13...............\left(1\right)$

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

$-a+b=3................\left(2\right)$

On solving (1) and (2), we get $a=5,b=8$

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

When p(x) is divided by(x-3),
Remainder = $p\left(3\right)={\left(3\right)}^4-2{\left(3\right)}^3+3{\left(3\right)}^2-5\left(3\right)+8$ [Using remainder theorem]

$p\left(3\right)=47$