Question

Apply the division algorithm to find the quotient and remainder on dividing $f\left(x\right)$ by $g\left(x\right)$ as given below:

$f\left(x\right)=x^4-5x+6,g\left(x\right)=2-x^2$ [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

We have,

$f\left(x\right)=x^4+0x^3+0x^2-5x+6$ and $g\left(x\right)=-x^2+2$

Here, degree $\left(f\right(x\left)\right)=4$ and degree $g\left(x\right)=2$

Therefore, quotient $q\left(x\right)$ and remainder $r\left(x\right)$ are of degree 2 and less than 2 respectively.

Let $q\left(x\right)=a{x}^2+bx+c$ and $r\left(x\right)=px+q$

By division algorithm, we have

$f\left(x\right)=g\left(x\right)\times q\left(x\right)+r\left(x\right)$

$\Rightarrow x^4+0x^3+0x^2-5x+6=(-x^2+2)(a{x}^2+bx+c)+px+q$

$\Rightarrow x^4+0x^3+0x^2-5x+6=-a{x}^4-b{x}^3+(2a-c)x^2+(2b+p)x+2c+q$

Equating the coefficients of various powers of x, we get

-a = 1 [On equating the coefficients of $x^4$]

-b = 0 [On equating the coefficients of $x^3$]

2a - c = 0 [On equating the coefficients of $x^2$]

2b + p = -5 [On equating the coefficients of $x$]

and, 2c + q = 6 [On equating the constant terms]

Solving these equations, we get

a = -1, b = 0, c = -2, p = -5 and q = 10

$\therefore$ Quotient $q\left(x\right)=-x^2-2$ and Remainder $r\left(x\right)=-5x+10$