Question

Apply the division algorithm to find the quotient and remainder on dividing f(x) by g(x) as given below:
(i) $f\left(x\right)=x^3-6x^2+11x-6,g\left(x\right)=x+2$

• A. $Q=x^2-8x-27$ , $R=6$
• B. $Q=x^2-8x-27$ , $R=60$
• C. $Q=x^2-8x+27$ , $R=-60$
• D. $Q=x^2-8x+27$ , $R=6$

Answer 1

Answer: (C)

$Q=x^2-8x+27$ , $R=-60$

We know that the division algorithm states that:

Dividend$=$Divisor$\times$Quotient$+$Remainder

It is given that the dividend is $x^3-6x^2+11x-6$, the divisor is $x+2$. And let the quotient be $a{x}^2+bx+c$ and the remainder be $k$. Therefore, using division algorithm we have:

$x^3-6x^2+11x-6=(x+2)(a{x}^2+bx+c)+k\Rightarrow x^3-6x^2+11x-6=\left[x\right(a{x}^2+bx+c)+2(a{x}^2+bx+c\left)\right]+k\Rightarrow x^3-6x^2+11x-6=(a{x}^3+b{x}^2+cx+2a{x}^2+2bx+2c)+k\Rightarrow x^3-6x^2+11x-6=a{x}^3+(b+2a)x^2+(c+2b)x+(2c+k)$

By comparing the coefficients of the variables and the constant term we get:

$a=1$

$b+2a=-6\Rightarrow b+(2\times 1)=-6\Rightarrow b+2=-6\Rightarrow b=-6-2\Rightarrow b=-8$

$c+2b=11\Rightarrow c+(2\times -8)=11\Rightarrow c-16=11\Rightarrow c=11+16\Rightarrow c=27$

$2c+k=-6\Rightarrow (2\times 27)+k=-6\Rightarrow 54+k=-6\Rightarrow k=-6-54\Rightarrow k=-60$

Substituting the values, we get the quotient and remainder as follows:

$q\left(x\right)=a{x}^2+bx+c=x^2-8x+27$

$r\left(x\right)=k=-60$

Hence, the quotient is $x^2-8x+27$ and the remainder is $-60$.