Question

Find the quotient and remainder on dividing $p\left(x\right)=x^3-6x^2+15x-8$ by $g\left(x\right)=x-2$

Answer 1

$p\left(x\right)=x^3-6x^2+15x-8$

$\therefore$ degree of $p\left(x\right)$ is $3$.

$g\left(x\right)=x-2$

$\therefore$ degree of $g\left(x\right)$ is $1$.

$\therefore$ degree of quotient $q\left(x\right)=3-1=2$,

and degree of remainder $r\left(x\right)$ is zero.

Let, $q\left(x\right)=a{x}^2+bx+c$ (Polynomial of degree $2$) and $r\left(x\right)=k$ (constant polynomial)

By using division algorithm, we have

$p\left(x\right)=\left[g\right(x)\times q(x)+r(x\left)\right]=x^3-6x^2+15x-8=(x-2)(a{x}^2+bx+c)+k$

$=a{x}^3+b{x}^2+cx-2a{x}^2-2bx-2c+k$

$\therefore x^3-6x^2+15x-8=a{x}^3+(b-2a)x^2+(c-2b)x-2c+k$
We have cubic polynomials on both the sides of the equation.

$\therefore$ Let us compare the coefficients of $x^3,x^2,x$ and

$k$ to get the values of $a,b,c$.

$1=a$, it is the coefficient of $x^3$ on both sides

$-6=b-2a$, it is the coefficient of $x^2$ on both sides

$15=c-2b$, it is the coefficient of $x$ on both sides

$-8=-2c+k$, it is the constant term on both sides
Let us solve these equations to get the values of $b,c$, and $k$.

$b-2a=-6$

$\therefore b=-6+2a$

$\therefore b=-6+2\left(1\right)=-4$

$c-2b=15$

$\therefore c=15+2b$

$\therefore c=15+2(-4)=7$

$-2c+k=-8$

$\therefore k=-8+2c$

$\therefore k=-8+2\left(7\right)=6$

Hence, $q\left(x\right)=a{x}^2+bx+c$

$=\left(1\right)x^2+(-4)x+7$

$=x^2-4x+7$

and $r\left(x\right)=k=6$

$\therefore$ the quotient is $x^2-4x+7$ and the remainder is $6$.