Question

If $x^3-5x^2+7$ is divisible by $(x+2)$, then the remainder is

• A. $-21$
• B. $-20$
• C. $-17$
• D. $-25$

Answer 1

Answer: (A)

$-21$

Let the given polynomial be $p\left(x\right)=x^3-5x^2+7$, the quotient be $a{x}^2+bx+c$ and the remainder be $d$.

We divide $p\left(x\right)$ by $(x+2)$ by equating coefficient approch to find the remainder as shown below:

$x^3-5x^2+7=(x+2)(a{x}^2+bx+c)+d\Rightarrow x^3-5x^2+7=\left[x\right(a{x}^2+bx+c)+2(a{x}^2+bx+c\left)\right]+d\Rightarrow x^3-5x^2+7=(a{x}^3+b{x}^2+cx+2a{x}^2+2bx+2c)+d\Rightarrow x^3-5x^2+7=a{x}^3+(b+2a)x^2+(c+2b)x+2c+d$

Now equating the coefficients, we get:

Coefficient of $x^3$:

$a=1$

Coefficient of $x^2$:

$b+2a=-5\Rightarrow b+(2\times 1)=-5\Rightarrow b+2=-5\Rightarrow b=-5-2\Rightarrow b=-7$

Coefficient of $x$:

$c+2b=0\Rightarrow c+(2\times -7)=0\Rightarrow c-14=0\Rightarrow c=14$

Constant terms:

$2c+d=7\Rightarrow (2\times 14)+d=7\Rightarrow 28+d=7\Rightarrow d=7-28\Rightarrow d=-21$

Hence, the remainder is $-21$.