Question

Given $f\left(x\right)$ is a cubic polynomial in $x$. If $f\left(x\right)$ is divided by $(x+3),(x+4),(x+5)$ and $(x+6)$ then it leaves the remainders $0,0,4$ and $6$ respectively. Find the remainder when $f\left(x\right)$ is divided by $x+7$.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

It is given that $f\left(x\right)$ leaves the remainder $0$ if divided by $(x+3)$ and $(x+4)$, which implies that $(x+3)$ and $(x+4)$ are factors of $f\left(x\right)$.

Let the other factor be $ax+p$, then $f\left(x\right)$ is given by:

$f\left(x\right)=(x+3)(x+4)(ax+p)$

Now, it is also given that $f\left(x\right)$ leaves the remainders $4$ and $6$ if divided by $(x+5)$ and $(x+6)$, which means that $f(-5)=4$ and $f(-6)=6$.

If $f(-5)=4$, then we have:

$f\left(x\right)=(x+3)(x+4)(ax+p)\Rightarrow f(-5)=(-5+3)(-5+4)\left(a\right(-5)+p)\Rightarrow 4=(-2)(-1)(-5a+p)\Rightarrow 4=2(-5a+p)\Rightarrow -5a+p=2........\left(1\right)$

And if $f(-6)=6$, then we have:

$f\left(x\right)=(x+3)(x+4)(ax+p)\Rightarrow f(-6)=(-6+3)(-6+4)\left(a\right(-6)+p)\Rightarrow 6=(-3)(-2)(-6a+p)\Rightarrow 6=6(-6a+p)\Rightarrow -6a+p=1........\left(2\right)$

Subtract eqn 2 from eqn 1 as follows:

$[-5a-(-6a\left)\right]+(p-p)=2-1\Rightarrow (-5a+6a)+0=1\Rightarrow a=1$

Substitute the value of $a$ in eqn 1:

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

Therefore,

$f\left(x\right)=(x+3)(x+4)\left[\right(1\times x+7\left)\right]\Rightarrow f\left(x\right)=(x+3)(x+4)(x+7)$

Thus, $f(-7)=0$

Hence, $f\left(x\right)$ leaves the remainder $0$ when divided by $x+7$.