Question

I: lf $(x-1)$ is a factor of $x^5-x^4-4x^3+4x^2+4x+k$, then $k=-4$.

II : If the remainders of the polynomial $f\left(x\right)$ when divided by $(x+1)$ and $(x-1)$ are $3,7$, then the remainder of $f\left(x\right)$ when divided by $(x^2-1)$ is $(2x+5).$
Which of the following statement is/are true?

• A. Only I
• B. Only II
• C. Both I and II
• D. Neither I nor II

Answer 1

The correct option is C Both I and II

Statement I:

Since, $(x-1)$ is a factor of $x^5-x^4-4x^3+4x^2+4x+k$
So, $x=1$ satisfies the polynomial
$1-1-4+4+4+k=0$
$\Rightarrow k=-4$
Statement I is correct.
Now, statement II:
Since, $f\left(x\right)$ when divided by $(x+1)$, remainder is $3$
$\Rightarrow f(-1)=3$ ....(1)
Also, $f\left(x\right)$ when divided by $(x-1)$, remainder is $7$
$\Rightarrow f\left(1\right)=7$ ......(2)
We have to find the remainder of $f\left(x\right)$ when divided by $(x+1)(x-1)$
Let $g\left(x\right)=(x+1)(x-1)=x^2-1$
Degree of divisor $g\left(x\right)$ is $2$.
Clearly, the remainder will be of the form $r\left(x\right)=ax+b$
We know
$f\left(x\right)=q\left(x\right)g\left(x\right)+r\left(x\right)$
$f\left(x\right)=q\left(x\right)(x^2-1)+ax+b$ ...(3)
Put, $x=1$ in (3), we get
$f\left(1\right)=0+a+b$
$\Rightarrow a+b=7$ ...(4)
Put $x=-1$ in (3), we get
Also, $f(-1)=0-a+b$
$\Rightarrow 3=-a+b$ ...(5)
Solving (4) and (5), we get
$b=5,a=2$
So, the remainder is $r\left(x\right)=2x+5$
Hence, statement II is correct.

$\therefore$ Both I and II are correct.