Question

A quadratic polynomial is exactly divisible by $(x+1)$ & $(x+2)$ and leaves the remainder $4$ after division by $(x+3)$ then that polynomial is

• A. $x^2+6x+4$
• B. $2x^2+6x+4$
• C. $2x^2+6x-4$
• D. $x^2+6x-4$

Answer 1

The correct option is B $2x^2+6x+4$

Since the quadratic polynomial is divisible by both $x+1$ and $x+2$, it will be of the form $k(x+1)(x+2),k$ being a constant.
Thus, it looks like $k{x}^2+3kx+2k.$
Now, when divided by $x+3$, it leaves remainder $4.$
So, $k{x}^2+3kx+2k-4$ is exactly divisible by $x+3.$
Dividing the two, first term in the quotient will be $kx.$
Now, $kx(x+3)$ is $k{x}^2+3kx$, which when subtracted from the dividend gives $2k-4$.
Since the remaining term is just a constant, it has to be zero for the polynomial to exactly divide it.
Hence, $2k-4=0$, implying $k$ to be $2.$
And the required polynomial to be $2x^2+6x+4.$