Question

If $(x-k)$ is the H.C.F of $(3x^2+14x+16)$ and $(6x^3+11x^2-4x-4)$, then find $"k"$.

• A. $-2$
• B. $2$
• C. $\frac{2}{3}$
• D. $\frac{-1}{2}$

Answer 1

The correct option is A $-2$

If $(x-k)$ is the HCF of $({3x}^2+14x+16)$ and $({6x}^3+{11x}^2-4x-4)$

then value of $k=?$

Now, we know HCF is the highest common factor of two numbers/polynomials.

The lowest degree polynomial is quadratic, here. We can find the factors of this quadratic polynomial and the factor must be either of the two among the the factors of ${3x}^2+14x+16=3x^2+6x+8x+16$

$=3x(x+2)+8(x+2)$

$=(3x+8)(x+2)$

So, since the factors are $(3x+8)\&(x+2)$ but the form is $(x-k)$ we can infer

$\Rightarrow (x-k)=(x+2)$

Comparing both sides, $k=-2.$

Hence, the answer is $-2.$