Question

Find $k$ so that $x^2+2x+k$ is a factor of $2x^4+x^3-14x^2+5x+6$. Also, find all the zeros of the two polynomials.

• A. $k=-3$, zeroes of $2x^4+x^3-14x^2+5x+6$ are $1,-3,2,-\frac{1}{2}$ and zeroes of $x^2+2x-3$ are $1,-3.$
• B. $k=-7$, zeroes of $2x^4+x^3-14x^2+5x+6$ are $1,4,-1,6$ and zeroes of $x^2+2x-3$ are $2,1.$
• C. $k=-1$, zeroes of $2x^4+x^3-14x^2+5x+6$ are $1,-4,1,-\frac{1}{2}$ and zeroes of $x^2+2x-3$ are $-5,-2.$
• D. $k=-4$, zeroes of $2x^4+x^3-14x^2+5x+6$ are $1,3,6,-\frac{1}{3}$ and zeroes of $x^2+2x-3$ are $-6,0$.

Answer 1

The correct option is A $k=-3$, zeroes of $2x^4+x^3-14x^2+5x+6$ are $1,-3,2,-\frac{1}{2}$ and zeroes of $x^2+2x-3$ are $1,-3.$

Since $x^2+2x+k$ is a factor of $2x^4+x^3-14x^2+5x+6$, so we dividing $2x^4+x^3-14x^2+5x+6$ by $x^2+2x+k$ should leave the remainder $0$ as shown below:

Comparing the coefficient of $x$, we get

$21+7k=0\Rightarrow 7k=-21\Rightarrow k=-\frac{21}{7}=-3$

The equation $x^2+2x+k$ becomes $x^2+2x-3$ and the factors of $x^2+2x-3=0$ are:

$x^2+3x-x-3=0$

$x(x+3)-1(x+3)=0$

$(x-1)(x+3)=0$

$x=1,-3$

Now, the equation $2x^2-3x-(8+2k)$ becomes $2x^2-3x-2$ and the factors of $2x^2-3x-2=0$ are:

$2x^2-4x+x-2=0$

$2x(x-2)+1(x-2)=0$

$(2x+1)(x-2)=0$

$x=-\frac{1}{2},2$

Hence, the zeros of the given two polynomials are $1,-3,-\frac{1}{2}$ and $2$.