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

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Given:

$x^{2}+2 x+k$ is a factor of the polynomial $f(x)=2 x^{4}+x^{3}-14 x^{2}+5 x+6$ when divided by $x^{2}+2 x+k$, the remainder is zero.

To do:

We have to find $k$ and find all the zeroes of the two polynomials.

Solution:

Divide $f(x)=2 x^{4}+x^{3}-14 x^{2}+5 x+6$ by $x^{2}+2 x+k$ by using long division method.

$x^2+2x+k$)$2x^4+x^3-14x^2+5x+6$($2x^2-3x-2(k+4)$

$2x^4+4x^3+2kx^2$

-----------------------------------

$-3x^3-2x^2(k+7)+5x+6$

$-3x^3-6x^2 -3kx$

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$-2x^2(k+4)+x(5+3k)+6$

$-2x^2(k+4)-4x(k+4)-2k(k+4)$

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$x(7k+21)+(2k^2+8k+6)$

Remainder $=x(7 k+21)+\left(2 k^{2}+8 k+6\right)$ and Quotient $=2 x^{2}-3 x-2(k+4)$. If it is a factor then the remainder $=0$

$\Rightarrow x(7 k+21)+2\left(k^{2}+4 k+3\right)=0$ for all $x$.

$\Rightarrow 7 k+21=0$ and $k^{2}+4 k+3=0$

$\Rightarrow 7(k+3)=0$ and $(k+1)(k+3)=0$

$\Rightarrow k+3=0$

$\Rightarrow k=-3$

Substituting the value of $k$ in $x^{2}+2 x+k$, we get,

$x^{2}+2 x-3=(x+3)(x-1)$ as the divisor.

Its zeros are $-3$ and 1.

Therefore, two zeros of $f(x)$ are $-3$ and $1 .$

For $k=-3$, we get,

Quotient $= 2x^{2}-3 x-2$

$=2 x^{2}-4 x+x-2$

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

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

Divisor $=x^{2}+2 x-3$

$=x^{2}+3 x-x-3$

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

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

Therefore, $f(x)=$Quotient$\times$Divisor

$\Rightarrow f(x)=2 x^{4}+x^{3}-14 x^{2}+5 x+6$

$=(x-2)(2 x+1)(x-1)(x+3)$

Hence, zeros of $f(x)$ are $2,\frac{-1}{2},1$ and $-3$.

Updated on 10-Oct-2022 13:27:09