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Find $ \mathrm{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.
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.
$x^2+2x+k$)$2x^4+x^3-14x^2+5x+6$($2x^2-3x-2(k+4)$
$2x^4+4x^3+2kx^2$
-----------------------------------
$-3x^3-2x^2(k+7)+5x+6$
$-3x^3-6x^2 -3kx$
--------------------------------------------
$-2x^2(k+4)+x(5+3k)+6$
$-2x^2(k+4)-4x(k+4)-2k(k+4)$
-------------------------------------------------
$x(7k+21)+(2k^2+8k+6)$
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 \( -3 \).
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