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.

AcademicMathematicsNCERTClass 10

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 \( -3 \).

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

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