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 the zeroes of the two polynomials.

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Solution

Given,
$x^{2} + 2 x + k$ is a factor of $2 x^{4} + x^{3} - 14 x^{2} + 5 x + 6$ .
Then we can write,
$2 x^{4} + x^{3} - 14 x^{2} + 5 x + 6 =$ $(x^{2} + 2 x + k) (2 x^{2} + a x + b)$
[Where we are to determine the values of $k, a, b$ ]
or, $2 x^{4} + x^{3} - 14 x^{2} + 5 x + 6 = 2 x^{4} + x^{3} (a + 4) + x^{2} (b + 2 a + 2 k) + x (2 b + k a) + k b$
Comparing the both sides we get,
$a + 4 = 1 \Rightarrow a = - 3$ ........(1) and $b + 2 a + 2 k = - 14$ ........(2) and $2 b + k a = 5$ ......(3) and $k b = 6$ ......(4).
Using (1) equation (2) and (3) can be written as
$b + 2 k = - 8$ and $2 b - 3 a = 5$
Solving these two equations we get,
$b = - 2$ and $k = - 3$ , which satisfies equation (4).
So, $k = - 3$ .
Now the factors are $(x^{2} + 2 x - 3) (2 x^{2} - 3 x - 2) = (x + 3) (x - 1) (2 x + 1) (x - 2)$ .
So, zeros of the $4^{t h}$ degree polynomial are $- 3, 1, - \frac{1}{2}, 2$ among which $- 3, 1$ are the zeros of the given second order polynomial.

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