Question

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

(a) k = -3, zeroes of $2{{x}^{4}}+{{x}^{3}}-14{{x}^{2}}+5x+6$ are 1, -3, 2, $-\dfrac{1}{2}$ and zeroes of ${{x}^{2}}+2x-3$ are 1, -3.
(b) k = -7, zeroes of $2{{x}^{4}}+{{x}^{3}}-14{{x}^{2}}+5x+6$ are 1, 4, -1, 6 and zeroes of ${{x}^{2}}+2x-3$ are 2, 1.
(c) k = -1, zeroes of $2{{x}^{4}}+{{x}^{3}}-14{{x}^{2}}+5x+6$ are 1, -4, 1, $-\dfrac{1}{2}$ and zeroes of ${{x}^{2}}+2x-3$ are -5, -2.
(d) k = -4, zeroes of $2{{x}^{4}}+{{x}^{3}}-14{{x}^{2}}+5x+6$ are 1, 3, 6, $-\dfrac{1}{3}$ and zeroes of ${{x}^{2}}+2x-3$ are -6, 0.

Answer 1

Hint: Divide the given two polynomials by normal division method and find quotient, remainder. As it is given as a factor, the division must leave the remainder to be zero. By equating the remainder to zero find the value of k. Substitute the value of k and then solve the quadratic equation to find the roots. The factorization method breaks the fourth degree equation into smaller degrees and then finds roots of each term to get all the roots of the equation.

Complete step-by-step answer:

Given factor in question: ${{x}^{2}}+2x+k=0$

Given polynomial: $2{{x}^{4}}+{{x}^{3}}-14{{x}^{2}}+5x+6$

Division:

${{x}^{2}}+2x+k\overset{2{{x}^{2}}-3x+\left( -8-2k \right)}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+{{x}^{3}}-14{{x}^{2}}+5x+6 \\

 & 2{{x}^{4}}+4{{x}^{3}}-2k{{x}^{2}} \\

\end{align}}\right.}}$

Step 1 : the term $2{{x}^{2}}$ is added. So,

By subtracting we get

$-3{{x}^{3}}+\left( -14-2k \right){{x}^{2}}+5x$

Step 2: $-3x$ term $-3{{x}^{3}}-6{{x}^{2}}-3kx$ must be subtracted

$\left( -8-2k \right){{x}^{2}}+\left( 5+3k \right)x+6$

Step 3: $\left( -8-2k \right)$ term is added $\left( -8-2k \right){{x}^{2}}+\left( -16-4k \right)x+\left( -8k-2{{k}^{2}} \right)$ must be subtracted

\[\left( 21-7k \right)x+\left( 6+8k+2{{k}^{2}} \right)\]

Hence the remainder = $\left( 21-7k \right)x+\left( 6+8k+2{{k}^{2}} \right)$

Given in the question it is a factor, so remainder must be zero. Hence equating the remainder to zero, we get: $21+7k=0$

By simplifying above equation, we get the value of k to be

$k=-3$

So, the value of k for which it is a factor is -3.

Next we need to find zeros of the both polynomials given:

First case: the given second degree polynomial in the question:

${{x}^{2}}+2x+k=0$

By substituting the value of k, the equation turns to be:

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

By factorization method above equation turns to be as follows:

$\left( x-1 \right)\left( x+3 \right)=0$

So, +1, -3 are the zeros of polynomial in the question

By division method we can write the fourth degree polynomial as:

$=\left( {{x}^{2}}+2x-3 \right)\left( 2{{x}^{2}}-3x+\left( -8-2k \right) \right)$

By substituting the value of k, the second part becomes:

$2{{x}^{2}}-3x-2=0$

By factorization method above equation turns to be as follows:

$\left( 2x+1 \right)\left( x-2 \right)=0$

So, $-\dfrac{1}{2},2$ , are the zeros of the polynomial in the question.

Therefore option(a) is correct.


Note: The observation that the division method turns the fourth degree polynomial into 2 second degree polynomials is crucial to solve carefully.