##### Find k, so that x2 + 2x + k is a factor of 2x4 + x3 - 14x2 + 5x + 6. Also, find all the zeroes of the two polynomials.

let g(x) = x2 + 2x + k and p(x) = 2x4 + x3 - 14x2 + 5x + 6

Then if g(x) is a factor of p(x)

By division algorithm

Dividend = (divisor) (quotient) + remainder

p(x) = g(x).q(x) + r(x)

Where q(x) is the quotient and

r(x) is the remainder which will be equal to zero.

i.e. r(x) = 0

Clearly q(x) = 2x2 - 3x - 8 - 2k and r(x) = (21 + 7k)x + (2k2 + 8k + 6) = 0

By comparing the coefficients, 21 + 7k = 0 and 2k2 + 8k + 6 = 0

2k2 + 8k + 6 = 0

k2 + 4k + 3 = 0

k2 + (k + 3k) + 3 = 0

k (k + 1) + 3(k + 1) = 0

(k + 1)(k + 3) = 0

k = - 1, - 3

Also, 21 + 7k = 0

Case1: k = - 1

21 + 7(- 1) = 0

= 21 - 7 = 14≠0

Hence k = - 1 is rejected.

Case2: k = - 3

21 + 7(- 3) = 0

= 21 - 21 = 0

the value of k is - 3

g(x) = x2 + 2x - 3

x2 + 2x - 3 = 0

x2 + (- x + 3x) - 3 = 0

x(x - 1) + 3(x - 1) = 0

(x - 1)(x + 3) = 0

x = 1, - 3

q(x) = 2x2 - 3x - 8 - 2(- 3)

= 2x2 - 3x - 2

2x2 - 3x - 2 = 0

2x2 - (4x - x) - 2 = 0

2x(x - 2) + (x - 2) = 0

(x - 2)(2x + 1) = 0

x = 2, - 1/2

Now, we know g(x) and q(x) are factors of p(x)

the zeroes of g(x) and q(x) will be the zeroes of p(x)

Hence, the zeroes of p(x) = - 3, - 1/2, 1, 2

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