Q. 44.1( 45 Votes )

Find k, so that x

Answer :

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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