Question

Q.5.
What is the value of k that (2x-1) may be a factor
of 4x^4-(k-1) x^3 + kx^2 - 6x + 1?
​

Answer 1

Given :

(2x-1) is factor of 4x² - (k-1)x³ + kx² - 6x + 1

To find :

k

Solution :

p(x) = 4x⁴ - (k-1)x³ + kx² - 6x + 1

if (2x-1) is factor of above given Equation,then

2x - 1 = 0

2x = 1

x = 1/2

Therefore , p(1/2) = 0

$\sf \: p\bigg( \dfrac{1}{2} \bigg) \: = \: 4 \times {\bigg( \dfrac{1}{2} \bigg)}^{4} \: - \: (k - 1) {\bigg( \dfrac{1}{2} \bigg)}^{3} + k \times \bigg( \dfrac{1}{2} \bigg) ^{2} \: - \: 6 \times \bigg( \dfrac{1}{2} \bigg) \: + \: 1 \\ \\ \\ \sf \implies \: 0 \: = \: 4 \times {\bigg( \dfrac{1}{16} \bigg)} \: - \: (k - 1) {\bigg( \dfrac{1}{8} \bigg)} + k \times \bigg( \dfrac{1}{4} \bigg) \: - \: 6 \times \bigg( \dfrac{1}{2} \bigg) \: + \: 1 \\ \\ \\ \sf \implies \: 0 \: = \: \dfrac{1}{4} \: - \: { \dfrac{(k - 1)}{8} } + \dfrac{k}{4} \: - \: 3 \: + \: 1 \\ \\ \\ \sf \implies \: 0 \: = \: \dfrac{1}{4} \: - \: { \dfrac{(k - 1)}{8} } + \dfrac{k}{4} \: - 2 \\ \\ \\ \sf \implies \: 2 - \dfrac{1}{4} \: = \: \dfrac{k}{4} \: - \: { \dfrac{(k - 1)}{8} } \\ \\ \sf \: take \: lcm \: \\ \\ \sf \implies \: \dfrac{(2 \times 4) - (1 \times 1)}{4} \: = \: \dfrac{(k \times 2) \: \: - (1\times (k - 1) \: \: )}{8} \\ \\ \sf \implies \: \dfrac{8 - 1}{4} \: = \: \dfrac{2k \: - (k - 1) }{8} \\ \\ \sf \implies \: \dfrac{7}{4} \times 8 = 2k - k + 1 \\ \\ \sf \implies \: 7 \times 2 = k + 1 \\ \\ \sf \implies \: 14 = k + 1 \\ \\ \sf \implies \: k \: = 14 - 1 \\ \\ \sf \implies \: k \: = 13$

ANSWER :

k = 13