Question

If (2x-5) is a factor of 6x^3-(k+6)x^2+2kx-25 find k

Answer 1

Answer :

$\tt k = \dfrac{125}{17}$

Given :

P(x) = 6x³ - (k +6) x² + 2kx -25

Factor = (2x-5)

Solution :

Since , (2x -5) is a factor of the given polynomial ,

$\therefore$ 2x - 5 = 0

2x = 5

x = $\dfrac{5}{2}$

Using this value of x in the cubic polynomial ,

$\implies \tt \: p(x) = 6 {x}^{3} - (k + 6) {x}^{2} + 2kx - 25$

$\implies \tt p(x) = 6 \times \left({ \frac{5}{2}}\right)^{3} - (k + 6) \left( { \frac{5}{2} } \right)^{2} + 2k \times \frac{5}{2} - 25$

Let P(x) = 0

$\implies \tt 0 = 6 \times \frac{125}{8} - (k + 6) \frac{25}{4} + 2k \times \frac{5}{2} - 25$

$\implies \tt 0 = \frac{375}{4} - (k + 6) \frac{25}{4} + 2k - 25$

Making the common denominator 4 for all the terms

$\implies\tt 0 = \frac{375}{4} -\frac{(25k + 150)}{4} + \frac{8k}{4} - \frac{100}{4}$

$\implies \tt 0 = 375 - (25 k+ 150) + 8k - 100$

$\implies \tt 0 = 275 - 25k + 8k - 150$

$\implies \tt 0 = 125 - 17k$

$\implies \tt 17k = 125$

$\implies \tt k = \dfrac{125}{17}$