Question

If x = - 4/ 3 is a root of polynomial p(x) = 6x3-11x2+kx-20 , find the value of k.

Answer 1

Given polynomial =

p(x)= 6x³-11x²+kx-20

Given root of p(x) = -⁴/3

It means if we divide p(x) by -⁴/3 then we will have the remainder = 0

So put x = -4/3 in p(x)

$6 {x}^{3} - 11 {x}^{2} + kx - 20 = 0 \\ \\ 6( \frac{ - 4}{3} ) ^{3} - 11( { \frac{ - 4}{3} }^{2} ) + k( \frac{ - 4}{3} ) - 20 = 0 \\ \\ 6 \times \frac{ - 64}{27} - 11 \times \frac{16}{9} \frac{ - 4k}{3} - 20 = 0 \\ \\ \frac{ - 128}{9} - \frac{176}{9 } - \frac{4k}{3} = 0 \\ \\ \frac{ - 128 - 176 - 12k}{9} = 0 \\ \\ - 304 - 12k = 0 \\ \\ - 12k = 304 \\ \\ k = - \frac{304}{12}$