# What is the value of the constant k if the real solutions to the equation x^4-kx^3+2kx^2+2x-20=0 are x=2 and x=-1?

Oct 3, 2017

$k = 7$

#### Explanation:

Given: ${x}^{4} - k {x}^{3} + 2 k {x}^{2} + 2 x - 20 = 0$

Substitute -1 for x:

${\left(- 1\right)}^{4} - k {\left(- 1\right)}^{3} + 2 k {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 20 = 0$

$1 + k + 2 k - 2 - 20 = 0$

$3 k - 21 = 0$

$k = 7$

Substitute 2 for x:

${\left(2\right)}^{4} - k {\left(2\right)}^{3} + 2 k {\left(2\right)}^{2} + 2 \left(2\right) - 20 = 0$

$16 - 8 k + 8 k + 4 - 20 = 0$

$0 = 0$

This means that all real values of k will give the polynomial a root of $x = 2$, therefore, well select the most restrictive, $k = 7$

Oct 3, 2017

See below.

#### Explanation:

According to the question

${x}^{4} - k {x}^{3} + 2 k {x}^{2} + 2 x - 20 = \left(x - 2\right) \left(x + 1\right) \left(a {x}^{2} + b x + c\right)$

or grouping coefficients

$\left\{\begin{matrix}2 c - 20 = 0 \\ 2 + 2 b + c = 0 \\ 2 a + b - c + 2 k = 0 \\ a - b - k = 0 \\ 1 - a = 0\end{matrix}\right.$

Solving for $a , b , c , k$ we obtain

$a = 1 , b = - 6 , c = 10 , k = 7$