# Can you show that p(x)=2x^3 + 7x^2 + kx - k is the produce of 3 linear factors, 2 of which are identical? Show that k can take 3 distinct values

Apr 4, 2017

See below.

#### Explanation:

Define $q \left(x\right) = a {\left(x + b\right)}^{2} \left(x + c\right)$ and consider

$p \left(x\right) = q \left(x\right)$ or

$p \left(x\right) = 2 {x}^{3} + 7 {x}^{2} + k x - k = a {\left(x + b\right)}^{2} \left(x + c\right)$

Grouping coefficients we get

$\left(2 - a\right) {x}^{3} + \left(7 - \left(2 a b + a c\right)\right) {x}^{2} + \left(k - \left(a {b}^{2} + 2 a b c\right)\right) x - \left(k + a {b}^{2} c\right) = 0$

Now solving for $a , b , c , k$

$\left\{\begin{matrix}a = 2 \\ 2 a b + a c = 7 \\ a {b}^{2} + 2 a b c = k \\ a {b}^{2} c = - k\end{matrix}\right.$

we obtain

$\left(\begin{matrix}a & b & c & k \\ 2 & - \frac{7}{4} & 7 & - \frac{343}{8} \\ 2 & 0 & \frac{7}{2} & 0 \\ 2 & 2 & - \frac{1}{2} & 4\end{matrix}\right)$