What is the factor? 2z^2+11z+6

Apr 16, 2018

$2 {z}^{2} + 11 z + 6 = 2 \left(z - \frac{- 11 + \sqrt{73}}{4}\right) \left(z - \frac{- 11 - \sqrt{73}}{4}\right)$

Explanation:

If we find a value $r$ such that $2 {r}^{2} + 11 r + 6 = 0$, then we can say that the binomial $z - r$ is a factor of $2 {z}^{2} + 11 z + 6$. In this case it turns out that there are two values of $r$ that meet this criteria, however neither of them are rational. I am going to assume that you do not know the quadratic formula and try to explain how to arrive at the two values of $r$ by completing the square.

We want to find out when

$2 {z}^{2} + 11 z + 6 = 0$.

Divide both sides of this equation by 2.

${z}^{2} + \frac{11}{2} z + 3 = 0$

Add 121/16 to both sides of this equation.

${z}^{2} + \frac{11}{2} z + \frac{121}{16} + 3 = \frac{121}{16}$

Factor the first three terms of the left-hand side of this equation.

${\left(z + \frac{11}{4}\right)}^{2} + 3 = \frac{121}{16}$

Subtract three from both sides of this equation.

${\left(z + \frac{11}{4}\right)}^{2} = \frac{121}{16} - 3$

Simplify.

${\left(z + \frac{11}{4}\right)}^{2} = \frac{73}{16}$

Take the square root of both sides of this equation.

$z + \frac{11}{4} = \pm \sqrt{\frac{73}{16}}$

Subtract $\frac{11}{4}$ from both sides of this equation.

$z = - \frac{11}{4} \pm \sqrt{\frac{73}{16}}$

Simplify

$z = \frac{- 11 \pm \sqrt{73}}{4}$

This means that $\left(z - \frac{- 11 + \sqrt{73}}{4}\right)$ and $\left(z - \frac{- 11 - \sqrt{73}}{4}\right)$ are both factors of $2 {z}^{2} + 11 z + 6$. Let's evaluate

$\left(z - \frac{- 11 + \sqrt{73}}{4}\right) \left(z - \frac{- 11 - \sqrt{73}}{4}\right)$

${z}^{2} - \frac{- 11 + \sqrt{73}}{4} z - \frac{- 11 - \sqrt{73}}{4} z + \frac{\left(- 11 + \sqrt{73}\right) \left(- 11 - \sqrt{73}\right)}{16}$

${z}^{2} + \frac{11}{2} z + \frac{121 - 73}{16}$

${z}^{2} + \frac{11}{2} z + 3$

Note that this is exactly 1/2 of our original expression. This means that the remaining factor is 2, and the fully-factored form is

$2 \left(z - \frac{- 11 + \sqrt{73}}{4}\right) \left(z - \frac{- 11 - \sqrt{73}}{4}\right)$.