What are the roots of (x+1)(2x+3)(2x+5)(x+3) = 945 ?

Sep 24, 2016

The roots are:

$x = 2$, $x = - 6$, $x = - 2 \pm \frac{\sqrt{59}}{2} i$

Explanation:

Given:

$\left(x + 1\right) \left(2 x + 3\right) \left(2 x + 5\right) \left(x + 3\right) = 945$

Notice that:

$\left(x + 1\right)$ and $\left(x + 3\right)$ differ by $2$

$\left(2 x + 3\right)$ and $\left(2 x + 5\right)$ differ by $2$

What are the factors of $945$?

The prime factorisation is:

$945 = {3}^{3} \cdot 5 \cdot 7$

So we have:

$945 = 3 \cdot 5 \cdot 7 \cdot 9 = \left(\textcolor{red}{2} + 1\right) \cdot \left(\textcolor{red}{2} + 3\right) \cdot \left(2 \cdot \textcolor{red}{2} + 3\right) \cdot \left(2 \cdot \textcolor{red}{2} + 5\right)$

So one solution is $x = 2$

Let $t = x + 2$

Then:

$0 = \left(x + 1\right) \left(2 x + 3\right) \left(2 x + 5\right) \left(x + 3\right) - 945$

$\textcolor{w h i t e}{0} = \left(t - 1\right) \left(2 t - 1\right) \left(2 t + 1\right) \left(t + 1\right) - 945$

$\textcolor{w h i t e}{0} = \left({t}^{2} - 1\right) \left(4 {t}^{2} - 1\right) - 945$

$\textcolor{w h i t e}{0} = 4 {t}^{4} - 5 {t}^{2} - 944$

We know that $t = 2 + 2 = 4$ is a root, so $t = - 4$ is also a zero $\left(t - 4\right) \left(t + 4\right) = {t}^{2} - 16$ is a factor:

$0 = 4 {t}^{4} - 5 {t}^{2} - 944 = \left({t}^{2} - 16\right) \left(4 {t}^{2} + 59\right)$

Hence the other two roots are $t = \pm \frac{\sqrt{59}}{2} i$

So the roots of our transformed quartic are:

$t = \pm 4$ and $t = \pm \frac{\sqrt{59}}{2} i$

Then $x = t - 2$ so the roots of the original equation are:

$x = 2$, $x = - 6$, $x = - 2 \pm \frac{\sqrt{59}}{2} i$