# Solve the following? (x−1)(x−3)(x−5)(x+1)=−12

Apr 22, 2018

x=−0.645751,0.267949,3.732051,4.645751

#### Explanation:

$\left(x - 1\right) \left(x - 3\right) \left(x - 5\right) \left(x + 1\right) = - 12$

first, we can expand each of the brackets using FOIL

$\left(x - 1\right) \left(x - 3\right) = {x}^{2} - 3 x - x + 3$

$\left(x - 5\right) \left(x + 1\right) = {x}^{2} + x - 5 x - 5$

Now we can simplify these and combine them back together into the original expression.

$\left({x}^{2} - 4 x + 3\right) \times \left({x}^{2} - 4 x - 5\right) = - 12$

Now we can factorise this.

$\left(\textcolor{\lim e}{{x}^{2} - 4 x + 3}\right) \left(\textcolor{s k y b l u e}{{x}^{2} - 4 x - 5}\right) = \left(\textcolor{\lim e}{{x}^{2}}\right) \left(\textcolor{s k y b l u e}{{x}^{2}}\right) + \left(\textcolor{\lim e}{{x}^{2}}\right) \left(\textcolor{s k y b l u e}{- 4 x}\right) + \left(\textcolor{\lim e}{{x}^{2}}\right) \left(\textcolor{s k y b l u e}{- 5}\right) + \left(\textcolor{\lim e}{- 4 x}\right) \left(\textcolor{s k y b l u e}{{x}^{2}}\right) + \left(\textcolor{\lim e}{- 4 x}\right) \left(\textcolor{s k y b l u e}{- 4 x}\right) + \left(\textcolor{\lim e}{- 4 x}\right) \left(\textcolor{s k y b l u e}{- 5}\right) + \left(\textcolor{\lim e}{3}\right) \left(\textcolor{s k y b l u e}{{x}^{2}}\right) + \left(\textcolor{\lim e}{3}\right) \left(\textcolor{s k y b l u e}{- 4 x}\right) + \left(\textcolor{\lim e}{3}\right) \left(\textcolor{s k y b l u e}{- 5}\right)$

$= \left({x}^{4}\right) + \left(- 4 {x}^{3}\right) + \left(- 5 {x}^{2}\right) + \left(- 4 {x}^{3}\right) + \left(16 {x}^{2}\right) + \left(20 x\right) + \left(3 {x}^{2}\right) + \left(- 12 x\right) + \left(- 15\right)$

Now we can collect like terms and simplify.

$= {x}^{4} - 4 {x}^{3} - 4 {x}^{3} - 5 {x}^{2} + 3 {x}^{2} + 16 {x}^{2} + 20 x - 12 x - 15$

${x}^{4} - 8 {x}^{3} + 14 {x}^{2} + 8 x - 15 = - 12$

Now we can find $x$

${x}^{4} - 8 {x}^{3} + 14 {x}^{2} + 8 x \textcolor{red}{\cancel{\textcolor{b l a c k}{- 15} + 15}} = - 12 \textcolor{red}{+ 15}$

${x}^{4} - 8 {x}^{3} + 14 {x}^{2} + 8 x = 3$

The only way that I could find to solve this is through using the Quartic formula.

color(blue)(x=−0.645751,0.267949,3.732051,4.645751)

Apr 27, 2018

$\pm \sqrt{3} + 2$ and $\pm \sqrt{7} + 2$

#### Explanation:

We have:

$\left(x - 1\right) \left(x - 3\right) \left(x - 5\right) \left(x + 1\right) = - 12$

Multiply the first two and the last two parentheses together.

$\implies \left({x}^{2} - 4 x + 3\right) \left({x}^{2} - 4 x - 5\right) = - 12$

Hmm... We see that the first quadratic expression is 8 larger than the second.

We let $s = \left({x}^{2} - 4 x + 3\right)$

We now have:

$s \left(s - 8\right) = - 12$

$\implies {s}^{2} - 8 s = - 12$ complete the square

$\implies \left({s}^{2} - 8 s + 16\right) - 16 = - 12$

$\implies {\left(s - 4\right)}^{2} = 4$

$\implies - 2 = s - 4 = 2$

$\implies 2 = s = 6$

We can use this to solve for $x$.

When $s = 2$...

${x}^{2} - 4 x + 3 = 2$

$\implies {x}^{2} - 4 x + 1 = 0$

$\implies \left({x}^{2} - 4 x + 4\right) - 4 + 1 = 0$

$\implies {\left(x - 2\right)}^{2} - 3 = 0$

$\implies {\left(x - 2\right)}^{2} = 3$

$\implies x - 2 = \pm \sqrt{3}$

$\implies x = \pm \sqrt{3} + 2$

When $s = 6$...

${x}^{2} - 4 x + 3 = 6$

$\implies {x}^{2} - 4 x - 3 = 0$

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

$\implies {\left(x - 2\right)}^{2} - 7 = 0$

$\implies {\left(x - 2\right)}^{2} = 7$

$\implies x - 2 = \pm \sqrt{7}$

$\implies x = \pm \sqrt{7} + 2$

The answers are $\pm \sqrt{3} + 2$ and $\pm \sqrt{7} + 2$