# How do you factor x^3-7x^(3/2)-8=0?

Apr 10, 2015

The key to factoring this is to notice (guess and verify) that the square of ${x}^{\frac{3}{2}}$ is ${x}^{3}$

So I can think of this as:

$\left[\text{something"]^2 - 7 ["something}\right] - 8 = 0$

Until you are more comfortable with the process, do the substitution: the "something" here is ${x}^{\frac{3}{2}}$ so we'll use a different variable to rename ${x}^{\frac{3}{2}}$. The traditional variable in this situation is $u$.

Let $u = {x}^{\frac{3}{2}}$. That makes ${u}^{2} = {\left({x}^{\frac{3}{2}}\right)}^{2} = {x}^{3}$

So we have:
${u}^{2} - 7 u - 8 = 0$

$\left(u - 8\right) \left(u + 1\right) = 0$

$u - 8 = 3$ or $u + 1 = 0$, so

$u = 8$ or $u = - 1$

Now go back to the original variable:

${x}^{\frac{3}{2}} = 8$ or ${x}^{\frac{3}{2}} = - 1$

$x = {8}^{\frac{2}{3}}$ or $x = {\left(- 1\right)}^{\frac{2}{3}}$

$x = {\left(\sqrt[3]{8}\right)}^{2} = 4$ or $x = {\left(\sqrt[3]{- 1}\right)}^{2} = - 1$