# How do you factor 10x^2-18x-4?

Mar 27, 2015

You factor a polynomial by writing it as a multiplication of lower-degree polynomials, and going on as long as this is possible.

The quadratic case is quite easy, as it only presents three possibilities: if the discriminant is (striclty) positive you can write $p \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$, where $p \left(x\right)$ is the original quadratic polynomial. If the discriminant equals zero, you have $p \left(x\right) = {\left(x - {x}_{0}\right)}^{2}$. If the discriminant is (strictly) negative, there is no factorization using real numbers.

So, given a quadratic polynomial $p \left(x\right) = a {x}^{+} b x + c$, the discriminant $\setminus \Delta$ is defined as
$\setminus \Delta = {b}^{2} - 4 a c$
In your case, $a = 10$, $b = - 18$, and $c = - 4$. Plugging these values into the formula, we obtain
$\setminus \Delta = {\left(- 18\right)}^{2} - 4 \cdot 10 \cdot \left(- 4\right) = 324 + 160 = 484$
We are in the first case, and the solutions ${x}_{1 , 2}$ are given by the formula
${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{\setminus \Delta}}{2 a}$
And plugging again the values into the formula gives the solutions ${x}_{1} = - \frac{1}{5}$ and ${x}_{2} = 2$. Recalling that we have $p \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$, we factor 10x^2−18x−4 as $\left(x + \frac{1}{5}\right) \left(x - 2\right)$