Factor a quadratic when the leading coefficient is not equal to 1 and you can't factor by grouping?

Question

If one has a quadratic, for example $5x^2-10x-2$, which has real roots which via the quadratic equation are $(5\pm \sqrt{35})/5$, can you find its factored form. As I understand it $5x^2-10x-2\ne(x-\text{root1})(x-\text{root2})$? How can you turn this type of quadratic into a factored form?

Answer 1

Take the coefficient of the squared term (In this quadratic the coefficient is 5). Multiply this coefficient multiply it by $(x-root_1)$ and $(x-root_2)$ where $root_1$ and $root_2$ are the two factors of the quadratic polynomial (In this quadratic the two roots are$\frac{5+\sqrt{35}}{5}$ and $(5-\sqrt{35})/5$). The final solution being:

$5\left(x-\frac{5+\sqrt{35}}{5}\right)\left(x-\frac{5-\sqrt{35}}{5}\right)$

Answer 2

You can use the quadratic formula. The roots of $5x^2-10x-2=0$ are $\frac{10\pm \sqrt{100+40}}{10}=\frac{5 \pm \sqrt{35}}5$ as you found. The assertion that $5x^2-10x-2 \neq \left(x-\frac{5 + \sqrt{35}}5\right)\left(x-\frac{5 - \sqrt{35}}5\right)$ is correct because you need to supply a factor of $a$, which here is $5$ on the right side. Multiply it out and see.