Question

How do you factor `15v^2-19v-10`?

Answer 1

`15v^2-19v-10 = (5v+2)(3v-5)`

Explanation:

Given:

`15v^2-19v-10`

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac`

where `a=15`, `b=-19` and `c=-10`

That is:

`Delta = (color(blue)(-19))^2-4(color(blue)(15))(color(blue)(-10))`

`color(white)(Delta) = 361+600`

`color(white)(Delta) = 961`

`color(white)(Delta) = 31^2`

Since `Delta > 0` is a perfect square, the given quadratic will factor perfectly using integers.

Let's use an AC method.

Look for a pair of factors of `AC=15*10 = 150` which differ by `B=19`. (We look for a pair of factors with this as a difference rather than a sum, since the sign of the constant term is negative.)

The pair `25, 6` works in that `25*6 = 150` and `25-6 = 19`

Use this pair to split the middle term and factor by grouping:

`15v^2-19v-10 = (15v^2-25v)+(6v-10)`

`color(white)(15v^2-19v-10) = 5v(3v-5)+2(3v-5)`

`color(white)(15v^2-19v-10) = (5v+2)(3v-5)`