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

1 Answer
George C.
Jun 5, 2017

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)

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