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

1 Answer
Jun 5, 2017

Answer:

#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)#