How do you factor the expression #15t^2 - 17t - 4#?

1 Answer
Mar 25, 2017

#15t^2-17t-4 = (5t+1)(3t-4)#

Explanation:

Given:

#15t^2-17t-4#

Use an AC method:

Find a pair of factors of #AC=15*4 = 60# which differ by #B=17#.

The pair #20, 3# works.

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

#15t^2-17t-4 = 15t^2-20t+3t-4#

#color(white)(15t^2-17t-4) = (15t^2-20t)+(3t-4)#

#color(white)(15t^2-17t-4) = 5t(3t-4)+1(3t-4)#

#color(white)(15t^2-17t-4) = (5t+1)(3t-4)#

#color(white)()#
Footnote

In the above example, we looked for a pair of factors that differed by #B=17#, since the coefficient of the constant term was negative.

If we were attempting to factor:

#15t^2-17t+4#

then we would instead look for a pair of factors of #AC=60# with sum #B=17#.

The pair #12, 5# works and hence we find:

#15t^2-17t+4 = (15t^2-12t)-(5t-4)#

#color(white)(15t^2-17t+4) = 3t(5t-4)-1(5t-4)#

#color(white)(15t^2-17t+4) = (3t-1)(5t-4)#