How do you factor #21r^2+32r-13#?

1 Answer
George C.
Jun 13, 2015

#21r^2+32r-13#

#= 21r^2-7r+39r-13#

#= (7r+13)(3r-1)#

Explanation:

The trick here is to find the split of the middle term into two numbers whose difference is #32# and whose product is #21*13 = 3*7*13 = 273#. There are not many choices of pairs of factors of #3*7*13# and we can quickly find #39# and #7#.

Then we can factor by grouping:

#21r^2+32r-13#

#= 21r^2-7r+39r-13#

#= (21r^2-7r)+(39r-13)#

#= 7r(3r-1)+13(3r-1)#

#= (7r+13)(3r-1)#

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