How do you factor #7n^2-13n-2#?

2 Answers
Jul 6, 2015

Answer:

#7(n - 2)(n + 1/7)#

Explanation:

#7n^2 - 13n - 2 = 0 \Rightarrow n = (13 ± sqrt{225})/14#

#n_1 = (13 + 15)/14 = 2#

#n_2 = (13 - 15)/14 = -1/7#

#ax^2 + bx + c = a(x - x_1)(x - x_2)#

Jul 6, 2015

Answer:

Alternatively, use a variant of the AC Method to find:

#7n^2-13n-2 = (7n+1)(n-2)#

Explanation:

Given #7n^2-13n-2#, let #A=7#, #B=13# and #C=2#.

Notice that the sign of the constant term is negative, so look for a pair of factors of #AC = 7*2 = 14#, whose difference is #B=13#.

It should not take long to spot that the pair #B1=1, B2=14# works.

Next, for each of the pairs #(A, B1)# and #(A, B2)# divide by the HCF (highest common factor) to find a pair of coefficients of a factor, choosing signs as appropriate:

#(7, 1) -> (7, 1) -> (7n+1)#
#(7, 14) -> (1, 2) -> (n-2)#

We choose #+# for the pair #(A, B1)# and #-# for the pair #(A, B2)# to get #(+B1)+(-B2) = -13# - the coefficient of the middle term.