How do you factor the expression #49p^2 + 63pq - 36q^2#?

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Answer 1 · 2017-01-23T00:09:19

#49p^2+63pq-36q^2 = (7p+12q)(7p-3q)#

Note that #49p^2 = (7p)^2# and #36q^2 = (6q)^2#

Further note that #63 = 3^2*7# is divisible by #7# and by #3# but not by #6#.

So let's look at this quadratic in terms of #7p# and #3q#...

#49p^2+63pq-36q^2 = (7p)^2+3(7p)(3q)-4(3q)^2#

Note that #4-1=3# and #4*1=4#, so we find:

#(7p)^2+3(7p)(3q)-4(3q)^2 = ((7p)+4(3q))((7p)-(3q))#

#color(white)((7p)^2+3(7p)(3q)-4(3q)^2) = (7p+12q)(7p-3q)#

Answer 2 · 2017-01-23T00:58:46

#-36(q+7/12p)(q-7/3p)#

Making #q=lambda p# and substituting

#49p^2 + 63 pq - 36q^2=(49 + 63 lambda - 36 lambda^2)p^2#

but #(49 + 63 lambda - 36 lambda^2)=-36(lambda+7/12)(lambda-7/3)#

so

#-36(lambda+7/12)(lambda-7/3)p^2=-36(lambdap+7/12p)(lambdap-7/3p) = -36(q+7/12p)(q-7/3p)#