How do you factor the trinomial #56r^2+121r+63#?

3 Answers
George C.
Dec 17, 2015

Use the quadratic formula to find zeros, hence factors:

#56r^2+121r+63 = (7r+9)(8r+7)#

Explanation:

#f(r) = 56r^2+121r+63# is of the form #ar^2+br+c# with #a=56#, #b=121# and #c=63#.

This has zeros for values of #r# given by the quadratic formula:

#r = (-b+-sqrt(b^2-4ac))/(2a)#

#=(-121+-sqrt(121^2-(4xx56xx63)))/(2xx56)#

#=(-121+-sqrt(14641-14112))/112#

#=(-121+-sqrt(529))/112#

#=(-121+-23)/112#

That is #r = -144/112 = -9/7# or #r = -98/112 = -7/8#

Hence:

#56r^2+121r+63 = (7r+9)(8r+7)#

George C.
Dec 17, 2015

Alternatively use prime factorisation and reasoning to find:

#56r^2+121r+63 = (7r+9)(8r+7)#

Explanation:

Alternatively, reason your way to the integer factorisation as follows:

#56r^2+121r+63#

The coefficients have the following prime factorisations:

#56 = 2*2*2*7#

#121= 11*11#

#63 = 3*3*7#

Notice that both the leading and trailing terms are divisible by #7# but the middle term is not. That means that if there is an exact factorisation with integer coefficients then it can take the form:

#56r^2+121r+63 = (7ar+b)(cr+7d)#

#= 7acr^2+(49ad+bc)r+7bd#

for some integers #a#, #b#, #c# and #d#, where we can take #a > 0# without loss of generality.

Notice also that the middle term is odd, so one of #a# and #c# is odd. Hence either #a=8# and #c=1# or #a=1# and #c=8#.

Similarly, notice that the middle term is not divisible by #3#, hence either #b=9# and #d=1# or #b=1# and #d=9#. #b# and #d# cannot be negative since the constant term is positive and the middle term is also positive.

This gives us #4# possibilities to try to match the middle term:

#49ad+bc = 121#

with #(a, b, c, d) = (8, 1, 9, 1), (1, 8, 9, 1), (8, 1, 1, 9) or (1, 8, 1, 9)#

We quickly find #a=1#, #c=8#, #d=1# and #b=9#

Hence:

#56r^2+121r+63 = (7r+9)(8r+7)#

George C.
Dec 17, 2015

Alternatively use an AC Method to find:

#56r^2+121r+63=(7r+9)(8r+7)#

Explanation:

Given #56r^2+121r+63# look for a pair of factors of #AC = 56*63 = 3528# whose sum is #B=121#.

First note that #56+63 = 119# is very close and to get a greater sum, the pair of factors needs to be slightly further apart.

The prime factorisation of #3528# is:

#3528 = 2*2*2*3*3*7*7#

Also note that #121=11*11# is not divisible by #2#, #3# or #7#, so all of the #2#'s must be in one of the factors, all of the #3#'s in one of the factors and all of the #7#'s in one of the factors.

We quickly find #49xx72 = 3528# with #49+72=121#

Hence we can factor by grouping:

#56r^2+121r+63#

#=(56r^2+49r)+(72r+63)#

#=7r(8r+7)+9(8r+7)#

#=(7r+9)(8r+7)#

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