How do you determine the binomial factors of #x^3-x^2-49x+49#?

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Answer 1 · 2016-12-04T08:38:57

The binomial factors of #x^3-x^2-49x+49# are:

#(x^2-49)#, #(x-1)#, #(x-7)#, #(x+7)#

The given cubic factors by grouping:

#x^3-x^2-49x+49 = (x^3-x^2)-(49x-49)#

#color(white)(x^3-x^2-49x+49) = x^2(x-1)-49(x-1)#

#color(white)(x^3-x^2-49x+49) = (x^2-49)(x-1)#

Note that both #color(blue)((x^2-49))# and #color(blue)((x-1))# are binomial factors.

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Hence we find:

#x^2-49 = x^2-7^2 = (x-7)(x+7)#

So #color(blue)((x-7))# and #color(blue)((x+7))# are also binomial factors.

Multiplying either of these by #(x-1)# will result in a trinomial, so there are no other binomial factors:

#(x-7)(x-1) = x^2-8x+7#

#(x+7)(x-1) = x^2+6x-7#

The complete list of polynomial factors of #x^3-x^2-49x+49# in descending degree is:

#x^3-x^2-49x+49#

#x^2-49#

#x^2-8x+7#

#x^2+6-7#

#x-7#

#x+7#

#x-1#

#1#