How do you multiply #\frac { b ^ { 2} - 4b } { b ^ { 2} + 9b + 20} \cdot ( b + 5) #?

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Answer 1 · 2017-06-12T01:46:00

#(b(b-4))/(b+4)#

1) Set the #b+5# on top of the fraction since it's really in the numerator anyways

#((b^2-4b)(b+5))/(b^2+9b+20)#

2) Factor out that #b^2-4b# to get #b(b-4)#

#(b(b-4)(b+5))/(b^2+9b+20)#

3) Factor the #b^2+9b+20# on the bottom to get (b+4)(b+5)

#((b-4)(b+5))/((b+4)(b+5))#

4) Cancel common factor of b+5

#(b(b-4))/(b+4)#

It can be tempting to want to multiply the second or third step out in the numerator in problems like this.

Take your time looking at the problem and seeing if you can factor out anything that might be eliminated later on to help "simplify" a little more.