How do you rewrite 3(x-5)+3x(x-5)3(x−5)+3x(x−5) as an equivalent product of two binomials?

1 Answer
Jim G.
May 28, 2017

3(x-5)(1+x)3(x−5)(1+x)

Explanation:

color(red)(3)(x-5)color(red)(+3x)(x-5)3(x−5)+3x(x−5)

"take out the "color(blue)"common factor"" of " (x-5)take out the common factor of (x−5)

rArr(x-5)(color(red)(3+3x))⇒(x−5)(3+3x)

=(x-5)3(1+x)larr" common factor of 3 in " (3+3x)=(x−5)3(1+x)← common factor of 3 in (3+3x)

=3(x-5)(1+x)=3(x−5)(1+x)

"we can check the 2 for equivalence"we can check the 2 for equivalence

3(x-5)+3x(x-5)3(x−5)+3x(x−5)

=3x-15+3x^2-15x=3x−15+3x2−15x

=3x^2-12x-15larrcolor(blue)" first expansion"=3x2−12x−15← first expansion

"and " 3(x-5)(1+x)larr" expand using FOIL"and 3(x−5)(1+x)← expand using FOIL

=3(x+x^2-5-5x)=3(x+x2−5−5x)

=3(x^2-4x-5)=3(x2−4x−5)

=3x^2-12x-15larrcolor(blue)" second expansion"=3x2−12x−15← second expansion

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