Question

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

Answer 1

`3(x-5)(1+x)`

Explanation:

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

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

`rArr(x-5)(color(red)(3+3x))`

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

`=3(x-5)(1+x)`

`"we can check the 2 for equivalence"`

`3(x-5)+3x(x-5)`

`=3x-15+3x^2-15x`

`=3x^2-12x-15larrcolor(blue)" first expansion"`

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

`=3(x+x^2-5-5x)`

`=3(x^2-4x-5)`

`=3x^2-12x-15larrcolor(blue)" second expansion"`