How do you factor #5x^2- 35x#?

3 Answers
Mar 30, 2018

#5x(x-7)#

Explanation:

take out common factors

#(1) " take " hcf(5,35) =color(red)(5)" out"#

#5x^2-35x= color(red)(5)(x^2-7x)#

#(2)" take "hcf(x^2,x)=color(blue)(x )" out"#

#5x^2-35x= color(red)(5)(x^2-7x)=color(red)(5)color(blue)(x)(x-7)#

Mar 30, 2018

#5x(x-7)#

Explanation:

We got:

#5x^2-35x#

Take the greatest common factor out, which is #5#.

#=5(x^2-7x)#

Notice how #x# still remains in both factors inside, and so we can take it out, and we get:

#=5x(x-7)#

Mar 30, 2018

#5x(x-7)#

Explanation:

Both terms are divisible by #5#, and #x#, so you factor those out of the polynomial. As for how this affects any solutions of this, because both factors equal #0# (or whatever is on the other side of the equation, but I'll use #0#), #5x=0#, so #x=0#, and #x-7=0#, so #x=7#