Question

Factor completely: 10ax^2-23ax-5a

Answer 1

Factoring the expression,10ax^2-23ax-5a, completely:

First, we find the GCF of all the terms in order to factor out a monomial. In this expression, 'a' is present in all of the terms, so we can pull it out.

a(10x^2 - 23x - 5)

From here, we have to factor by grouping. We multiply the quadratic term's coefficient by the constant term and find what factors of that add up to the linear term's coefficient.

10(-5) = -50
Factors of -50 that can add up to a negative (so, the largest factor is negative):
1 * -50 => -50 + 1 = -49
2 * -25 => -25 + 2 = -23

Since 2 and -25 satisfy our condition, we can rewrite the expression replacing our linear term's coefficient with this.

a(10x^2 + (-25 + 2)x - 5) -25+2 = -23, this is still the same expression!
a(10x^2 - 25x + 2x - 5) distributing the x
a((10x^2 - 25x) + (2x - 5)) we are focusing on those pairs for our grouping
a(5x(2x - 5) + 1(2x - 5)) we find the GCF of each pair. this should give us one common binomial between the two to factor out!
a(2x - 5)(5x + 1)

Then we are done! This is completely factored!
We can then check to make sure by multiplying through, which will help us see the symmetry between our actions as well.

a(2x - 5)(5x + 1) = a(5x + 1)(2x - 5) rewriting
= a(5x(2x - 5) + 1(2x - 5)) distribute the binomial to each term
= a(10x^2 - 25x + 2x - 5) distribute again
= a(10x^2 - 23x - 5) simplify
= 10ax^2 - 23ax - 5a distribute once more


In case you are ignoring the long explanation, the answer is " a(2x - 5)(5x + 1)". I just hope that you are able to understand why. :)