Factorise copletely
1) 2ax+3ay-2bx-3by
2)5x^2-20
3)3x^2+4x-15 is the ans to this one = 7x^3-15
first one is "grouping"
2ax+3ay-2bx-3by
= a(2x+3y) - b(2x+3y)
= (2x+3y)(a-b)
second, common factor then difference of squares
5x^2-20
= 5(x^2 - 4)
= 5(x+2)(x-2)
third, general quadratic
3x^2+4x-15
= (3x-5)(x+3)
looks like you got your answer by
adding 3x^2 + 4x to get 7x^3
that is just unforgivable!
2ax+3ay-2bx-3by factorize
Sure! I can help you factorize these expressions.
1) 2ax+3ay-2bx-3by
To factorize this expression, we can group the terms with similar variables together. Notice that we have the variables "ax" and "ay" in the first two terms, and "bx" and "by" in the last two terms.
Rearranging the terms, we get: (2ax - 2bx) + (3ay - 3by)
Now, we can factor out common factors from each group: 2x(a - b) + 3y(a - b)
Finally, we can factor out the common factor "(a - b)" which appears in both terms: (a - b)(2x + 3y)
So, the expression 2ax+3ay-2bx-3by can be completely factorized as (a - b)(2x + 3y).
2) 5x^2 - 20
To factorize this expression, we can first look for common factors. In this case, 5 is a common factor that can be factored out.
Rearranging the terms, we get: 5(x^2 - 4)
Now, we can further factorize the expression inside the parentheses. It is a difference of squares, which can be factored as: (x + 2)(x - 2)
So, the expression 5x^2 - 20 can be completely factorized as 5(x + 2)(x - 2).
3) 7x^3 - 15
It seems like you have already written the answer to this expression. If you meant to ask for the factorization of 7x^3 - 15, then the given answer "3x^2 + 4x - 15" does not appear to be correct.
To factorize 7x^3 - 15, we need to look for common factors, but in this case, there are no common factors that can be factored out.
So, the expression 7x^3 - 15 is already in its simplest form and cannot be further factorized.
You mean factor
a(2x+3y)-b(2x+3y)
= (a-b)(2x+3y)
5(x^2-4)
= 5 (x-2)(x+2)
(3x-5)(x+3)