To factor the given expression [tex]\(2x^2 + 4xy - 30y\)[/tex] completely, we can follow these steps:
1. Identify common factors:
First, look for common factors in all terms of the expression. In this case, each term in the polynomial [tex]\(2x^2 + 4xy - 30y\)[/tex] has a common factor of 2.
[tex]\[
2x^2 + 4xy - 30y = 2(x^2 + 2xy - 15y)
\][/tex]
2. Factor the quadratic trinomial:
Now we need to factor the quadratic expression [tex]\(x^2 + 2xy - 15y\)[/tex]. To do this, we look for two numbers that multiply to give the constant term [tex]\(-15y\)[/tex] and add up to the coefficient of the linear term (2y).
Let's consider the quadratic trinomial in the form [tex]\(x^2 + bx + c\)[/tex], where [tex]\(b = 2y\)[/tex] and [tex]\(c = -15y\)[/tex]. We need to find two numbers whose product is [tex]\(c = -15y\)[/tex] and whose sum is [tex]\(b = 2y\)[/tex].
3. Find the factors:
The two numbers that multiply to [tex]\(-15y\)[/tex] and add up to [tex]\(2y\)[/tex] are [tex]\(5y\)[/tex] and [tex]\(-3y\)[/tex]:
[tex]\[
(5y) \cdot (-3y) = -15y^2
\][/tex]
[tex]\[
(5y) + (-3y) = 2y
\][/tex]
4. Rewrite the middle term:
Using the numbers [tex]\(5y\)[/tex] and [tex]\(-3y\)[/tex], we can split the middle term [tex]\(2xy\)[/tex]:
[tex]\[
x^2 + 2xy - 15y = x^2 + 5xy - 3xy - 15y
\][/tex]
5. Factor by grouping:
Factor by grouping terms:
[tex]\[
x^2 + 5xy - 3xy - 15y = x(x + 5y) - 3y(x + 5y)
\][/tex]
6. Factor out the common binomial:
Notice that both groups contain the common factor [tex]\((x + 5y)\)[/tex]:
[tex]\[
x(x + 5y) - 3y(x + 5y) = (x + 5y)(x - 3y)
\][/tex]
7. Combine the factors:
Substituting this back into our original expression, we get:
[tex]\[
2(x^2 + 2xy - 15y) = 2(x + 5y)(x - 3y)
\][/tex]
Thus, the completely factorized form of [tex]\(2x^2 + 4xy - 30y\)[/tex] is:
[tex]\[
2(x^2 + 2xy - 15y) = 2(x + 5y)(x - 3y)
\][/tex]
So, the complete factorization of the expression [tex]\(2x^2 + 4xy - 30y\)[/tex] is:
[tex]\[
2(x + 5y)(x - 3y)
\][/tex]
