## Introductory Algebra for College Students (7th Edition)

$$(2x - 3)(2x + 5)$$
We want to use the factorization method by grouping. First, we want to multiply the leading coefficient $a$ with the constant $c$: $$(4)(-15) = -60$$ Now, we find the factors of $ac$ that sum up to $b$. Here are the factors: $-60$ and $1$ or $60$ and $-1$ $-15$ and $4$ or $15$ and $-4$ $-20$ and $3$ or $20$ and $-3$ $-10$ and $6$ or $10$ and $-6$ $-30$ and $2$ or $30$ and $-2$ $-12$ and $5$ or $12$ and $-5$ The factors that sum up to $b$ would be $10$ and $-6$. We can now rewrite the middle term as the difference of the two factors: $$4x^2 + (10x - 6x) - 15$$ We can now group them together: $$(4x^2 + 10x) + (-6x - 15)$$ We factor out the common factor for each binomial group: $$2x(2x + 5) - 3(2x + 5)$$ Now we factor by grouping: $$(2x - 3)(2x + 5)$$