How to factor a trinomial with leading coefficient 2

Key takeaway: Trinomials with a leading coefficient greater than 1 usually need the split-the-middle-term method. Once you find the pair that multiplies to $ac$ and adds to $b$, the rest is routine.

We want to factor

$2b^2+3b-9$

Step 1: Multiply $a\cdot c$

Here, $a=2$ and $c=-9$.

$ac=2(-9)=-18$

Step 2: Find two numbers

We need two numbers that:

• multiply to $-18$
• add to $3$

Those numbers are $6$ and $-3$.

Step 3: Split the middle term

$2b^2+6b-3b-9$

Step 4: Factor by grouping

$2b(b+3)-3(b+3)$

Step 5: Factor out the common binomial

$(2b-3)(b+3)$

So the factorization is:

$\boxed{(2b-3)(b+3)}$

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Pitfalls the pros know 👇 Do not try to guess factors from the constant term alone. The middle coefficient matters. A fast check at the end helps: $(2b-3)(b+3)=2b^2+6b-3b-9=2b^2+3b-9$.

What if the problem changes? If the trinomial were $2b^2-3b-9$, the pair would need to multiply to $-18$ and add to $-3$, which gives $3$ and $-6$. That would factor as $(2b+3)(b-3)$.

Tags: factor by grouping, split the middle term, trinomial factoring