Factoring reverses multiplication. For a monic trinomial \(x^2+bx+c\), seek two integers whose sum is \(b\) and product is \(c\). When \(a\ne1\), the AC method creates a grouping structure without guessing binomial coefficients.
Case A: monic trinomials
Both conditions must hold. A pair with the correct product but wrong sum does not factor the trinomial.
| Product | Required sum | Signs of pair | Illustration |
|---|---|---|---|
| \(c>0\) | \(b>0\) | Both positive | \(x^2+9x+20=(x+4)(x+5)\) |
| \(c>0\) | \(b<0\) | Both negative | \(x^2-9x+20=(x-4)(x-5)\) |
| \(c<0\) | Either sign | One positive, one negative | \(x^2+x-20=(x+5)(x-4)\) |
Use both sum and product
Factor \(x^2-3x-28\).
- Sign pattern
The product is negative, so the pair has opposite signs.
- Find the pair
\(4(-7)=-28\) and \(4+(-7)=-3\).
- Write and verify
\((x+4)(x-7)=x^2-3x-28\).
Case B: the AC method
For \(ax^2+bx+c\) with \(a\ne\pm1\), find integers \(m,n\) satisfying \(m+n=b\) and \(mn=ac\). Replace \(bx\) with \(mx+nx\), then factor by grouping.
Factor with the AC method
A seven-step algebra flow from a trinomial through middle-term splitting and grouping to verified binomial factors.
- Start\[6x^2+13x+6\]
Identify all three coefficients.
- Compute the product\[ac=6(6)=36\]
Use the leading coefficient and constant.
- Find the pair\[9+4=13,\quad9\cdot4=36\]
The pair must satisfy both conditions.
- Split the middle term\[6x^2+9x+4x+6\]
Replace the original middle term without changing value.
- Group\[(6x^2+9x)+(4x+6)\]
Make two factorable pairs.
- Factor each pair\[3x(2x+3)+2(2x+3)\]
The same binomial must appear.
- Factor and verify\[(3x+2)(2x+3)=6x^2+13x+6\]
Expansion reproduces the original trinomial.
Factor a nonmonic trinomial
Factor \(8x^2-14x-15\).
- Compute
\(ac=8(-15)=-120\).
- Find the pair
\(-20+6=-14\) and \((-20)(6)=-120\).
- Split and group
\(8x^2-20x+6x-15=4x(2x-5)+3(2x-5)\).
- Factor and check
\((4x+3)(2x-5)=8x^2-14x-15\).
Factor a GCF before the trinomial
Not every trinomial factors over integers
If no integer pair satisfies both the sum and product conditions, report that the trinomial is not factorable over integers. Do not manufacture a pair from only one condition.
Common mistakes
- For \(a\ne1\), use the product \(ac\), not just \(c\).
- A correct product is insufficient unless the pair also sums to \(b\).
- After splitting the middle term, grouping signs must produce the same binomial.
- Factor the greatest common factor before applying the trinomial method.
- Verify the final factors by expansion.
Choose an AC pair
Which pair splits the middle term of \(10x^2+17x+3\)?
- \(15,2\)
- \(10,3\)
- \(30,1\)
- \(-15,-2\)
Show answer and explanation
Answer: \(15,2\)
The pair must multiply to \(ac=30\) and add to \(17\).
Key takeaways
What to remember
- For \(x^2+bx+c\), find a pair with sum \(b\) and product \(c\).
- For \(ax^2+bx+c\), split the middle term using a pair with product \(ac\).
- Extract a GCF first and factor until no supported pattern remains.
- Expansion is the decisive equivalence check.
Put these notes into practice
Apply the ideas with SAT-style questions, then reinforce key details with flashcards.