Algebra Lessons
How to Factor Trinomials: Step by Step, With Practice
Factor any trinomial on a placement test: common factors, a sign table, the ac method, special patterns, and eight practice questions.
Factoring a trinomial means writing it as a product of two binomials, the reverse of multiplying them out. For example:
\[x^2 + 7x + 12 = (x + 3)(x + 4)\]
because \(3 \times 4 = 12\) and \(3 + 4 = 7\). That one idea, two numbers that multiply to the last term and add to the middle term, handles most trinomials you will see on placement tests. This lesson covers it step by step, then the harder case where \(x^2\) has a number in front, the special patterns, and eight practice questions.
Step 0: take out a common factor
Before anything else, check whether every term shares a factor. In \(3x^2 - 3x - 36\), all three terms are divisible by 3:
\[\begin{gathered}3x^2 - 3x - 36 = 3(x^2 - x - 12) \\ = 3(x - 4)(x + 3)\end{gathered}\]
Skipping this step makes the numbers bigger and the factoring harder. On a test, the answer choices usually show the common factor out front.
When the first term is \(x^2\)
For \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\). The signs of \(b\) and \(c\) tell you the signs of the numbers before you start searching:
| If \(c\) is | and \(b\) is | the two numbers are | Example |
|---|---|---|---|
| positive | positive | both positive | \(x^2 + 7x + 12 = (x + 3)(x + 4)\) |
| positive | negative | both negative | \(x^2 - 9x + 20 = (x - 4)(x - 5)\) |
| negative | either | opposite signs; the larger one has the sign of \(b\) | \(x^2 - 2x - 15 = (x - 5)(x + 3)\) |
Then list the factor pairs of \(|c|\) and pick the pair with the right sum. For \(x^2 - 2x - 15\): the pairs of 15 are 1 and 15, 3 and 5. Opposite signs that add to \(-2\) are \(-5\) and \(+3\).
When there is a number in front of \(x^2\): the ac method
For \(ax^2 + bx + c\) with \(a \ne 1\), the "multiply and add" search uses \(a \times c\) instead of \(c\). Take \(2x^2 + 7x + 3\):
- Multiply \(a \times c = 2 \times 3 = 6\).
- Find two numbers that multiply to 6 and add to 7: they are 6 and 1.
- Split the middle term with them: \(2x^2 + 6x + x + 3\).
- Group in pairs and factor each pair:
\[2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)\]
The two groups must leave the same binomial, here \((x + 3)\). If they don't, check your signs or try the numbers in the other order.
Two patterns worth recognizing
- Difference of squares: \(a^2 - b^2 = (a - b)(a + b)\). Example: \(x^2 - 49 = (x - 7)(x + 7)\). There is no middle term.
- Perfect square trinomial: \(a^2 + 2ab + b^2 = (a + b)^2\). Example: \(x^2 + 10x + 25 = (x + 5)^2\), because \(25 = 5^2\) and \(10 = 2 \times 5\).
A sum of squares like \(x^2 + 49\) does not factor using whole numbers. That fact shows up as a trick answer surprisingly often.
Always check by multiplying back
Multiply your two binomials using FOIL (first, outer, inner, last). If you get the original trinomial, you are done. Checking \((2x + 1)(x + 3)\): \(2x^2 + 6x + x + 3 = 2x^2 + 7x + 3\). It takes 15 seconds and catches nearly every sign mistake.
What if nothing works?
Some trinomials can't be factored with whole numbers, for example \(x^2 + 3x + 1\). If you have tried every factor pair and none adds to \(b\), the trinomial is prime. To solve an equation with it, use the quadratic formula instead; it is printed on the GED formula sheet, and you can practice it in our GED math practice test.
Practice: factoring trinomials
Factoring is part of the Advanced Algebra and Functions test on the ACCUPLACER and of algebraic reasoning on the TSIA2. These eight questions go from the easiest case to the hardest.
1.Factor: \(x^2 + 9x + 14\)
- A\((x + 1)(x + 14)\)
- B\((x - 2)(x - 7)\)
- C\((x + 2)(x + 7)\)
- D\((x + 3)(x + 6)\)
Show solution
Two numbers that multiply to 14 and add to 9: 2 and 7. Both positive, because \(c\) and \(b\) are positive.
\[x^2 + 9x + 14 = (x + 2)(x + 7)\]Common trap: Picking \((x + 3)(x + 6)\) because 3 + 6 = 9. The numbers must also multiply to 14, and \(3 \times 6 = 18\).
Answer: C (\((x + 2)(x + 7)\))
2.Factor: \(x^2 - 4x - 21\)
- A\((x - 7)(x + 3)\)
- B\((x + 7)(x - 3)\)
- C\((x - 7)(x - 3)\)
- D\((x - 21)(x + 1)\)
Show solution
\(c = -21\) is negative, so the numbers have opposite signs, and the larger one takes the sign of \(b = -4\): \(-7\) and \(+3\).
\[x^2 - 4x - 21 = (x - 7)(x + 3)\]Common trap: Putting the minus on the wrong number: \((x + 7)(x - 3)\) multiplies out to \(x^2 + 4x - 21\). Multiply your answer back to check the middle term.
Answer: A (\((x - 7)(x + 3)\))
3.Factor: \(x^2 - 11x + 24\)
- A\((x + 3)(x + 8)\)
- B\((x - 4)(x - 6)\)
- C\((x - 2)(x - 12)\)
- D\((x - 3)(x - 8)\)
Show solution
\(c = 24\) is positive and \(b = -11\) is negative, so both numbers are negative. Factor pairs of 24 that add to 11: 3 and 8.
\[x^2 - 11x + 24 = (x - 3)(x - 8)\]Common trap: Choosing \((x - 4)(x - 6)\): it multiplies to 24 but adds to \(-10\). List the factor pairs of 24 and check each sum.
Answer: D (\((x - 3)(x - 8)\))
4.Factor: \(3x^2 + 10x + 8\)
- A\((3x + 2)(x + 4)\)
- B\((3x + 4)(x + 2)\)
- C\((3x + 8)(x + 1)\)
- D\((3x + 1)(x + 8)\)
Show solution
Use the \(ac\) method: \(ac = 3 \times 8 = 24\). Two numbers that multiply to 24 and add to 10: 6 and 4. Split the middle term and group.
\[\begin{gathered}3x^2 + 6x + 4x + 8 \\ = 3x(x + 2) + 4(x + 2) \\ = (3x + 4)(x + 2)\end{gathered}\]Common trap: Choosing \((3x + 2)(x + 4)\). Its middle term is \(12x + 2x = 14x\), not \(10x\). With \(a \ne 1\), where each number sits changes the middle term.
Answer: B (\((3x + 4)(x + 2)\))
5.Factor: \(2x^2 - 5x - 12\)
- A\((2x - 3)(x + 4)\)
- B\((2x + 4)(x - 3)\)
- C\((2x + 3)(x - 4)\)
- D\((2x - 4)(x + 3)\)
Show solution
\(ac = 2 \times (-12) = -24\). Two numbers that multiply to \(-24\) and add to \(-5\): \(-8\) and \(3\).
\[\begin{gathered}2x^2 - 8x + 3x - 12 \\ = 2x(x - 4) + 3(x - 4) \\ = (2x + 3)(x - 4)\end{gathered}\]Common trap: Answering \((2x - 3)(x + 4)\), which gives \(+5x\) in the middle. Switching the signs flips the middle term, so always check it by multiplying back.
Answer: C (\((2x + 3)(x - 4)\))
6.Factor completely: \(5x^2 - 20\)
- A\(5(x - 2)(x + 2)\)
- B\(5(x - 2)^2\)
- C\((5x - 2)(x + 10)\)
- D\(5(x - 4)(x + 4)\)
Show solution
Take out the common factor 5 first, then use the difference of squares.
\[\begin{gathered}5x^2 - 20 = 5(x^2 - 4) \\ = 5(x - 2)(x + 2)\end{gathered}\]Common trap: Stopping at \(5(x^2 - 4)\), or writing \(5(x - 4)(x + 4)\). After the 5 comes out, what is left is \(x^2 - 4\), and \(4 = 2^2\).
Answer: A (\(5(x - 2)(x + 2)\))
7.Factor: \(x^2 - 14x + 49\)
- A\((x + 7)^2\)
- B\((x - 7)(x + 7)\)
- C\((x - 49)(x + 1)\)
- D\((x - 7)^2\)
Show solution
49 is \(7^2\) and the middle term is \(2 \times 7 = 14\), so this is a perfect square trinomial. The middle term is negative, so the sign inside is minus.
\[x^2 - 14x + 49 = (x - 7)^2\]Common trap: Choosing \((x - 7)(x + 7)\). That product is \(x^2 - 49\), with no middle term at all.
Answer: D (\((x - 7)^2\))
8.Factor completely: \(4x^2 + 8x - 32\)
- A\(4(x - 4)(x + 2)\)
- B\(4(x + 4)(x - 2)\)
- C\((4x + 8)(x - 4)\)
- D\(4(x + 8)(x - 1)\)
Show solution
Every term is divisible by 4. Take it out first, then factor the simpler trinomial.
\[4(x^2 + 2x - 8) = 4(x + 4)(x - 2)\]Common trap: Skipping the common factor and trying to factor \(4x^2 + 8x - 32\) directly. It can be done, but the numbers get large and errors creep in. Always look for a common factor first.
Answer: B (\(4(x + 4)(x - 2)\))
Teacher's note: Before hunting for numbers, write the sign pattern down: "both plus," "both minus," or "one of each." It takes two seconds, cuts the search in half, and stops the most common factoring mistake, which is finding the right numbers with the wrong signs.
Keep practicing
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