How to Solve a Quadratic Equation by Factoring?
Solving a Quadratic Equation by Factoring
Factoring a quadratic rewrites \(x^2 + bx + c\) as a product of two binomials, \((x + p)(x + q)\), by finding two numbers that multiply to \(c\) and add to \(b\). It’s the quickest route to a quadratic’s roots when the numbers cooperate. We’ll practice the pattern, with a solver and a worksheet maker a tap away.
Solve a Quadratic Equation by Factoring: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
A reliable path
- Organize by degreeWrite terms from highest power to lowest power.
- Look for structureTry GCF, special products, grouping, or division depending on the expression.
- Check with featuresZeros, multiplicity, and end behavior should agree with your algebra.
Worked examples
Combine like terms
- Group x squared terms.
- Group x terms.
- Combine each group.
Factor a difference of squares
- Recognize a squared term minus a squared term.
- Use a^2 – b^2.
- Write conjugate factors.
Try one before moving on
Solve a Quadratic Equation by Factoring: pop-up practice
Factoring a quadratic rewrites \(x^2 + bx + c\) as a product of two binomials, \((x + p)(x + q)\). It’s the reverse of FOIL, and it’s the fastest way to find a quadratic’s roots when the numbers cooperate. The whole task boils down to one search: find two numbers that multiply to \(c\) and add to \(b\).
In short: to factor \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\); those numbers are \(p\) and \(q\) in \((x + p)(x + q)\). For example, \(x^2 + 7x + 12 = (x + 3)(x + 4)\).
Two Numbers: Product \(c\), Sum \(b\)
When you FOIL \((x + p)(x + q)\) you get \(x^2 + (p + q)x + pq\). So the middle coefficient is the sum of your two numbers and the constant is their product. Factoring just runs that backwards: hunt for the pair that fits both conditions.
How to factor (3 steps):
- List factor pairs of \(c\).
- Find the pair that adds to \(b\) (mind the signs).
- Write \((x + p)(x + q)\), and check by FOILing.
Worked Examples
Each area model shows how the two binomials build the trinomial: the four cells are the products, and they add back to the original.
Example A — Both numbers positive
Factor \(x^2 + 7x + 12\).
- List factor pairs of \(12\): \(1\cdot12,\ 2\cdot6,\ 3\cdot4\).
- Which pair adds to \(7\)? \(3 + 4 = 7\). ✓
- Write the factors: \((x + 3)(x + 4)\).
- Check with the area model → \(x^2 + 4x + 3x + 12 = x^2 + 7x + 12\).
Answer: \((x + 3)(x + 4)\)
Example B — Both numbers negative
Factor \(x^2 – 5x + 6\).
- Need a product of \(+6\) and a sum of \(-5\) — both numbers negative.
- \((-2)(-3) = 6\) and \(-2 + -3 = -5\). ✓
- Write the factors: \((x – 2)(x – 3)\).
Answer: \((x – 2)(x – 3)\)
Example C — Opposite signs
Factor \(x^2 + 2x – 15\).
- Product is \(-15\) (negative → opposite signs); sum is \(+2\).
- \(5\) and \(-3\): \(5\cdot(-3) = -15\), \(5 + (-3) = 2\). ✓
- Write the factors: \((x + 5)(x – 3)\).
Answer: \((x + 5)(x – 3)\)
Example D — Factor, then find the roots
Solve \(x^2 – x – 6 = 0\).
- Factor: product \(-6\), sum \(-1\) → \(-3\) and \(2\), giving \((x – 3)(x + 2)\).
- Set each factor to zero: \(x – 3 = 0\) or \(x + 2 = 0\).
- Solve: \(x = 3\) or \(x = -2\).
Answer: \(x = 3\) or \(-2\)
Where You’ll Use It
Factoring is the quickest route from a quadratic to its roots, which is why it’s central to solving quadratic equations and graphing parabolas (the roots are the x-intercepts). It also simplifies rational expressions and shows up throughout Algebra 2 and beyond.
Slip-Ups That Cost Easy Points
- Wrong signs. The product and sum signs must both match. \(x^2 + 2x – 15\) needs \(+5\) and \(-3\), not \(-5\) and \(+3\).
- Not checking with FOIL. Multiply your factors back out; it takes seconds and catches sign slips.
- Assuming everything factors. If no integer pair works, the quadratic may need the formula instead.
- Forgetting a common factor first. Pull out a GCF (like \(2x^2 + 6x = 2x(x+3)\)) before looking for binomials.
Your Turn: Factor
Factor each completely, then reveal the answers.
- \(x^2 + 5x + 4\)
- \(x^2 – 8x + 15\)
- \(x^2 + x – 12\)
- \(x^2 – 9x + 20\)
- \(x^2 – x – 20\)
- \(x^2 + 6x + 9\)
Show answers
- \(\color{blue}{(x+1)(x+4)}\)
- \(\color{blue}{(x-3)(x-5)}\)
- \(\color{blue}{(x+4)(x-3)}\)
- \(\color{blue}{(x-4)(x-5)}\)
- \(\color{blue}{(x-5)(x+4)}\)
- \(\color{blue}{(x+3)^2}\)
Make Your Own Factoring Worksheet
Generate fresh factoring problems with a full answer key — print or save as a PDF.
Frequently Asked Questions
How do I factor a quadratic \(x^2 + bx + c\)?
Find two numbers that multiply to \(c\) and add to \(b\). Those numbers fill in \((x + p)(x + q)\). Check by FOILing back to the original.
How do the signs work?
If \(c\) is positive, both numbers share \(b\)’s sign. If \(c\) is negative, the numbers have opposite signs, and the larger absolute value carries \(b\)’s sign.
What if it won’t factor?
If no integer pair multiplies to \(c\) and adds to \(b\), the quadratic doesn’t factor over the integers — use the quadratic formula or completing the square instead.
How does factoring give the roots?
Set each factor to zero (zero-product property). \((x-3)(x+2)=0\) means \(x = 3\) or \(x = -2\).
Related Topics
Continue Your Study
Ready for the next step? Pick up right where this lesson leaves off:
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