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
What is factoring quadratic equations?
Factoring a quadratic equation means rewriting ax² + bx + c = 0 as a product of two binomials equal to zero, then using the zero product property. For x² − x − 12 = 0, the factors are (x − 4)(x + 3) = 0, so x − 4 = 0 or x + 3 = 0, and the solutions are x = 4 and x = −3.
What is the formula for factoring quadratics?
There is no single formula, but there is a reliable pattern. For x² + bx + c, find two numbers p and q with p × q = c and p + q = b; then x² + bx + c = (x + p)(x + q). Also learn the special forms a² − b² = (a − b)(a + b) and a² + 2ab + b² = (a + b)², which save time on tests.
How do you factor quadratics when a is not 1?
Use the AC method. For 3x² + 10x + 8, multiply a × c = 24 and find two numbers that multiply to 24 and add to 10: 6 and 4. Split the middle term: 3x² + 6x + 4x + 8. Group: 3x(x + 2) + 4(x + 2). Factor out the common binomial to get (3x + 4)(x + 2).
How do you factor quadratics with a leading coefficient of 1?
When the quadratic starts with x², you only need two numbers that multiply to c and add to b. For x² + 7x + 12, the pairs that multiply to 12 are 1 and 12, 2 and 6, and 3 and 4; only 3 and 4 add to 7, so the answer is (x + 3)(x + 4). If c is negative, the two numbers have opposite signs.
What is factored form of a quadratic?
Factored form is y = a(x − r)(x − s), where r and s are the zeros, the x-intercepts of the parabola. For y = 2(x − 1)(x + 5), the graph crosses the x-axis at x = 1 and x = −5. The axis of symmetry is halfway between them, at x = −2, which helps you find the vertex quickly.
What grade is factoring quadratics taught in?
In the US, students usually learn to factor quadratics in Algebra 1, often in 8th or 9th grade, and revisit it in Algebra 2 with harder cases. Students start with x² + bx + c, then move to leading coefficients other than 1, grouping and special products like the difference of squares.
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