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MathematicsGrade 9· U.S. National — Common Core & NGSS
Aligned to:Common Core State Standards (Math)

Solving Quadratic Equations by Factoring

Students use the zero-product property to solve factorable quadratic equations and verify their solutions.

Solving Quadratic Equations by Factoring

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Recognizing Standard Form

A quadratic equation is in standard form when it is written as ax² + bx + c = 0, where a, b, and c are numbers and a is not zero. Before factoring, move every term to one side so that the other side equals zero. Then combine any like terms and arrange the powers of x from greatest to least. For example, consider 2x² + 7x = 15. Subtract 15 from both sides to obtain 2x² + 7x − 15 = 0. In this form, a = 2, b = 7, and c = −15. Standard form makes the expression easier to factor and prepares the equation for the zero-product property. Always check that the equation equals zero before using that property.

A before-and-after equation diagram shows all terms moved to one side and the coefficients identified.
A before-and-after equation diagram shows all terms moved to one side and the coefficients identified.Source: Illustrated for this lesson

Factoring Quadratic Expressions

To factor a quadratic expression, rewrite it as a product of two binomials. For 2x² + 7x − 15, multiply the leading coefficient and constant: 2 times −15 equals −30. Find two integers whose product is −30 and whose sum is 7. The numbers 10 and −3 work. Split the middle term to get 2x² + 10x − 3x − 15. Then factor by grouping: 2x(x + 5) − 3(x + 5). Because both groups contain x + 5, the expression factors as (2x − 3)(x + 5). You can confirm the factorization by multiplying the binomials. The product returns 2x² + 7x − 15, so the factors are correct.

A factoring flow diagram shows the number pair, split middle term, grouping, and final binomial product.
A factoring flow diagram shows the number pair, split middle term, grouping, and final binomial product.Source: Illustrated for this lesson

Applying the Zero-Product Property

The zero-product property states that if the product of two factors equals zero, then at least one factor must equal zero. After factoring 2x² + 7x − 15 = 0, the equation becomes (2x − 3)(x + 5) = 0. This product can equal zero only if 2x − 3 = 0 or x + 5 = 0. The word or is important because either factor may be zero. Do not set the factors equal to each other, and do not divide by a factor containing x because that could remove a valid solution. Instead, create one equation from each factor. The zero-product property connects the factored form of a quadratic equation to simpler linear equations that can be solved separately.

A branching equation diagram shows the factored product splitting into two equations that each equal zero.
A branching equation diagram shows the factored product splitting into two equations that each equal zero.Source: Illustrated for this lesson

Solving for Both Roots

A quadratic equation can have two real solutions, so solve every linear equation produced by the factors. Consider x² − 9x + 20 = 0. The expression factors as (x − 4)(x − 5) = 0 because −4 and −5 multiply to 20 and add to −9. Apply the zero-product property: x − 4 = 0 or x − 5 = 0. Solving gives x = 4 or x = 5. These values are called roots, zeros, or solutions of the quadratic equation. Record both values in the solution set, {4, 5}. Stopping after solving only one factor would give an incomplete answer. In some equations, both factors may produce the same root, but each distinct factor should still be examined.

A two-branch solution diagram shows each factor producing one root and both roots entering the solution set.
A two-branch solution diagram shows each factor producing one root and both roots entering the solution set.Source: Illustrated for this lesson

Checking Solutions

Check each proposed solution by substituting it into the original equation. For 2x² + 7x − 15 = 0, factoring gives the possible solutions x = 3/2 and x = −5. Substitute x = 3/2: 2(3/2)² + 7(3/2) − 15 = 9/2 + 21/2 − 15 = 15 − 15 = 0. Next substitute x = −5: 2(−5)² + 7(−5) − 15 = 50 − 35 − 15 = 0. Both substitutions produce a true equation, so both values are solutions. Checking can reveal arithmetic mistakes, incorrect signs, or missing roots. A value is verified only when substitution makes the left side equal the right side of the original equation.

A substitution chart shows both proposed values making the original quadratic expression equal zero.
A substitution chart shows both proposed values making the original quadratic expression equal zero.Source: Illustrated for this lesson