What if the constant term $c$ is zero?

The equation is $ax^2 + bx = 0$. Factor out the greatest common factor, which is $x$, yielding $x(ax + b) = 0$. The roots are $x = 0$ and $x = -b/a$.

What if no two numbers multiply to $ac$ and add to $b$?

This indicates the roots are irrational or complex. The factoring method fails here. You must use the quadratic formula or complete the square.

Can I divide the equation by a common constant before factoring?

Yes. Dividing all terms by a non-zero constant greatest common factor reduces the magnitude of $a$, $b$, and $c$, making the factoring steps mathematically simpler without altering the roots.

How to solve a quadratic equation by factoring

Factoring a quadratic equation requires expressing the standard form $ax^2 + bx + c = 0$ as a product of two linear binomials. This method applies exclusively when the roots are rational numbers, which occurs when the discriminant $b^2 - 4ac$ is a perfect square.

The setup

Move all terms to one side of the equation to achieve the standard form $ax^2 + bx + c = 0$ . Ensure the opposite side is exactly zero. If $a$ is negative, multiplying the entire equation by $-1$ often simplifies the factoring process.

The steps

Identify coefficients $a$ , $b$ , and $c$ . 2. Find two numbers that multiply to $ac$ and add to $b$ . 3. Rewrite the middle term $bx$ using these two numbers. 4. Factor by grouping to produce the form $(px + q)(rx + s) = 0$ . 5. Apply the zero product property by setting each binomial factor to zero. 6. Solve each resulting linear equation for $x$ .

Checking the result

Substitute each calculated root back into the original equation. Evaluate both sides to verify they yield a true identity, such as $0 = 0$ . Verify both roots independently.

Common errors

Attempting to factor before setting the equation to zero is a critical error. Another frequent mistake is dividing the entire equation by a variable expression, which permanently deletes one of the valid roots.

Worked example

Solve $2x^2 = 5x + 3$ by factoring.

First, rewrite in standard form: $2x^2 - 5x - 3 = 0$ . Here, $a=2$ , $b=-5$ , $c=-3$ . The product $ac = -6$ . We need two numbers that multiply to $-6$ and add to $-5$ . These numbers are $-6$ and $1$ . Rewrite the middle term: $2x^2 - 6x + x - 3 = 0$ . Factor by grouping: $2x(x - 3) + 1(x - 3) = 0$ . Factor out the common binomial: $(2x + 1)(x - 3) = 0$ . Set each factor to zero: $2x + 1 = 0$ yields $x = -1/2$ . $x - 3 = 0$ yields $x = 3$ . The solutions are $x = -1/2$ and $x = 3$ .

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References: OpenStax College Algebra, Chapter 2.5 · Khan Academy, Quadratics: Solving by factoring