Math FoundationsEquationsQuadratic Equations

Factoring quadratics

Rewriting quadratic trinomials and quadratic equations as a product of factors instead of a sum of terms. Lets you solve equations quickly, simplify messy expressions, and spot the roots immediately.
Key elements:
Connections:
Completing the square is required to factor a quadratic trinomial in the general case, so you absolutely need to know how to use this method.
Adding and compensating
Factoring a quadratic trinomial uses the discriminant and the general root formula for quadratic equations. You need all of that, no excuses.
The quadratic formula
Discriminant
Biquadratic equation
Statistics:
Statement1
Important1
Problem9
Article
Summary
Practice
Possible factorization cases for a quadratic trinomial

Why factor at all?

  • 1
    You can simplify expressions
    Writing a quadratic trinomial as a product of factors often lets you simplify complicated expressions:
    5⋅(x2−2x−24)(x+4)⋅(x−6)⋅10=5⋅(x+4)⋅(x−6)10⋅(x+4)⋅(x−6)=510=12=0.5\frac{5 \cdot (x^2 - 2x - 24)}{(x+4) \cdot (x-6) \cdot 10} = \frac{5 \cdot \cancel{(x+4)} \cdot \cancel{(x-6)}}{10 \cdot \cancel{(x+4)} \cdot \cancel{(x-6)}} = \frac{5}{10} = \frac{1}{2} = 0.5
  • 2
    Getting back to the sum is easy
    Once a quadratic trinomial is factored, getting the sum form back is just a matter of expanding the brackets:
    2(x−1)(4+x)=2(4x+x2−4−x)=2(x2+3x−4)=2x2+6x−82(x-1)(4+x) = 2(4x + x^2 - 4 -x) = 2(x^2 + 3x - 4) = \boxed{2x^2 + 6x - 8}
    But going the other way, from the sum form to factors, is much harder!
    2x2+6x−8=… ? …=2(x−1)(4+x)2x^2 + 6x - 8 = \ldots \text{ ? } \ldots = \boxed{2(x-1)(4+x)}
  • 3
    The roots become visible immediately
    Writing a quadratic trinomial as a product of factors lets you immediately see the roots of its “equation.” In that form it falls under Zero factors — you just set each factor equal to zero separately and get a true equality 0 = 0:
    In factor form, the roots are the numbers next to x, but with the opposite sign!
  • 4
    It’s a way to solve quadratic equations
    Factoring is one more way to solve quadratic equations, alongside the methods you already know: completing the square and the general root formula. We rewrite the quadratic trinomial as multiplication and immediately see the roots if that expression were a quadratic equation. A big plus is that for simple quadratics, this lets you find the roots fast, sometimes right in your head!

    • Factoring by hand

    Simple quadratic trinomials can sometimes be factored by hand. To do that, you rewrite coefficient B as the sum of two numbers, and coefficient C as the product of those exact same two numbers. Geometrically, that means you take a few smaller shapes and assemble one big rectangle out of them.

    Examples of hand factoring

    👀
    Example
    Factor the quadratic trinomial and find the roots of the corresponding quadratic equation:
    x2+5x+6x^2 + 5x + 6

    Factoring in the general case

    Here is the direct derivation of the general factoring formula using completing the square and the difference of squares formula. If you want the full blow-by-blow version, the article breaks down every step in detail.
    StepExpression
    1Write it in the general form.Ax2+Bx+C\displaystyle Ax^2 + Bx + C
    2Factor out A.A(x2+BAx+CA)\displaystyle A\left(x^2 + \frac{B}{A}x + \frac{C}{A}\right)
    3Add and compensate the 2 to set up a perfect square.A(x2+2⋅xBA⋅12+CA)\displaystyle A\left(x^2 + \yellow{2} \cdot x \frac{B}{A} \cdot \yellow{\frac{1}{2}} + \frac{C}{A}\right)
    4Add and subtract b2b^2 to complete the square.A(x2+2⋅xBA⋅12+(B2A)2⏟a2+2ab+b2=(a+b)2−(B2A)2+CA)\displaystyle A\left(\underbrace{x^2 + 2 \cdot x \frac{B}{A} \cdot \frac{1}{2} + \yellow{\left(\frac{B}{2A}\right)^2}}_{\small a^2 + 2ab + b^2 = (a+b)^2} - \yellow{\left(\frac{B}{2A}\right)^2} + \frac{C}{A}\right)
    5Complete the square.A([x+B2A]2−B24A2+CA)\displaystyle A\left(\left[x + \frac{B}{2A}\right]^2 - \frac{B^2}{4A^2} + \frac{C}{A}\right)
    6Bring the right-hand side to a common denominator.A([x+B2A]2−B2−4AC4A2)\displaystyle A\left(\left[x + \frac{B}{2A}\right]^2 - \frac{B^2 - 4AC}{4A^2}\right)
    7Introduce the discriminant.A([x+B2A]2−D4A2)\displaystyle A\left(\left[x + \frac{B}{2A}\right]^2 - \frac{\brand{D}}{4A^2}\right)
    8Factor it using the difference of squares formula.A(x+B2A−D2A)(x+B2A+D2A)\displaystyle A\left(x + \frac{B}{2A} - \frac{\sqrt{D}}{2A}\right)\left(x + \frac{B}{2A} + \frac{\sqrt{D}}{2A}\right)
    9Bring everything to a common denominator.A(x−−B−D2A⏟Root 1)(x−−B+D2A⏟Root 2)\displaystyle A\left(x - \underbrace{\frac{-B - \sqrt{D}}{2A}}_{\text{Root 1}}\right)\left(x - \underbrace{\frac{-B + \sqrt{D}}{2A}}_{\text{Root 2}}\right)
    10Replace the fractions with the notation from the quadratic formula.A(x−x1)(x−x2)\displaystyle A(x - x_1)(x - x_2)

    Factoring a quadratic trinomial

    If a quadratic trinomial has roots (let’s call them x1x_1 and x2x_2), then that trinomial can always be factored:
    These are two different notations, one through addition and one through multiplication, but they mean the same thing, just like 10 + 6 and 2⋅82\cdot8 denote the same number. You can switch a quadratic trinomial into factor form and back again in absolutely any situation!

    Examples of factoring through roots

    👀
    Example
    Factor the quadratic trinomial by solving the corresponding quadratic equation:
    4x2+15x−44x^2 + 15x - 4
    Factoring quadratics