Algebra I

Factoring Polynomials

Big Picture

Factoring polynomials uses the same concept of factoring integers - we look for simpler monomials or binomials whose product is equal to the binomial/trinomial we’re factoring. Some techniques used in factoring polynomials include looking for common factors and using special factoring patterns.

Key Terms

Factor: : A number or term that is multiplied by another factor. To factor a number or polynomial is to find all of the factors for that number or polynomial.

Common Factor: A factor that appears in all terms of the polynomial. It can be a number, a variable, or a combination of numbers and variables.

Quadratic Polynomial: A polynomial of the 2nd degree.

Factoring Binomials

  • Look for the greatest common factor (can be a number or a variable) in both terms
  • Divide both terms in the binomial by the common factor.
  • Put parentheses around the terms that have been divided by the common factor.
  • Put the common factor outside the parentheses.

Example: Factor \(12+4x\).

  • 2 and 4 are both common factors, and 4 is the greatest common factor.
  • The factored form is \(4(3 + x)\)
X or .
\(\div\) or /
\(\sqrt{}\) or \(\sqrt[n]{}\)
square root, nth root
| |
Absolute value
Not equal
Approximately equal
<, ≤
Less than, less than or equal to
>, ≥
Greater than, greater than or equal to
{  }
Set symbol
An element of a set
( ), [ ]
Group symbols

Special Products Polynomials

Recognizing these special products polynomials can make factoring easier.

  • Difference of two squares: \(a^2 - b^2 = (a + b)(a - b)\)
  • Square of a binomial: \(a^2 +2ab + b^2 = (a + b)^2\) or \(a^2 -2ab+b^2 = (a-b)^2\)

Whenever we recognize these patterns, just figure out the values of \(a\) and \(b\) and we’re done!

Factoring Quadratic Polynomials

A quadratic polynomial has the form \(ax^2 + bx + c\).

When \(a = 1\), the polynomial looks like \(x^2 + bx + c\).

  • The two factors are \((x + m)\) and \((x + n)\).
  • \((x + m)(x + n) = x^2 + (m + n)x + mn\)
  • To factor, find the values of \(m\) and \(n\) so that \((m + n)\) = \(b\) and \(mn = c\).

When \(a = 1\), \(b > 0\), and \(c > 0\), both \(m\) and \(n\) are positive.

When \(a = 1\), \(b < 0\), and \(c > 0\), both \(m\) and \(n\) are negative.

When \(a = 1\) and \(c < 0\), either \(m\) or \(n\) (only one of them) will be negative.

When \(a = -1\), factor out the negative, then factor as usual.

  • Example: \(-x^2 + bx + c = -(x^2-bx-c)\), then continue to factor.

      Factoring Polynomials Completely

      A polynomial is factored completely when it can't be factored anymore.

      Tips for factoring completely:

      • Factor common monomials and binomials first.
      • See if there are any special products, such as difference of squares or the square of a binomial. Factor according to the formula.
      • If there are no special products, factor using the methods in this guide.
      • Look at each factor and see if any of these can be factored further.

        Algebra I

        Factoring Polynomials cont.

        Factoring Polynomials Completely (cont.)

        Factoring by Groups

        If there are four or more terms, sometimes we can only factor a common monomial from some of the terms. Go ahead and factor the common monomials from groups of terms, then see if there is a common binomial factor.

        Example: Factor \(2x+2y+ax+ay\).

        • There is no monomial common to all the terms, but there is an \(x\) common to one pair of terms and a \(y\) common to another pair. Factor them out: \(x(2 + a) + y(2 + a)\)
        • There is a \((2 + a)\) common to both terms, so factor it out: \((x + y)(2 + a)\)

        This can be used to solve quadratic expressions \(ax^2+bx+c\) where \(a ≠ 1\).

        • Find the product \(ac\).
        • Look for two numbers that multiply to \(ac\) and add up to \(b\).
        • Rewrite the \(bx\) term into two terms using the numbers we found in step 2
        • Factor the expression by grouping.

        Example: Factor \(3x^2 + 8x + 4\).

        • \(ac = 12\)
        • \(2 ∙ 6 = 12\) and \(2 + 6 = 8\)
        • Rewrite \(8x\) as \(2x + 6x: 3x^2 + 2x + 6x + 4\).
        • Factor by grouping:
          \(x(3x + 2) + 2(3x + 2)
          (3x + 2)(x + 2)\)

        It might be helpful to make an organized list when looking for two numbers that multiply to ac and add up tob. For the problem above, the list might look like this:

        Factors of 12
        Sum of Factors
        Sum Equal 8?
        \(1 ∙ 12\)
        \(1+12 = 13\)
        \(2 ∙ 6\)
        \(2+6 = 8\)
        \(3 ∙ 4\)
        \(3+4 = 7\)

        Solving Problems by Factoring

        Zero-Product Property: If two numbers multiply to zero, then at least one of those numbers must be zero

        If \(a ∙ b = 0\), then \(a = 0\) or \(b = 0\) (or both \(a\) and \(b\) equal \(0\)).

        Factoring can be used to solve polynomial equations

        • If necessary, rewrite the equation in standard form so that the right side equals zero.
        • Factor the polynomial completely.
        • Use the zero-product property to set each factor equal to zero.
        • Solve each equation.

        Example: Solve \(x^2+7x = -6\).

        • Rewrite the equation so that the right side equals zero: \(x^2 + 7x + 6 = 0\).
        • Factor completely: \((x + 6)(x + 1) = 0\).
        • Use the zero-product property: \((x + 6) = 0\) or \((x + 1) = 0\)
        • Solve each equation: \(x = -6\) or \(x = -1\)

        It is very easy to make a sign error. Always check your answers by plugging them back into the equation!