Factoring by Grouping (Factoring a Trinomial with Leading Coefficient of Not 1)
Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the [latex]x[/latex] term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression.
Factoring Trinomial (Leading Coefficient [latex]\ne 1[/latex])
Factoring by grouping is a method used to decompose a trinomial of the form [latex]ax^2+bx+c[/latex] into a product of two binomials.
This method involves finding two numbers that combine to give the product of the leading coefficient and the constant term ([latex]a \times c[/latex]) and the sum of the middle coefficient ([latex]b[/latex]). We use these numbers to divide the [latex]x[/latex] term into the sum of two terms and factor each portion of the expression separately then factor out the GCF of the entire expression.
- List factors of [latex]a \times c[/latex].
- Find [latex]p[/latex] and [latex]q[/latex], a pair of factors of [latex]a \times c[/latex] with a sum of [latex]b[/latex].
- Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[/latex].
- Pull out the GCF of [latex]a{x}^{2}+px[/latex].
- Pull out the GCF of [latex]qx+c[/latex].
- Factor out the GCF of the expression.
- List factors of [latex]a \times c[/latex].
- Calculate [latex]a \times c[/latex]: [latex]a = 5[/latex] and [latex]c = -6[/latex], so [latex]a \times c = -30[/latex].
- Factors of [latex]-30[/latex]:
Factors of [latex]-30[/latex] Sum of Factors [latex]1,-30[/latex] [latex]-29[/latex] [latex]-1,30[/latex] [latex]29[/latex] [latex]2,-15[/latex] [latex]-13[/latex] [latex]-2,15[/latex] [latex]13[/latex] [latex]3,-10[/latex] [latex]-7[/latex] [latex]-3,10[/latex] [latex]7[/latex]
- Find [latex]p[/latex] and [latex]q[/latex], a pair of factors of [latex]a \times c[/latex] with a sum of [latex]b[/latex].
- Based on the table above, the correct pair is [latex]p = -3[/latex] and [latex]q = 10[/latex].
- Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[/latex].
[latex]5{x}^{2}+7x - 6 = 5{x}^{2} -3x+10x-6[/latex]
- Group the first 2 terms and the last 2 terms. Then, pull out the GCF of each group.
[latex]5{x}^{2}+7x - 6 = (5{x}^{2} -3x)+(10x-6) = x(5x-3)+2(5x-3)[/latex]
- Factor out the GCF of the expression.
[latex]5{x}^{2}+7x - 6 = (x+2)(5x-3)[/latex]
- [latex]2{x}^{2}+9x+9[/latex]
- [latex]6{x}^{2}+x - 1[/latex]
Factoring a Perfect Square Trinomial
A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.
perfect square trinomial
A perfect square trinomial can be written as the square of a binomial:
- Confirm that the first and last term are perfect squares.
- Confirm that the middle term is twice the product of [latex]ab[/latex].
- Write the factored form as [latex]{\left(a+b\right)}^{2}[/latex].
To factor the quadratic expression [latex]25{x}^{2}+20x+4[/latex], recognizing it as a perfect square trinomial will streamline the process. This type of expression comes from squaring a binomial and has a special format, [latex]a^2 +2ab+b^2[/latex], where it can be rewritten as [latex](a+b)^2[/latex].[latex]\begin{align*} \text{Original expression:} & \quad 25x^2 + 20x + 4 \\ \text{Identify square terms:} & \quad 25x^2 = (5x)^2 \quad \text{and} \quad 4 = 2^2 \\ \text{Check middle term:} & \quad 2 \times 5x \times 2 = 20x \\ \text{Write as a square of a binomial:} & \quad (5x + 2)^2 \end{align*}[/latex]
Factoring a Difference of Squares
A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.
We can use this equation to factor any differences of squares.
difference of squares
A difference of squares can be rewritten as two factors containing the same terms but opposite signs.
- Confirm that the first and last term are perfect squares.
- Write the factored form as [latex]\left(a+b\right)\left(a-b\right)[/latex].
To factor the quadratic expression [latex]9{x}^{2}-25[/latex], we recognize that it is a difference of squares.[latex]\begin{align*} \text{Original expression:} & \quad 9x^2 - 25 \\ \text{Identify square terms:} & \quad 9x^2 = (3x)^2 \quad \text{and} \quad 25 = 5^2 \\ \text{Apply difference of squares formula:} & \quad (3x + 5)(3x - 5) \end{align*}[/latex]