1.5 Factoring Polynomials

Learning Objectives

• Identify and Factor out the greatest common factor (GCF) from a polynomial
• Understand the process of Factoring By Grouping a 4 term polynomial
• Factor a trinomial with a leading coefficient of 1
• Factor a trinomial using factoring by group or the AC Method
• Factor a perfect square trinomial.
• Factor a difference of squares.
• Factor the sum and difference of cubes.
• Factor expressions using fractional or negative exponents.

Factor out the Greatest Common Factor (GCF)

Let’s look at the polynomial below to see what the greatest common factor (GCF) is.

[latex]4xy^2+8zb+2q^3[/latex]

This polynomial has three terms and each one of the terms can be divided by 2 evenly. None of the variable parts of the terms share anything in common. In fact, only 2 can evenly divide into each term. This is the GCF of this polynomial. We factor it out by putting the GCF in front of the polynomial and what is left of the polynomial after dividing by 2 in parentheses as shown below:

[latex]2(2xy^2+4zb+q^3)[/latex]

Let’s look at some additional examples that involve both the coefficients and variables of the given polynomials.

Example 1.5.1

Find the GCF for [latex]4xy^2+ 2xy+16x[/latex]

We can easily determine the GFC by inspection. The coefficients of each of the three terms of the polynomial are all divisible by 2. Examination of the variable parts of each term shows that each term has only an [latex]x[/latex] in common. So, we factor out the GFC by writing [latex]2x[/latex] outside the parentheses to multiply the result of dividing the given polynomial by[latex]2x[/latex].

[latex]2x(y^2+y+8)[/latex]

Example 1.5.2

Find the GCF for [latex]-12p^3t+16p^2t-48p^2t^3[/latex]

This example has some interesting twists. The leading coefficient is negative. And, it may not be immediately apparent what the largest coefficient that can be divided evenly into each term of the polynomial.

When this is the case, we create a prime factor tree for each coefficient and then use the largest factor in common.

[latex]12\to3 \times 4 \to 3 \times 2 \times 2 \to (3)(2^2)[/latex]

[latex]16\to4\times4\to 2\times2\times2\times2\to(2^2)(2^2)[/latex]

[latex]48\to16\times4\to(2^2)(2^2)(2^2)[/latex]

The prime factors make it easier to see that the greatest common factor for 12, 16, and 48 is [latex]2^2[/latex] or 4. However, we notice that the leading term is -12. Whenever the leading term is negative, we will factor out the negative factor.

Next, we turn to the variables and can see that each of the three terms do have a p and a t. We search for the greatest exponent of each variable that is common among the three terms. For the p variable, it is [latex]p^2[/latex]. For the t variable, it is t.

[latex]-4p^2t(3p-4+12t^2)[/latex]

Example 1.5.3

Find the GCF for [latex]x(x+4) + 3(x+4)[/latex]

In this case, the largest factor that divides into each term is the binomial factor [latex]x+4[/latex]. We factor out the GCF by writing (x+4) outside of the parentheses and what remains of the polynomial after dividing each term by (x+4) inside the parentheses.

[latex](x+4)(x+3)[/latex]

Note that since the factor is more than one term, it is placed in parentheses as well so it is properly grouped.

Factoring a Four Term Polynomial By Grouping

Let’s look at how a four term polynomial can be fully factored by grouping terms and factoring out the greatest common factors.

[latex]2x+2y+ax+ay[/latex]

We start by grouping the first two terms in a set of parentheses and the second two terms in a set of parentheses.

[latex](2x+2y)+(ax+ay)[/latex]

Now, we factor out the GCF of each group.

[latex]2(x+y) + a(x+y)[/latex]

There is now a binomial GCF that can be factored out as was done in Example 1.5.3

[latex](x+y)(2+a)[/latex]

And, there we have completely factored the four term polynomial by grouping.

Example 1.5.4

Factor [latex]2x^3z-4x^2z+32xz-64z[/latex]

This polynomial has four terms. It is a good candidate to try factoring by grouping. A first step should always be to examine all terms for a GCF first. We note that the GCF is 2z.

[latex]2z[x^3-2x^2+16x-32][/latex]

Next, we group and factor the GCF from each group.

[latex]2z[(x^3-2x^2)+(16x-32)][/latex]

[latex]2z[x^2(x-2)+16(x-2)][/latex]

The last step is to factor out the binomial GCF.

[latex]2z(x-2)(x^2+16)[/latex]

Factoring Trinomials with a Leading Coefficient of 1

Factoring by grouping can be used to factor trinomials. To see how this can be done, let’s start with our desired result of two binomials factors. We’ll use the FOIL technique to multiple them.

1. [latex](x+m)(x+n)[/latex]

2. [latex]x^2+nx+mx+mn[/latex]

Before the two middle terms are added together there are four total terms. We can easily return to the binomial factors by using the factoring by grouping method.

3. [latex](x^2+nx) + (mx+mn)[/latex]

4. [latex]x(x+n)+ m(x+n)[/latex]

5. [latex](x+n)(x+m)[/latex]

Let’s return to step 2 again. If we factor the GCF from the two middle terms, we get

[latex]x^2+(m+n)x+mn[/latex]

The expression above shows us a way to factor, a trinomial with a leading coefficient of 1. The middle term of the trinomial adds to the two factors of the constant.

Example 1.5.5

Factor [latex]x^2+8x+15[/latex]

From the shortcut above, we have

[latex]x^2+(m+n)x+mn[/latex]

We need to find two factors that multiply (mn) to 15 and add (m+n) to 8.

[latex]x^2+(3+5)x+(3\cdot5)[/latex]

[latex]x^2+8x+15[/latex]

Example 1.5.6

Factor [latex]x^2+7x+12[/latex]

From the shortcut above, we have

[latex]x^2+(m+n)x+mn[/latex]

We need to find two factors that multiply (mn) to 12 and add (m+n) to 7.

[latex]x^2+(3+4)x+(3\cdot4)[/latex]

[latex]x^2+7x+12[/latex]

Factoring a Trinomial Using Factoring By Grouping

A trinomial of the form [latex]ax^2+bx+c[/latex]. When the leading coefficient, a,   is other than 1, we can still use the factoring by grouping technique to find its binomial factors.

Let’s examine how to do this with [latex]12a^2-4a-16[/latex]. We use this trinomial to further demonstrate that the first step to factoring should always be to check for a GCF. Each of the terms can be divided evenly by [latex]4[/latex] so this is the GCF.

[latex]4[3a^2-a-4][/latex]

We now factor the trinomial in brackets using factoring by grouping. To see how this is done, we will start with the factors of[latex]4[3a^2-a-4][/latex].

[latex](3a-4)(a+1)[/latex]

When we multiply the factors using the foil technique, we get

[latex]3a^2-4a+3a-4[/latex]

[latex]3a^2+(-4+3)a-4[/latex]

We can see that the middle term, b, is the sum of the factors of ac. The signs of the factors are chosen carefully. Since the middle term is negative, the larger factor is chosen to be negative and the smaller positive.

This is how we can factor trinomials with leading coefficients other than 1. We may need to examine several possibilities of factors that add to b and multiply to ac. Note: while the two factors turned out to be the values of a and c, this may not always be the case.

Let’s lay out the steps to follow.

Example 1.5.7

Factor [latex]2x^2+9x+9[/latex]

We follow the steps outlined.

Step 1: We see there is now GCF for the overall expression.

Step 2: We multiply ac to get (2)(9)=18

Step 3: We must find two factors, mn, that multiply to 18 but add to the middle term, 9.

Step 4: We rewrite the trinomial to have four terms by splitting the middle term into bx=mx+nx.

[latex]2x^2+3x+6x+9[/latex]

Step 5: We continue with the factoring by grouping method.

Step 5.1 Group the terms.

[latex](2x^2+3x)+(6x+9)[/latex]

Step 5.2 Factor GCF from first group.

[latex]x(2x+3)+(6x+9)[/latex]

Step 5.3 Factor GCF from second group.

[latex]x(2x+3)+3(2x+3)[/latex]

Step 5.4 Factor GCF from the whole expression.

[latex](2x+3)(x+3)[/latex]

Example 1.5.8

Factor [latex]6y^4+14y^3-40y^2[/latex]

Step 1: Factor the GCF

[latex]2y^2[3y^2+7y-20[/latex]

Step 2: We multiply ac of the trinomial in brackets to get [latex](3)(-20)=-60[/latex].

Step 3: We must find two factors, mn, that multiply to -60 by add to b, 7. Note: the constant is negative which means that m and n must be opposite signs. With b being positive that means the larger factor should be positive and the smaller should be negative.

Step 4: We rewrite the trinomial to have four terms by splitting the middle term into bx=mx+nx.

[latex]3y^2+(12-5)y-20[/latex]

Step 5: We continue with the factoring by grouping method.

Step 5.1 Group the terms.

[latex](3y^2+12y)+(-5y-20)[/latex]

Step 5.2 Factor GCF from first group.

[latex]3y(y+4)+(-5y-20)[/latex]

Step 5.3 Factor GCF from second group.

[latex]3y(y+4)-5(y+4)[/latex]

Step 5.4 Factor GCF from the whole expression.

[latex](y+4)(3y-5)[/latex]

[latex]2y^2(y+4)(3y-5)[/latex]

The GCF from Step 1 was included for the complete factored expression.

Bonus: A Short-Cut AC Method for Factoring

1. Start with a trinomial of the form, [latex]ax^2+bx+c[/latex]
2. Rewrite the trinomial as [latex]x^2+bx+ac[/latex]
3. Now factor as a trinomial with a leading coefficient of 1 by finding two factors, m & n, where [latex]mn=ac[/latex] and [latex]m+n=b[/latex]
4. Now, divide the second term of each binomial factor by a. [latex](x+\frac{m}{a})(x+\frac{n}{a})[/latex]
5. If a divides evenly, simplify the term. If a does not divide evenly, multiply the first term in the binomial factor by a.

We’ll use the trinomial,[latex]3x^2+7x-20[/latex] to demonstrate.

Step 1:[latex]3x^2+7x-20[/latex]

Step 2:[latex]x^2+7x-60[/latex]

Step 3: [latex](x+12)(x-5)[/latex]

Step 4: [latex](x+\frac{12}{3})(x-\frac{5}{3})[/latex]

Step 5: [latex](x+4)(3x-5)[/latex]

Factor A Perfect Trinomial Square

A perfect square trinomial results from squaring a binomial. Let’s look below at two examples.

[latex](a+b)^2 = (a+b)(a+b)= a^2+2ab+b^2[/latex]

[latex](a-b)^2 = (a-b)(a-b)= a^2-2ab+b^2[/latex]

If we can recognize a trinomial as a perfect square, we can factor them quickly.

Example 1.5.9

Recognize if these trinomials are Perfect Square Trinomials?

Difference of Squares

A difference of squares is exactly what it sounds like.  It is a perfect square subtracted from a perfect square. The binomial factors of a difference of squares have the same terms but the signs are opposite as shown below:

[latex]a^2-b^2=(a+b)(a-b)[/latex]

The factors are known as conjugates. Difference of Squares are easy to spot since there’s only two terms and they’re both perfect squares.

Example 1.5.10

Factor the Difference of Squares

Factor Sum and Difference of Cubes

Admittedly, the sums and differences of cubes seem to be the trickiest special factoring pattern to remember. Let’s look at the pattern:

[latex]a^3+b^3=(a+b)(a^2-ab+b^2)[/latex]

The first factor is a binomial and the second is a trinomial. For the binomial factor, we take the cube root of each term. If we are factoring a sum of cubes, the sign in the middle is positive. If we are factoring a difference of cubes the sign in the middle is negative.

Here’s how to build the second trinomial factor:

First term: square the first term of the binomial factor

Third term: square the second term of the binomial factor

Middle term: product of first and second term of the binomial factor

Now for the signs, if the binomial factor is a sum, then there should be a negative sign between the first and second term of the trinomial factor. If the binomial factor is a difference, then there should be a positive sign between the first and second term. The sign between the second and third terms is always positive.

Example 1.5.11

Factor Sum or Difference of Cubes

Factor Expressions Using Fractional or Negative Exponents

While expressions with negative exponents or fractional exponents may seem a bit intimidating, we can factor using the same techniques as we have been using in this section.  Let’s look at the expression below that has variables with negative exponents.

[latex]2x^2+4x^{-3}+x^{-2}[/latex]

In this expression, each term has a variable, x, raised to different powers. To factor out the GCF,   we actually look for the term that has the smallest exponent. For this expression, we chose the most negative value, -3.

[latex]x^{-3}(2x^5+4+x)[/latex]

Let’s look a little deeper into how the GFC was factored out.

When a GFC is being factored out, we’re actually multiplying the expression by GFC/GFC.

[latex](\frac{x^{-3}}{x^{-3}}(2x^2+4x^{-3}+x^{-2}))[/latex]

Then, the GCF in the numerator is left outside the parentheses and the GCF in the denominator is multiplied through each term to give the following:

[latex](x^{-3}(\frac{2x^2}{x^{-3}}+\frac{4x^{-3}}{x^{-3}}+\frac{x^-2}{x^{-3}}))[/latex]

Now, we carry out the division by using our exponent rules which tell us that when dividing exponents of the same base, we subtract the exponents.

[latex]x^{-3}(x^{2-(-3)}+4x^{-3-(-3)}+x^{-2-(-3)})[/latex]

[latex]x^{-3}(x^5+4+x)[/latex]

Example 1.5.12

Factor [latex](x+1)^{\frac{-1}{3}}+(x+1)^{\frac{2}{3}}[/latex]

We take the same approach to factor out the GCF. The smallest exponent between the two terms is [latex]\frac{-1}{3}[/latex]. The intermediate step before simplifying gives

[latex](x+1)^{-\frac{1}{3}}[(x+1)^{-\frac{1}{3}-(-\frac{1}{3})}+(x+1)^{\frac{2}{3}-(-\frac{1}{3})}][/latex]

Subtracting the exponents gives us the factored expression.

[latex](x+1)^{-\frac{1}{3}}[1+(x+1)][/latex]

[latex](x+1)^{-\frac{1}{3}}(2+x)[/latex]