ORCCA Factoring Strategies

Section 7.6 Factoring Strategies

Deciding which method to use when factoring a random polynomial can seem like a daunting task. Understanding all of the techniques that we have learned and how they fit together can be done using a decision tree.

A decision tree to help factoring. — Figure 7.6.1. Factoring Decision Tree

Using the decision tree can guide us when we are given an expression to factor.

Example 7.6.2.

Factor the expression \(4k^2+12k-40\) completely.

Explanation

Start by noting that the GCF is \(4\text{.}\) Factoring this out, we get

\begin{equation*} 4k^2+12k-40=4\left(k^2+3k-10\right)\text{.} \end{equation*}

Following the decision tree, we now have a trinomial where the leading coefficient is \(1\) and we need to look for factors of \(-10\) that add to \(3\text{.}\) We find that \(-2\) and \(5\) work. So, the full factorization is:

\begin{align*} 4k^2+12k-40\amp=4\left(k^2+3k-10\right)\\ \amp=4(k-2)(k+5) \end{align*}

Example 7.6.3.

Factor the expression \(64d^2+144d+81\) completely.

Explanation

Start by noting that the GCF is \(1\text{,}\) and there is no GCF to factor out. We continue along the decision tree for a trinomial. Notice that both \(64\) and \(81\) are perfect squares and that this expression might factor using the pattern \(A^2+2AB+B^2=(A+B)^2\text{.}\) To find \(A\) and \(B\text{,}\) determine what can be squared to obtain the first and last terms. So \(A=8d\text{,}\) since \((8d)^2=64d^2\) and \(B=9\text{,}\) since \(9^2=81\text{.}\) Then, we have to check that the middle term is correct: since \(2AB=2(8d)(9)=144d\) matches our middle term, the expression must factor as

\begin{equation*} 64d^2+144d+81=(8d+9)^2\text{.} \end{equation*}

Example 7.6.4.

Factor the expression \(10x^2y-12xy^2\) completely.

Explanation

Start by noting that the GCF is \(2xy\text{.}\) Factoring this out, we get

\begin{equation*} 10x^2y-12xy^2=2xy(5x-6y)\text{.} \end{equation*}

Since we have a binomial inside the parentheses, the only options on the decision tree for a binomial involve squares or cubes. Since there are none, we conclude that \(2xy(5x-6y)\) is the complete factorization.

Example 7.6.5.

Factor the expression \(9b^2-25y^2\) completely.

Explanation

Start by noting that the GCF is \(1\text{,}\) and there is no GCF to factor out. We continue along the decision tree for a binomial and notice that we now have a difference of squares, \(A^2-B^2=(A-B)(A+B)\text{.}\) To find the values for \(A\) and \(B\) that fit the patterns, just find what needs to be squared to obtain the first and last terms. So \(A=3b\) since \((3b)^2=9b^2\) and \(B=5y\) since \((5y)^2=25y^2\text{.}\) So, the expression must factor as

\begin{equation*} 9b^2-25y^2=(3b-5y)(3b+5y)\text{.} \end{equation*}

Example 7.6.6.

Factor the expression \(24w^3+6w^2-9w\) completely.

Explanation

Start by noting that the GCF is \(3w\text{.}\) Factoring this out, we get

\begin{equation*} 24w^3+6w^2-9w=3w\left(8w^2+2w-3\right)\text{.} \end{equation*}

Following the decision tree, we now have a trinomial inside the parentheses where \(a\neq1\text{.}\) We should try the AC method because neither \(8\) nor \(-3\) are perfect squares. In this case, \(ac=-24\) and we must find two factors of \(-24\) that add to be \(2\text{.}\) The numbers \(6\) and \(-4\) work in this case. The rest of the factoring process is:

\begin{align*} 24w^3+6w^2-9w\amp=3w\left(8w^2\overbrace{{}+2w}-3\right)\\ \amp=3w\left(8w^2\overbrace{{}+6w-4w}-3\right)\\ \amp=3w\left(\left(8w^2+6w\right)+\left(-4w-3\right)\right)\\ \amp=3w\left(2w\highlight{(4w+3)}-1\highlight{(4w+3)}\right)\\ \amp=3w\highlight{(4w+3)}(2w-1) \end{align*}

Example 7.6.7.

Factor the expression \(q^5+q^2\) completely.

Explanation

Start by noting that the GCF is \(q^2\text{.}\) Factoring this out, we find

\begin{equation*} q^5+q^2=\highlight{q^2}\left(q^3+1\right)\text{.} \end{equation*}

Following the decision tree, we now have a binomial with a sum of cubes. (Notice that \(1^3=1\text{.}\)) So using the sum of cubes formula, we have the complete factorization:

\begin{align*} q^5+q^2\amp=q^2\left(q^3+1\right)\\ \amp=q^2(q+1)\left(q^2-q+1\right)\text{.} \end{align*}

Example 7.6.8.

Factor the expression \(-6xy+9y+2x-3\) completely.

Explanation

Start by noting that the GCF is \(1\text{,}\) and so there is no GCF to factor out. However, the leading coefficient is negative and when this is the case, it is usually helpful to factor out the negative sign, which we will do first. We continue along the decision tree. Since we have a four-term polynomial, we should try to factor by grouping. The full process is:

\begin{align*} -6xy+9y+2x-3\amp=-(6xy-9y-2x+3)\\ \amp=-((6xy-9y)+(-2x+3))\\ \amp=-(3y\highlight{(2x-3)}-1\highlight{(2x-3)})\\ \amp=-\highlight{(2x-3)}(3y-1) \end{align*}

Example 7.6.9.

Factor the expression \(4w^3-20w^2+24w\) completely.

Explanation

Start by noting that the GCF is \(4w\text{.}\) Factoring this out, we get

\begin{equation*} 4w^3-20w^2+24w=4w\left(w^2-5w+6\right)\text{.} \end{equation*}

Following the decision tree, we now have a trinomial with \(a=1\) inside the parentheses. So, we can look for factors of \(6\) that add up to \(-5\text{.}\) Since \(-3\) and \(-2\) fit the requirements, the full factorization is:

\begin{align*} 4w^3-20w^2+24w\amp=4w\left(w^2-5w+6\right)\\ \amp=4w(w-3)(w-2) \end{align*}

Example 7.6.10.

Factor the expression \(9-24y+16y^2\) completely.

Explanation

Start by noting that the GCF is \(1\text{,}\) and there is no GCF to factor out. It is also good to note that the terms are not written in descending powers of the variable \(y\text{,}\) so we'll start off by reordering the terms. This will give us the equivalent expression, \(16y^2-24y+9\text{.}\) Continue along the decision tree. We now have a trinomial where both the first term, \(16y^2\text{,}\) and last term, \(9\text{,}\) look like perfect squares. To use the perfect squares difference pattern, \(A^2-2AB+B^2=(A-B)^2\text{,}\) recall that to find \(A\) and \(B\text{,}\) we need to mentally determine what is being squared to get the first and last terms. So, \(A=4y\) since \((4y)^2=16y^2\text{,}\) and \(B=3\) since \(3^2=9\text{.}\) Now we have to check that \(2AB\) matches \(24y\text{:}\)

\begin{equation*} 2AB=2(4y)(3)=24y\text{.} \end{equation*}

So the full factorization is:

\begin{equation*} 16y^2-24y+9=(4y-3)^2\text{.} \end{equation*}

Example 7.6.11.

Factor the expression \(9-25y+16y^2\) completely.

Explanation

Start by noting that the GCF is \(1\text{,}\) and there is no GCF to factor out. Next, note that the terms are not written in descending powers of the variable \(y\text{,}\) so we'll start off by reordering the terms. This will give us the equivalent expression, \(16y^2-25y+9\text{.}\) Notice that we now have a trinomial where both the first term and last term are perfect squares in exactly the same way as in Example 10. However, we cannot apply the perfect squares method to this problem because \(2AB=-24y\) and our middle term is \(-25y\text{.}\) So we can be certain that it is not a perfect square trinomial.

Continuing on with the decision tree, our next option is to use the AC method. In this case, \(ac=144\) and we need to come up with two factors of \(144\) that add to be \(-25\text{.}\) After a brief search, we conclude that those values are \(-16\) and \(-9\text{.}\) The remainder of the factorization is:

\begin{align*} 16y^2\overbrace{{}-25y}+9\amp=16y^2\overbrace{{}-16y-9y}+9\\ \amp=\left(16y^2-16y\right)+\left(-9y+9\right)\\ \amp=16y\highlight{\left(y-1\right)}-9\highlight{\left(y-1\right)}\\ \amp=\highlight{\left(y-1\right)}(16y-9) \end{align*}

Example 7.6.12.

Factor the expression \(20x^4+13x^3-21x^2\) completely.

Explanation

Start by noting that the GCF is \(x^2\text{.}\) Factoring this out, we get

\begin{equation*} 20x^4+13x^3-21x^2=x^2\left(20x^2+13x-21\right)\text{.} \end{equation*}

Following the decision tree, we now have a trinomial inside the parentheses where \(a\neq1\) and we should try the AC method. In this case, \(ac=-420\) and we need factors of \(-420\) that add to \(13\text{.}\)

Factor Pair	Sum
\(1\cdot-420\)	\(-419\)
\(2\cdot-210\)	\(-208\)
\(3\cdot-140\)	\(-137\)
\(4\cdot-105\)	\(-101\)

Factor Pair	Sum
\(5\cdot-84\)	\(-79\)
\(6\cdot-70\)	\(-64\)
\(7\cdot-60\)	\(-53\)
\(10\cdot-42\)	\(-32\)

Factor Pair	Sum
\(12\cdot-35\)	\(-23\)
\(14\cdot-30\)	\(-16\)
\(15\cdot-28\)	\(-13\)
\(20\cdot-21\)	\(-1\)

In the table of the factor pairs of \(-420\) we find \(15+(-28)=-13\text{,}\) the opposite of what we want, so we want the opposite numbers: \(-15\) and \(28\text{.}\) The rest of the factoring process is shown:

\begin{align*} 20x^4+13x^3-21x^2\amp=x^2\left(20x^2\overbrace{{}+13x}-21\right)\\ \amp=x^2\left(20x^2\overbrace{-15x+28x}-21\right)\\ \amp=x^2\left(\left(20x^2-15x\right)+\left(28x-21\right)\right)\\ \amp=x^2\left(5x\highlight{(4x-3)}+7\highlight{(4x-3)}\right)\\ \amp=x^2\highlight{(4x-3)}(5x+7) \end{align*}