ORCCAOpen Resources for Community College Algebra

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

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:

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 \(B\text{,}\) take the square roots of the first and last terms, so \(A=8d\) and \(B=9\text{.}\) 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

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

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.

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 take the square roots. So \(A=3b\) since \((3b)^2=9b^2\) and \(B=5y\) since \((5y)^2=25y^2\text{.}\) So, the expression must factor as

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

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:

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

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:

Start by noting that the GCF is \(1\text{,}\) and there is no GCF to factor out. We continue along the decision tree. Since we have a four-term polynomial, we should try to factor by grouping. The full process is:

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

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:

Start by noting that the GCF is \(1\text{,}\) and there is no GCF to factor out. Continue along the decision tree. We now have a trinomial where both the first term, \(9\text{,}\) and last term, \(16y^2\text{,}\) look like perfect squares. To use the perfect squares difference pattern, \(A^2-2AB+B^2=(A-B)^2\text{,}\) recall that we need to mentally take the square roots of these two terms to find \(A\) and \(B\text{.}\) So, \(A=3\) since \(3^2=9\text{,}\) and \(B=4y\) since \((4y)^2=16y^2\text{.}\) Now we have to check that \(2AB\) matches \(24y\text{:}\)

So the full factorization is:

Start by noting that the GCF is \(1\text{,}\) and there is no GCF to factor out. Since we now have a trinomial where both the first term and last term are perfect squares in exactly the same way as in Example 11. However, we cannot apply the perfect squares method to this problem because it worked when \(2AB=24y\text{.}\) Since our middle term is \(25y\text{,}\) we can be certain that it won't be a perfect square.

Continuing on with the decision tree, our next option is to use the AC method. You might be tempted to rearrange the order of the terms, but that is unnecessary. 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:

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

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 \(13\text{.}\)

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: