ORCCA Factoring Chapter Review

Section 10.8 Factoring Chapter Review

Subsection 10.8.1 Factoring out the GCF

In Section 1 we covered how to factor out the greatest common factor. Recall that the greatest common factor between two expressions is the largest factor that goes in evenly to both expressions.

Example 10.8.1. Finding the Greatest Common Factor.

What is the greatest common factor between \(12x^3y\) and \(42x^2y^2\text{?}\)

Explanation.

Break down each of these into its factors:

\begin{align*} 12x^3y \amp =(2\cdot 2)\cdot 3\cdot (x\cdot x\cdot x) \cdot y \amp 42x^2y^2 \amp =2\cdot 3\cdot 7\cdot (x\cdot x)\cdot (y \cdot y)\\ \end{align*}

Identify the common factors:

\begin{align*} 12x^3y \amp =\attention{2}\cdot 2\cdot \attention{3}\cdot \attention{x}\cdot \attention{x}\cdot x \cdot \attention{y} \amp 42x^2y^2 \amp =\attention{2}\cdot \attention{3}\cdot 7\cdot \attention{x}\cdot \attention{x}\cdot \attention{y} \cdot y \end{align*}

With \(2\text{,}\) \(3\text{,}\) two \(x\)’s and a \(y\) in common, the greatest common factor is \(6x^2y\text{.}\)

Example 10.8.2.

What is the greatest common factor between \(18c^3y^2\) and \(27y^3c\text{?}\)

Explanation.

Break down each into factors. You can definitely do this mentally with practice.

\begin{align*} 18c^3y^2\amp=2\cdot 3\cdot 3\cdot c\cdot c\cdot c\cdot y\cdot y \amp 27y^3c\amp= 3\cdot 3\cdot 3\cdot y\cdot y\cdot y\cdot c\\ \end{align*}

And take note of the common factors.

\begin{align*} 18c^3y^2\amp=2\cdot \attention{3}\cdot \attention{3}\cdot \attention{c}\cdot c\cdot c\cdot \attention{y}\cdot \attention{y} \amp 27y^3c\amp= \attention{3}\cdot \attention{3}\cdot 3\cdot \attention{y}\cdot \attention{y}\cdot y\cdot c \end{align*}

And so the GCF is \(9y^2c\)

Example 10.8.3. Factoring out the Greatest Common Factor.

Factor out the GCF from the expression \(32mn^2-24m^2n-12mn\text{.}\)

Explanation.

To factor out the GCF from the expression \(32mn^2-24m^2n-12mn\text{,}\) first note that the GCF to all three terms is \(4mn\text{.}\) Begin by writing that in front of a blank pair of parentheses and fill in the missing pieces.

\begin{align*} 32mn^2-24m^2n-12mn\amp=4mn(\phantom{8n}-\phantom{6m}-\phantom{3})\\ \amp=4mn(8n-6m-3) \end{align*}

Example 10.8.4.

Factor out the GCF from the expression \(14x^3-35x^2\text{.}\)

Explanation.

First note that the GCF of the terms in \(14x^3-35x^2\) is \(7x^2\text{.}\) Factoring this out, we have:

\begin{align*} 14x^3-35x^2\amp=7x^2\left(\phantom{2x}-\phantom{5}\right)\\ \amp=7x^2\left(2x-5\right) \end{align*}

Example 10.8.5.

Factor out the GCF from the expression \(36m^3n^2-18m^2n^5+24mn^3\text{.}\)

Explanation.

First note that the GCF of the terms in \(36m^3n^2-18m^2n^5+24mn^3\) is \(6mn^2\text{.}\) Factoring this out, we have:

\begin{align*} 36m^3n^2-18m^2n^5+24mn^3\amp=6mn^2\left(\phantom{6m^2}-\phantom{3mn^3}+\phantom{4n}\right)\\ \amp=6mn^2\left(6m^2-3mn^3+4n\right) \end{align*}

Example 10.8.6.

Factor out the GCF from the expression \(42f^3w^2-8w^2+9f^3\text{.}\)

Explanation.

First note that the GCF of the terms in \(42f^3w^2-8w^2+9f^3\) is \(1\text{,}\) so we call the expression prime. The only way to factor the GCF out of this expression is:

\begin{equation*} 42f^3w^2-8w^2+9f^3=1\left(42f^3w^2-8w^2+9f^3\right) \end{equation*}

Subsection 10.8.2 Factoring by Grouping

In Section 2 we covered how to factor by grouping. Recall that factoring using grouping is used on four-term polynomials, and also later in the AC method in Section 4. Begin by grouping two pairs of terms and factoring out their respective GCF; if all is well, we should be left with two matching pieces in parentheses that can be factored out in their own right.

Example 10.8.7.

Factor the expression \(2x^3+5x^2+6x+15\) using grouping.

Explanation.

\begin{align*} 2x^3+5x^2+6x+15\amp=\left(2x^3+5x^2\right)+\left(6x+15\right)\\ \amp=\highlight{x^2}\left(2x+5\right)\highlight{{}+3}\left(2x+5\right)\\ \amp=\highlight{\left(x^2+3\right)}(2x+5) \end{align*}

Example 10.8.8.

Factor the expression \(2xy-3x-8y+12\) using grouping.

Explanation.

\begin{align*} 2xy-3x+8y-12\amp=\left(2xy-3x\right)+\left(-8y+12\right)\\ \amp=\highlight{x}\left(2y-3\right)\highlight{{}-4}\left(2y-3\right)\\ \amp=\highlight{(x-4)}(2y-3) \end{align*}

Example 10.8.9.

Factor the expression \(xy-2-2x+y\) using grouping.

Explanation.

This is a special example because if we try to follow the algorithm without considering the bigger context, we will fail:

\begin{equation*} xy-2-2x+y=\left(xy-2\right)+\left(-2x+y\right) \end{equation*}

Note that there is no common factor in either grouping, besides \(1\text{,}\) but the groupings themselves don’t match. We should now recognize that whatever we are doing isn’t working and try something else. It turns out that this polynomial isn’t prime; all we need to do is rearrange the polynomial into standard form where the degrees decrease from left to right before grouping.

\begin{align*} xy-2-2x+y\amp=xy-2x+y-2\\ \amp=\left(xy-2x\right)+\left(y-2\right)\\ \amp=\highlight{x}\left(y-2\right)\highlight{{}+1}\left(y-2\right)\\ \amp=\highlight{(x+1)}(y-2) \end{align*}

Example 10.8.10.

Factor the expression \(15m^2-3m-10mn+2n\) using grouping.

Explanation.

\begin{align*} 15m^2-3m-10mn+2n\amp=\left(15m^2-3m\right)+\left(-10mn+2n\right)\\ \amp=\highlight{3m}\left(5m-1\right)\highlight{{}-2n}\left(5m-1\right)\\ \amp=\highlight{(3m-2n)}(5m-1) \end{align*}

Subsection 10.8.3 Factoring Trinomials with Leading Coefficient \(1\)

In Section 3 we covered factoring expressions that look like \(x^2+bx+c\text{.}\) The trick was to look for two numbers whose product was \(c\) and whose sum was \(b\text{.}\) Always remember to look for a greatest common factor first, before looking for factor pairs.

Example 10.8.11.

Answer the questions to practice for the factor pairs method.

What two numbers multiply to be \(6\) and add to be \(5\text{?}\)
What two numbers multiply to be \(-6\) and add to be \(5\text{?}\)
What two numbers multiply to be \(-6\) and add to be \(-1\text{?}\)
What two numbers multiply to be \(24\) and add to be \(-10\text{?}\)
What two numbers multiply to be \(-24\) and add to be \(2\text{?}\)
What two numbers multiply to be \(-24\) and add to be \(-5\text{?}\)
What two numbers multiply to be \(420\) and add to be \(44\text{?}\)
What two numbers multiply to be \(-420\) and add to be \(-23\text{?}\)
What two numbers multiply to be \(420\) and add to be \(-41\text{?}\)

Explanation.

What two numbers multiply to be \(6\) and add to be \(5\text{?}\) The numbers are \(2\) and \(3\text{.}\)
What two numbers multiply to be \(-6\) and add to be \(5\text{?}\) The numbers are \(6\) and \(-1\text{.}\)
What two numbers multiply to be \(-6\) and add to be \(-1\text{?}\) The numbers are \(-3\) and \(2\text{.}\)
What two numbers multiply to be \(24\) and add to be \(-10\text{?}\) The numbers are \(-6\) and \(-4\text{.}\)
What two numbers multiply to be \(-24\) and add to be \(2\text{?}\) The numbers are \(6\) and \(-4\text{.}\)
What two numbers multiply to be \(-24\) and add to be \(-5\text{?}\) The numbers are \(-8\) and \(3\text{.}\)
What two numbers multiply to be \(420\) and add to be \(44\text{?}\) The numbers are \(30\) and \(14\text{.}\)
What two numbers multiply to be \(-420\) and add to be \(-23\text{?}\) The numbers are \(-35\) and \(12\text{.}\)
What two numbers multiply to be \(420\) and add to be \(-41\text{?}\) The numbers are \(-20\) and \(-21\text{.}\)

Note that for parts g–i, the factors of 420 are important. Below is a table of factors of 420 which will make it much clearer how the answers were found. To generate a table like this, we start with \(1\text{,}\) and we work our way up the factors of 420.

Factor Pair

\(1\cdot420\)

\(2\cdot210\)

\(3\cdot140\)

\(4\cdot105\)

Factor Pair

\(5\cdot84\)

\(6\cdot70\)

\(7\cdot60\)

\(10\cdot42\)

Factor Pair

\(12\cdot35\)

\(14\cdot30\)

\(15\cdot28\)

\(20\cdot21\)

It is now much easier to see how to find the numbers in question. For example, to find two numbers that multiply to be \(-420\) and add to be \(-23\text{,}\) look in the table for two factors that are \(23\) apart and assign a negative sign appropriately. As we found earlier, the numbers that are \(23\) apart are \(12\) and \(35\text{,}\) and making the larger one negative, we have our answer: \(12\) and \(-35\text{.}\)

Example 10.8.12.

Factor the expression \(x^2-3x-28\)

Explanation.

To factor the expression \(x^2-3x-28\text{,}\) think of two numbers that multiply to be \(-28\) and add to be \(-3\text{.}\) In the Section 3, we created a table of all possibilities of factors, like the one shown, to be sure that we never missed the right numbers; however, we encourage you to try this mentally for most problems.

Factor Pair	Sum of the Pair
\(-1\cdot28\)	\(27\)
\(-2\cdot14\)	\(12\)
\(-4\cdot7\)	\(3\) (close; wrong sign)

Factor Pair	Sum of the Pair
\(1\cdot(-28)\)	\(-27\)
\(2\cdot(-14)\)	\(-12\)
\(4\cdot(-7)\)	\(-3\) (what we wanted)

Since the two numbers in question are \(4\) and \(-7\) that means that

\begin{equation*} x^2-3x-28=(x+4)(x-7) \end{equation*}

Remember that you can always multiply out your factored expression to verify that you have the correct answer. We will use the FOIL expansion.

\begin{align*} (x+4)(x-7)\amp=x^2-7x+4x-28\\ \amp\confirm{=}x^2-3x-28 \end{align*}

Example 10.8.13. Factoring in Stages.

Completely factor the expression \(4x^3-4x^2-120x\text{.}\)

Explanation.

Remember that some expressions require more than one step to completely factor. To factor \(4x^3-4x^2-120x\text{,}\) first, always look for any GCF; after that is done, consider other options. Since the GCF is \(4x\text{,}\) we have that

\begin{equation*} 4x^3-4x^2-120x=4x\left(x^2-x-30\right)\text{.} \end{equation*}

Now the factor inside parentheses might factor further. The key here is to consider what two numbers multiply to be \(-30\) and add to be \(-1\text{.}\) In this case, the answer is \(-6\) and \(5\text{.}\) So, to completely write the factorization, we have:

\begin{align*} 4x^3-4x^2-120x\amp=4x\left(x^2-x-30\right)\\ \amp=4x(x-6)(x+5) \end{align*}

Example 10.8.14. Factoring Expressions with Higher Powers.

Completely factor the expression \(p^{10}-6p^5-72\text{.}\)

Explanation.

If we have a trinomial with an even exponent on the leading term, and the middle term has an exponent that is half the leading term exponent, we can still use the factor pairs method. To factor \(p^{10}-6p^5-72\text{,}\) we note that the middle term exponent \(5\) is half of the leading term exponent \(10\text{,}\) and that two numbers that multiply to be \(-72\) and add to be \(-6\) are \(-12\) and \(6\text{.}\) So the factorization of the expression is

\begin{equation*} p^{10}-6p^5-72=\left(p^5-12\right)\left(p^5+6\right) \end{equation*}

Example 10.8.15. Factoring Expressions with Two Variables.

Completely factor the expression \(x^2-3xy-70y^2\text{.}\)

Explanation.

If an expression has two variables, like \(x^2-3xy-70y^2\text{,}\) we pretend for a moment that the expression is \(x^2-3x-70\text{.}\) To factor this expression we ask ourselves “what two numbers multiply to be \(-70\) and add to be \(-3\text{?}\)” The two numbers in question are \(7\) and \(-10\text{.}\) So \(x^2-3x-70\) factors as \((x+7)(x-10)\text{.}\)

To go back to the original problem now, make the two numbers \(7y\) and \(-10y\text{.}\) So, the full factorization is

\begin{equation*} x^2-3xy-70y^2=(x+7y)(x-10y) \end{equation*}

With problems like this, it is important to verify the your answer to be sure that all of the variables ended up where they were supposed to. So, to verify, FOIL your answer.

\begin{align*} (x+7y)(x-10y)\amp=x^2-10xy+7yx-70y^2\\ \amp=x^2-10xy+7\highlight{xy}-70y^2\\ \amp\confirm{=}x^2-3xy-70y^2 \end{align*}

Example 10.8.16.

Completely factor the expressions.

\(\displaystyle x^2-11x+30\)
\(\displaystyle -s^2+3s+28\)
\(\displaystyle g^2-3g-24\)
\(\displaystyle w^2-wr-30r^2\)
\(\displaystyle z^8+2z^4-63\)

Explanation.

\(\displaystyle x^2-11x+30=(x-6)(x-5)\)
\(\displaystyle \begin{aligned}[t] -s^2+3s+28\amp=-\mathopen{}\left(s^2-3s-28\right)\mathclose{}\\ \amp=-(s-7)(s+4)\end{aligned}\)
\(g^2-3g-24\) is prime. No two integers multiply to be \(-24\) and add to be \(-3\text{.}\)
\(\displaystyle w^2-wr-30r^2=(w-6r)(w+5r)\)
\(\displaystyle z^8+2z^4-63=\left(z^4-7\right)\left(z^4+9\right)\)

Subsection 10.8.4 Factoring Trinomials with Non-Trivial Leading Coefficient

In Section 4 we covered factoring trinomials of the form \(ax^2+bx+c\) when \(a\neq1\) using the AC method.

Example 10.8.17. Using the AC Method.

Completely factor the expression \(9x^2-6x-8\text{.}\)

Explanation.

To factor the expression \(9x^2-6x-8\text{,}\) we first find \(ac\text{:}\)

\(9\cdot(-8)=-72\text{.}\)

Examine factor pairs that multiply to \(-72\text{,}\) looking for a pair that sums to \(-6\text{:}\)

Factor Pair	Sum of the Pair
\(1\cdot-72\)	\(-71\)
\(2\cdot-36\)	\(-34\)
\(3\cdot-24\)	\(-21\)
\(4\cdot-18\)	\(-14\)
\(6\cdot-12\)	\(-6\)
\(8\cdot-9\)	(no need to go this far)

Factor Pair	Sum of the Pair
\(-1\cdot72\)	(no need to go this far)
\(-2\cdot36\)	(no need to go this far)
\(-3\cdot24\)	(no need to go this far)
\(-4\cdot18\)	(no need to go this far)
\(-6\cdot12\)	(no need to go this far)
\(-8\cdot9\)	(no need to go this far)

Intentionally break up the \(-6\) as \(6+(-12)\) and then factor using grouping:

\begin{align*} 9x^2\overbrace{{}-6x}-8\amp=9x^2\overbrace{{}+6x-12x}-8\\ \amp=\left(9x^2+6x\right)+(-12x-8)\\ \amp=3x\highlight{(3x+2)}-4\highlight{(3x+2)}\\ \amp=\highlight{(3x+2)}(3x-4) \end{align*}

Example 10.8.18.

Completely factor the expression \(3x^2+5x-6\text{.}\)

Explanation.

First note that there is no GCF besides \(1\) and that \(ac=-18\text{.}\) To look for two factors of \(-18\) that add up to \(5\text{,}\) we will make a factor pair table.

Factor Pair	Sum of the Pair
\(1\cdot-18\)	\(-17\)
\(2\cdot-9\)	\(-7\)
\(3\cdot-6\)	\(-3\)

Factor Pair	Sum of the Pair
\(-1\cdot18\)	\(17\)
\(-2\cdot9\)	\(7\)
\(-3\cdot6\)	\(3\)

Since none of the factor pairs of \(-18\) sum to \(5\text{,}\) we must conclude that this trinomial is prime. The only way to factor it is \(3x^2+5x-6=1\left(3x^2+5x-6\right)\text{.}\)

Example 10.8.19.

Completely factor the expression \(3y^2+20y-63\text{.}\)

Explanation.

First note that \(ac=-189\text{.}\) Looking for two factors of \(-189\) that add up to \(20\text{,}\) we find \(27\) and \(-7\text{.}\) Breaking up the \(+20\) into \(+27-7\text{,}\) we can factor using grouping.

\begin{align*} 3y^2\overbrace{{}+20y}-63\amp=3y^2\overbrace{{}+27y-7y}-63\\ \amp=\left(3y^2+27y\right)+\left(-7y-63\right)\\ \amp=3y\highlight{\left(y+9\right)}-7\highlight{\left(y+9\right)}\\ \amp=\highlight{\left(y+9\right)}(3y-7) \end{align*}

Example 10.8.20. Factoring in Stages with the AC Method.

Completely factor the expression \(8y^3+54y^2+36y\text{.}\)

Explanation.

Recall that some trinomials need to be factored in stages: the first stage is always to factor out the GCF. To factor \(8y^3+54y^2+36y\text{,}\) first note that the GCF of the three terms in the expression is \(2y\text{.}\) Then apply the AC method:

\begin{align*} 8y^3+54y^2+36y\amp=2y\left(4y^2+27y+18\right)\\ \end{align*}

Now we find \(ac=4\cdot18=72\text{.}\) What two factors of \(72\) add up to 27? After checking a few numbers, we find that \(3\) and \(24\) fit the requirements. So:

\begin{align*} \phantom{8y^3+54y^2+36y}\amp=2y\left(4y^2\overbrace{+27y}+18\right)\\ \amp=2y\left(4y^2\overbrace{+3y+24y}+18\right)\\ \amp=2y\left(\left(4y^2+3y\right)+\left(24y+18\right)\right)\\ \amp=2y\left(y\highlight{\left(4y+3\right)}+6\highlight{\left(4y+3\right)}\right)\\ \amp=2y\highlight{\left(4y+3\right)}\left(y+6\right) \end{align*}

Example 10.8.21.

Completely factor the expression \(18x^3+26x^2+4x\text{.}\)

Explanation.

First note that there is a GCF of \(2x\) which should be factored out first. Doing this leaves us with \(18x^3+26x^2+8x=2x\left(9x^2+13x+4\right)\text{.}\) Now we apply the AC method on the factor in the parentheses. So, \(ac=36\text{,}\) and we must find two factors of \(36\) that sum to be \(13\text{.}\) These two factors are \(9\) and \(4\text{.}\) Now we can use grouping.

\begin{align*} 18x^3+26x^2+8x\amp=2x\left(9x^2\overbrace{{}+13x}+4\right)\\ \amp=2x\left(9x^2\overbrace{{}+9x+4x}+4\right)\\ \amp=2x\left(\left(9x^2+9x\right)+\left(4x+4\right)\right)\\ \amp=2x\left(9x\highlight{\left(x+1\right)}+4\highlight{\left(x+1\right)}\right)\\ \amp=2x\highlight{(x+1)}(9x+4) \end{align*}

Subsection 10.8.5 Factoring Special Forms

In Section 5 we covered how to factor binomials and trinomials using formulas. Using these formulas, when appropriate, often drastically increased the speed of factoring. Below is a summary of the formulas covered. For each, consider that \(A\) and \(B\) could be any algebraic expressions.

Difference of Squares: \(A^2-B^2=(A+B)(A-B)\)
Perfect Square Sum: \(A^2+2AB+B^2=(A+B)^2\)
Perfect Square Difference: \(A^2-2AB+B^2=(A+B)^2\)

Example 10.8.22. Factoring the Form \(A^2-2AB+B^2\).

Completely factor the expression \(16y^2-24y+9\text{.}\)

Explanation.

To factor \(16y^2-24y+9\) we notice that the expression might be of the form \(A^2-2AB+B^2\text{.}\) To find \(A\) and \(B\text{,}\) we mentally take the square root of both the first and last terms of the original expression. The square root of \(16y^2\) is \(4y\) since \((4y)^2=4^2y^2=16y^2\text{.}\) The square root of \(9\) is \(3\text{.}\) So, we conclude that \(A=4y\) and \(B=3\text{.}\) Recall that we now need to check that the \(24y\) matches our \(2AB\text{.}\) Using our values for \(A\) and \(B\text{,}\) we indeed see that \(2AB=-2(4y)(3)=24y\text{.}\) So, we conclude that

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

Example 10.8.23. Mixed Special Forms Factoring.

Completely factor the expression \(9w^2+12w+4\text{.}\)
Completely factor the expression \(4q^2-81\text{.}\)
Completely factor the expression \(9p^2+25\text{.}\)
Completely factor the expression \(121b^2-36\text{.}\)
Completely factor the expression \(25u^2-70u+49\text{.}\)

Explanation.

The first step for each problem is to try to fit the expression to one of the special factoring forms.

To factor \(9w^2+12w+4\) we notice that the expression might be of the form \(A^2+2AB+B^2\) where \(A=3w\) and \(B=2\text{.}\) With this formula we need to check the value of \(2AB\) which in this case is \(2AB=2(3w)(2)=12w\text{.}\) Since the value of \(2AB\) is correct, the expression must factor as

\begin{equation*} 9w^2+12w+4=(3w+2)^2 \end{equation*}
To factor \(4q^2-81\) we notice that the expression is of the form \(A^2-B^2\) where \(A=2q\) and \(B=9\text{.}\) Thus, the expression must factor as

\begin{equation*} 4q^2-81=(2q-9)(2q+9) \end{equation*}
To factor \(9p^2+25\) we notice that the expression is of the form \(A^2+B^2\text{.}\) This is called a sum of squares. If you recall from the section, the sum of squares is always prime. So \(9p^2+25\) is prime.
To completely factor the expression \(121b^2-36\) first note that the expression is of the form \(A^2-B^2\) where \(A=11b\) and \(B=6\text{.}\) So, the expression factors as

\begin{equation*} 121b^2-36=(11b+6)(11b-6)\text{.} \end{equation*}
To completely factor the expression \(25u^2-70u+49\) first note that the expression might be of the form \(A^2-2AB+B^2\) where \(A=5u\) and \(B=7\text{.}\) Now, we check that \(2AB\) matches the middle term: \(2AB=2(5u)(7)=70u\text{.}\) So, the expression factors as

\begin{equation*} 25u^2-70u+49=(5u-7)^2\text{.} \end{equation*}

Subsection 10.8.6 Factoring Strategies

In Section 6 we covered a factoring decision tree to help us decide what methods to try when factoring a given expression. Remember to always factor out the GCF first.

Example 10.8.24.

Factor the expressions using an effective method.

\(24xy-20x-18y+15\text{.}\)
\(12t^2+36t+27\text{.}\)
\(8u^2+14u-9\text{.}\)
\(18c^2-98p^2\text{.}\)

Explanation.

To factor the expression \(24xy-20x-18y+15\text{,}\) we first look for a GCF. Since the GCF is \(1\text{,}\) we can move further on the flowchart. Since this is a four-term polynomial, we will try grouping.

\begin{align*} 24xy-20x-18y+15\amp=24xy+(-20x)+(-18y)+15\\ \amp=(24xy-20x)+(-18y+15)\\ \amp=\highlight{4x}(6y-5)+\highlight{(-3)}(6y-5)\\ \amp=\highlight{4x}\overbrace{(6x-5)}\highlight{{}-3}\overbrace{(6x-5)}\\ \amp=(6x-5)\highlight{(4x-3)} \end{align*}
To factor the expression \(12t^2+36t+27\text{,}\) we first look for a GCF. Since the GCF is \(\highlight{3}\text{,}\) first we will factor that out.

\begin{align*} 12t^2+36t+27\amp=\highlight{3}\left(4t^2+12t+9\right)\\ \end{align*}
Next, we can note that the first and last terms are perfect squares where \(A^2=4t^2\) and \(B=9\text{;}\) so \(A=2t\) and \(B=3\text{.}\) To check the middle term, \(2AB=12t\text{.}\) So the expression factors as a perfect square.
\begin{align*} 12t^2+36t+27\amp=\highlight{3}\left(4t^2+12t+9\right)\\ \amp=\highlight{3}(2t+3)^2 \end{align*}
To factor the expression \(8u^2+14u-9\text{,}\) we first look for a GCF. Since the GCF is \(1\text{,}\) we can move further on the flowchart. Since the expression is a trinomial with leading coefficient other than \(1\text{,}\) we should try the AC method. Note that \(AC=-72\) and factor pairs of \(-72\) that add up to \(14\) are \(\highlight{18}\) and \(\highlight{-4}\text{.}\)

\begin{align*} 8u^2+\firsthighlight{14u}-9\amp=8u^2+\firsthighlight{18u-4u}-9\\ \amp=\left(8u^2+18\right)+(-4u-9)\\ \amp=\secondhighlight{2u}(4u+9)\secondhighlight{{}-1}(4u+9)\\ \amp=\secondhighlight{(2u-1)}(4u+9) \end{align*}
To factor the expression \(18c^2-98p^2\text{,}\) we first look for a GCF. Since the GCF is \(\highlight{2}\text{,}\) first we will factor that out.

\begin{align*} 18c^2-98p^2\amp=\highlight{2}\left(9c^2-49p^2\right)\\ \end{align*}
Now we notice that we have a binomial where both the first and second terms can be written as squares: \(9c^2=(3c)^2\) and \(49p^2=(7p)^2\text{.}\)
\begin{align*} 18c^2-98p^2\amp=\highlight{2}\left(9c^2-49p^2\right)\\ \amp=\highlight{2}(3c-7p)(3c+7p) \end{align*}

Subsection 10.8.7 Solving Quadratic Equations by Factoring

In Section 7 we covered the zero product property and learned an algorithm for solving quadratic equations by factoring.

Example 10.8.25. Solving Using Factoring.

Solve the quadratic equations using factoring.

\(\displaystyle x^2-2x-15=0\)
\(\displaystyle 4x^2-40x=-96\)
\(\displaystyle 6x^2+x-12=0\)
\(\displaystyle (x-3)(x+2)=14\)
\(\displaystyle x^3-64x=0\)

Explanation.

Use factor pairs.

\begin{align*} x^2-2x-15\amp=0\\ (x-5)(x+3)\amp=0 \end{align*}

\begin{align*} x-5\amp=0 \amp\text{ or }\amp\amp x+3\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=-3 \end{align*}

So the solution set is \(\{5,-3\}\text{.}\)
Start by putting the equation in standard form and factoring out the greatest common factor.

\begin{align*} 4x^2-40x\amp=-96\\ 4x^2-40x+96\amp=0\\ 4\left(x^2-10x+24\right)\amp=0\\ 4(x-6)(x-4)\amp=0 \end{align*}

\begin{align*} x-6\amp=0 \amp\text{ or }\amp\amp x-4\amp=0\\ x\amp=6 \amp\text{ or }\amp\amp x\amp=4 \end{align*}

So the solution set is \(\{4,6\}\text{.}\)
Use the AC method.

\begin{align*} 6x^2+x-12\amp=0\\ \end{align*}
Note that \(a\cdot c=-72\) and that \(\firsthighlight{9\cdot-8}=-72\) and \(\firsthighlight{9-8}=1\)
\begin{align*} 6x^2\mathbin{\firsthighlight{+}}\firsthighlight{9x-8x}-12\amp=0\\ \left(6x^2+\firsthighlight{9x}\right)+\left(\mathbin{\firsthighlight{-}}\firsthighlight{8x}-12\right)\amp=0\\ \secondhighlight{3x}\left(2x+3\right)\mathbin{\secondhighlight{-4}}\left(2x+3\right)\amp=0\\ \left(2x+3\right)\secondhighlight{\left(3x-4\right)}\amp=0 \end{align*}

\begin{align*} 2x+3\amp=0 \amp\text{ or }\amp\amp \highlight{3x-4}\amp=0\\ x\amp=-\frac{3}{2} \amp\text{ or }\amp\amp x\amp=\highlight{\frac{4}{3}} \end{align*}

So the solution set is \(\left\{-\frac{3}{2},\frac{4}{3}\right\}\text{.}\)
Start by putting the equation in standard form.

\begin{align*} (x-3)(x+2)\amp=14\\ x^2-x-6\amp=14\\ x^2-x-20\amp=0\\ (x-5)(x+4)\amp=0 \end{align*}

\begin{align*} x-5\amp=0 \amp\text{ or }\amp\amp x+4\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=-4 \end{align*}

So the solution set is \(\{5,-4\}\text{.}\)
Even though this equation has a power higher than \(2\text{,}\) we can still find all of its solutions by following the algorithm. Start by factoring out the greatest common factor.

\begin{align*} x^3-64x\amp=0\\ x\left(x^2-64\right)\amp=0\\ x(x-8)(x+8)\amp=0 \end{align*}

\begin{align*} x\amp=0 \amp\text{ or }\amp\amp x-8\amp=0 \amp\text{ or }\amp\amp x+8\amp=0\\ x\amp=0 \amp\text{ or }\amp\amp x\amp=8 \amp\text{ or }\amp\amp x\amp=-8 \end{align*}

So the solution set is \(\{0,8,-8\}\text{.}\)