Lane ORCCA (2020–2021): Open Resources for Community College Algebra

Break down each of these into its factors:

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

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

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

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.

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

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:

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

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

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.

1. What two numbers multiply to be \(6\) and add to be \(5\text{?}\) The numbers are \(2\) and \(3\text{.}\)

2. What two numbers multiply to be \(-6\) and add to be \(5\text{?}\) The numbers are \(6\) and \(-1\text{.}\)

3. What two numbers multiply to be \(-6\) and add to be \(-1\text{?}\) The numbers are \(-3\) and \(2\text{.}\)

4. What two numbers multiply to be \(24\) and add to be \(-10\text{?}\) The numbers are \(-6\) and \(-4\text{.}\)

5. What two numbers multiply to be \(-24\) and add to be \(2\text{?}\) The numbers are \(6\) and \(-4\text{.}\)

6. What two numbers multiply to be \(-24\) and add to be \(-5\text{?}\) The numbers are \(-8\) and \(3\text{.}\)

7. What two numbers multiply to be \(420\) and add to be \(44\text{?}\) The numbers are \(30\) and \(14\text{.}\)

8. What two numbers multiply to be \(-420\) and add to be \(-23\text{?}\) The numbers are \(-35\) and \(12\text{.}\)

9. 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 \(1\) and we work our way up the factors of 420.

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 \(-23\text{,}\) simply 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 \(-35\text{.}\)

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 7.3, 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.

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

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

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

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:

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

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, simply make the two numbers \(7y\) and \(-10y\text{.}\) So, the full factorization is

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

Note that when two or more variables are multiplied together in a term, it is common practice to list the variables in alphabetical order, which is why we rewrote \(7yx\) as \(7xy\text{.}\) This makes it easier to identify like terms.

1. \(\displaystyle x^2-11x+30=(x-6)(x-5)\)

2. \(\displaystyle -s^2+3s+28=-\left(s^2-3s-28\right)=-(s-7)(s+4)\)

3. \(g^2-3g-24\) is prime. No two integers multiply to be \(-24\) and add to be \(-3\text{.}\)

4. \(\displaystyle w^2-wr-30r^2=(w-6r)(w+5r)\)

5. \(\displaystyle z^8+2z^4-63=\left(z^4-7\right)\left(z^4+9\right)\)

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

1. \(9\cdot(-8)=-72\text{.}\)

2. 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)

3. 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*}

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.

Since none of the factor pairs of \(-18\) sum to \(5\text{,}\) we must conclude that this trinomial is prime.

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.

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:

First, note that there is a GCF of \(2x\text{,}\) 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, \(ac=36\) 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.

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 find what needs to be squared to obtain both the first and last terms of the original expression. We determine that \(A=4y\) since \((4y)^2=4^2y^2=16y^2\) and \(B=3\) since \(3^2=9\) . 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

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

1. To factor \(v^3-27\) we notice that the expression is of the form \(A^3-B^3\text{.}\) To find values for \(A\) and \(B\text{,}\) mentally determine what needs to be cubed to obtain the first and last terms. So, \(A=v\) and \(B=3\text{,}\) and using the form \(A^3-B^3=(A-B)(A^2+AB+B^2)\text{,}\) we have that

   \begin{align*} v^3-27\amp=(v-3)\left(v^2+(v)(3)+3^2\right)\\ \amp=(v-3)\left(v^2+3v+9\right) \end{align*}

2. 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 \(2AB\text{,}\) 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*}

3. 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*}

4. 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.

5. 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*}

6. 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*}

7. To completely factor the expression \(64q^3-27y^3\) first note that the expression is of the form \(A^3-B^3\) where \(A=4q\) and \(B=3y\text{.}\) So, the expression factors as

   \begin{align*} 64q^3-27y^3\amp=(4q-3y)\left((4q)^2+(4q)(3y)+(3y)^2\right)\\ \amp=(4q-3y)\left(16q^2+12qy+9y^2\right) \end{align*}

1. 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=4x\highlight{\overbrace{(6x-5)}}-3\highlight{\overbrace{(6x-5)}}\\ \amp=\highlight{(6x-5)}(4x-3) \end{align*}

2. 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*}

3. 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+\lighthigh{14u}-9\amp=8u^2+\lighthigh{18u-4u}-9\\ \amp=\left(8u^2+18\right)+(-4u-9)\\ \amp=2u\highlight{(4u+9)}-1\highlight{(4u+9)}\\ \amp=\highlight{(4u+9)}(2u-1) \end{align*}

4. 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*}

1. 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 \(\{5,-3\}\text{.}\)

2. 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 \(\{4,6\}\text{.}\)

3. Use the AC method.

   \begin{align*} 6x^2+x-12\amp=0\\ \end{align*}

   Note that \(a\cdot c=-72\) and that \(\highlight{9\cdot-8}=-72\) and \(\highlight{9-8}=1\)

   \begin{align*} 6x^2\substitute{+9x-8x}-12\amp=0\\ \left(6x^2+9x\right)+\left(-8x-12\right)\amp=0\\ \highlight{3x}\left(2x+3\right)\mathbin{\highlight{-4}}\left(2x+3\right)\amp=0\\ \left(2x+3\right)\highlight{\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 \(\left\{-\frac{3}{2},\frac{4}{3}\right\}\text{.}\)

4. 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 \(\{5,-4\}\text{.}\)

5. 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 \(\{0,8,-8\}\text{.}\)