ORCCA (Draft)Open Resources for Community College Algebra

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

And take note of the common factors.

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

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:

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 \(1\text{,}\) 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 \(-35\text{.}\)

1. \(x^2-11x+30=(x-6)(x-5)\)
2. \begin{align*} -s^2+3s+28\amp=-\left(s^2-3s-28\right)\\ \amp=-(s-7)(s+4) \end{align*}
3. \(g^2-3g-24\) is prime.
4. \(w^2-wr-30r^2=(w-6r)(w+5r)\)
5. \(z^8+2z^4-63=\left(z^4-7\right)\left(z^4+9\right)\)

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.

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.

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. The only way to factor it is \(3x^2+5x-6=1\left(3x^2+5x-6\right)\text{.}\)

1. 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*}
2. 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*}
3. 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 \(-6xy+9y+2x-3\text{,}\) start by noting that the GCF is \(1\text{.}\) Since we don't have to factor a \(1\) out, we continue following the flow chart. We have a four-term polynomial and we should use factoring by grouping. The full process is:

   \begin{align*} -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*}

   We should note that the negative sign in front of the \(3y\) can be factored out if you wish. That would look like:

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

2. To factor the expression \(4w^3-20w^2+24w\text{,}\) 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 flow chart, 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*}

3. To factor the expression \(9-24y+16y^2\text{,}\) start by noting that the GCF is \(1\text{.}\) Since we don't have to factor a \(1\) out, we continue following the flow chart. We now have a trinomial where both the first term, \(9\text{,}\) and last term, \(16y^2\) 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 \(24y\text{:}\)

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

   So the full factorization is:

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

4. To factor the expression \(9-25y+16y^2\text{,}\) start by noting that the GCF is \(1\text{.}\) Since we don't have to factor a \(1\) out, we continue following the flow chart. Since we now have a trinomial where both the first term and last term are perfect squares in exactly the same way as in the solution to part 3. However, we note that if 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 flow chart, 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:

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

5. To factor the expression \(20x^4+13x^3-21x^2\text{,}\) 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 flow chart, 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

   \(1\cdot420\)

   \(2\cdot210\)

   \(3\cdot140\)

   \(4\cdot105\)

   Factor Pair

   \(5\cdot84\)

   \(6\cdot70\)

   \(7\cdot60\)

   \(10\cdot42\)

   Factor Pair

   \(12\cdot35\)

   \(14\cdot30\)

   \(15\cdot28\)

   \(20\cdot21\)

   In the table of the factors of \(420\) we can fairly easily see that the correct factors of \(-420\) in question are \(15\) 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*}