Adding, Multiplying and Using FOIL
Sums and differences:
𝐄𝐱𝐚𝐦𝐩𝐥𝐞: (5𝑟 2 + 7𝑟 − 1) + (𝑟 + 4 − 5𝑟 2 ) =
5𝑟 2 + 7𝑟 − 1 + 𝑟 + 4 − 5𝑟 2 =
5𝑟 2 − 5𝑟 2 + 7𝑟 + 𝑟 − 1 + 4 =
0𝑟 2 + 8𝑟 + 3 = 8𝑟 + 3
𝐄𝐱𝐚𝐦𝐩𝐥𝐞: (𝑟 2 + 7𝑟 3 ) − (3𝑟 2 + 4𝑟 − 12) =
(𝑟 2 + 7𝑟 3 ) − 1(3𝑟 2 + 4𝑟 − 12) =
𝑟 2 + 7𝑟 3 − 3𝑟 2 − 4𝑟 + 12 =
7𝑟 3 + 𝑟 2 − 3𝑟 2 − 4𝑟 + 12 = 7𝑟 3 − 2𝑟 2 − 4𝑟 + 12
Distributive Property
𝐄𝐱𝐚𝐦𝐩𝐥𝐞: 6𝑥(𝑥 − 5) − 4(𝑥 2 − 𝑥) =
6𝑥 2 − 30𝑥 − 4𝑥 2 + 4𝑥 = 2𝑥 2 − 26𝑥
Remember, everything on the outside of a set of parentheses needs to be multiplied by everything on the
inside of that set of parentheses. This means that the 6x is multiplied by the x to get 6𝑥 2 and by the -5 to
get -30x. Use this same pattern for all groups of parentheses.
When factoring out the GCF (Greatest Common Factor), find a number, letter, or number/letter
combination that is a factor of all the terms.
Example: Factor 4𝑥 2 + 8𝑥 − 2
If you break each term into its prime factors, you will be able to find the
numbers or letters that are common to all three terms.
4𝑥 2 = 2 ∙ 2 ∙ 𝑥 ∙ 𝑥 = 2(2𝑥 2 )
8𝑥 = 2 ∙ 2 ∙ 2 ∙ 𝑥 = 2(4𝑥)
−2 = −1 ∙ 2 = 2(−1)
I have highlighted the number (2) that is common to all three terms. So this factors to 2(2𝑥 2 + 4𝑥 − 1).
Notice that if you multiply this out you get the original equation. The numbers inside the parentheses are
the three terms inside the parentheses that are left over after you factor out the GCF.
Example: Factor 3𝑥 4 + 6𝑥 3 − 12𝑥
Break down each term into its prime factors.
3𝑥 4 = 3 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 = 3𝑥(𝑥 3 )
6𝑥 3 = 2 ∙ 3 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 = 3𝑥(2𝑥 2 )
−12𝑥 = −1 ∙ 2 ∙ 2 ∙ 3 ∙ 𝑥 = 3𝑥(−4)
I have shown the common factors of all three terms in red, so 3x is the GCF and this factors to
3𝑥(𝑥 3 + 2𝑥 2 − 4).
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Over time, this will become much easier and you will be able to see the GCF fairly quickly and simply pull
it out of the parentheses. Just remember that when you multiply (distribute) across the parentheses, you
need to get the original expression.
Example: Factor 18𝑥 2 𝑎3 + 6𝑥𝑎2 − 2𝑥𝑎.
Notice that all three terms have a common factor of 2xa. This factors to 2𝑥𝑎(9𝑥𝑎 + 3𝑎 − 1).
If you multiply this out, you get 18𝑥 2 𝑎3 + 6𝑥𝑎2 − 2𝑥𝑎 back again.
Example of how the distributive property works with two sets of parentheses using only numbers.
The first method adds the numbers in the parentheses first, then multiplies the result. This is normal
order of operations. The second method takes the first number from the first set of parentheses and
multiplies it individually by both numbers in the second set of parentheses. This is repeated for the
second number in the first set of parentheses.
Method 1 (regular order of
operations)
(2 + 3)(6 − 4) =
5 ∙ 2 = 10
Method 2 (using the distributive rule for
multiplying binomials)
𝑜𝑟 ⟹
(2 + 3)(6 − 4) =
2 ∙ 6 + 2(−4) + 3 ∙ 6 + 3(−4) =
12 − 8 + 18 − 12 = 10
Notice that you get the same answer either way. Method 2 leads us to FOIL.
FOIL = First Outer Inner Last
𝐄𝐱𝐚𝐦𝐩𝐥𝐞: (𝑥 + 2)(𝑥 − 3)
𝐅 = 𝑥 ∙ 𝑥 𝐎 = 𝑥 ∙ (−3) 𝐈 = 2 ∙ 𝑥 𝐋 = 2 ∙ (−3)
(𝑥 + 2)(𝑥 − 3) = 𝑥 ∙ 𝑥 + 𝑥(−3) + 2𝑥 + 2(−3)
= 𝑥 2 − 3𝑥 + 2𝑥 − 6
= 𝑥2 − 𝑥 − 6
𝐄𝐱𝐚𝐦𝐩𝐥𝐞: (5𝑥 − 4)(2𝑥 + 3)
𝐅 = 5𝑥 ∙ 2𝑥 𝐎 = 5𝑥 ∙ (3) 𝐈 = −4 ∙ 2𝑥 𝐋 = −4 ∙ (3)
(5𝑥 − 4)(2𝑥 + 3) = 5𝑥 ∙ 2𝑥 + 5𝑥(3) + (−4) ∙ 2𝑥 + (−4)(3)
= 10𝑥 2 + 15𝑥 − 8𝑥 − 12
= 10𝑥 2 + 7𝑥 − 12
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This method also works with more than two terms inside the parentheses. Just take each term in
the first set of parentheses and multiply it by each term in the second set of parentheses.
Example: (4𝑥 2 − 7𝑥 + 3)(5𝑥 − 2)
= (4𝑥 2 )(5𝑥) + (4𝑥 2 )(−2) + (−7𝑥)(5𝑥) + (−7𝑥)(−2) + 3(5𝑥) + 3(−2)
= 20𝑥 3 − 8𝑥 2 − 35𝑥 2 + 14𝑥 + 15𝑥 − 6
= 20𝑥 3 − 43𝑥 2 + 29𝑥 − 6
It also works with the imaginary number 𝒊, 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝒊 = √−𝟏. Since 𝑖 = √−1, then 𝑖 2 = −1. Check
the “i packet” on my website for more information.
𝐄𝐱𝐚𝐦𝐩𝐥𝐞: (𝑥 − 8𝑖)(𝑥 + 𝑖) = 𝑥 ∙ 𝑥 + 𝑥 ∙ 𝑖 + (−8𝑖)𝑥 + (−8𝑖)𝑖
= 𝑥 2 + 𝑥𝑖 − 8𝑥𝑖 − 8𝑖 2
= 𝑥 2 − 7𝑥𝑖 − 8(−1)
= 𝑥 2 − 7𝑥𝑖 + 8
= (𝑥 2 + 8) − 7𝑥𝑖
Notice on this last one that after turning 𝑖 2 into − 1 and combining 𝑥𝑖 and − 8𝑥𝑖, there was no real way to
combine like terms. The rule with imaginary numbers is to combine all the real numbers (including
variables) together on the left hand side of the expression and all the imaginary numbers (ones with i in
them) on the right hand side of the expression.
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