Algebra 2: Chapter 5 Guideline on Polynomials, Factoring, and Multiplying
Example 1: Finding the Value of a Polynomial Function
A polynomial function is a function that can be
defined as a polynomial. For this one, all you have
to do is plug the ‘x’ value back into the equation.
P(2)  (2) 2  2  1
What just happened was that plugged back 2 into
the equation and then simplified it.
P(2)  4  2  1
P(2)  7
Example 2: Adding and Subtracting Polynomials
2a) Add 5x 5  3x 3  7x  1 and 4x 5  3x 3  6x 2  5 When adding polynomial functions, you’re just
going to be combining like terms
5x 5  3x 3
 7x  1
+ 4x 5  3x 3  6x 2
5
9x 5  6x 2  7x  4
Notice how I realigned the terms with their
appropriate degree. The x3 terms cancel out and
those empty spaces are filled with invisible zeroes.
2b) Subtract 5x  7xy  6x 2 y and 2x  3xy  4x 2 y
6x 2 y  7xy  5x
4x 2 y  3xy  2x
When subtracting, you will always find the additive inverse of
the second set of terms. The additive inverse of the second set
will be -4x2y - 3xy - 2x, which was done ahead of time.
2x 2 y  4xy  3x
Example 3: Multiplying Polynomials (Vertical Multiplication)
Multiply 2a 2  ab  b 2 and a  2b
2a 2  ab  b 2
a  2b
2a  a 2 b  ab 2
3
 4a 2 b  2ab 2  2b 3
2a 3  5a 2 b  3ab 2  2b 3
When multiplying a polynomial with a binomial, using the
method of vertical multiplication is the best one to use. You
would treat this problem like when you multiply whole
numbers, as you can see.
Example 4: Methods to Multiply Binomials
4a) The F.O.I.L. Method
O
F
(A  B)(C  D)  AC  AD  BC  BD
The F.O.I.L method is used commonly
when you have four unlike terms. Before,
students were only taught to use this
method for all types of binomials even
those with same two terms ‘A’ and ‘B’.
I
L
(3x  2y)(5x  y)
(3x)(5x)  (3x)(y)  (2y)(5x)  (2y)(y)
15x 2  3xy  10xy  2y 2
15x 2  13xy  2y 2
4b) Square of Binomials
The square of a binomial is the square of the first expression, plus or minus twice the
produce of the expressions, plus the square of the second expression.
(A  B) 2
(A  B) 2
A 2  2AB  B 2
A 2  2AB  B 2
(x  4) 2
(2x  3) 2
(x) 2  2(4)(x)  (4) 2
(2x) 2  2(3)(2x)  (3) 2
x 2  8x  16
4x 2  12x  9
*Pay attention to the “skeletons” of each equation. The only difference between the
formulas is the operation used in the second term of the equation.
4c) Sum and Difference
The product of the sum and difference of two expressions is the square of the first
expression minus the square of the second.
(A  B)(A  B)  A 2  B 2
(x  6)(x  6)
(3x  4y)(3x  4y)
(x) 2  (6) 2
(3x) 2  (4y) 2
x 2  36
9x 2  16y 2
4d) Cubing Binomials
The cube of a binomial is the cube of the first expression, plus or minus three times the
product of the first expression squared times the second plus three times the first
expression times the second squared plus or minus the cube of the second expression.
(A  B) 3
(A  B) 3
A 3  3A 2 B  3AB 2  B 3
A 3  3A 2 B  3AB 2  B 3
(2x  3y) 3
(t  7) 3
(2x) 3  3(2x) 2 (3y)  3(2x)(3y) 2  (3y) 3
(t) 3  3(t) 2 (7)  3(t)(7) 2  (7) 3
8x 3  36x 2 y  54xy 2  27y 3
t 3  21t 2  147t  343
Just like squaring binomials, pay attention to the change of signs on the second and fourth
terms when cubing binomials.
Example 5: Factoring Methods
Factoring is the process simplifying an expanded form equation. It is basically thought of
as the opposite of multiplication.
5a) Common term factoring is one method that is used when a polynomial has a common
term that is factorable.
5x 4  20x 3
3x 2  12
6u 2 v 3  21uv 2
5x 3 (x  4)
3(x 2  4)
3uv 2 (2uv  7)
Sometimes when there are more than two terms, it gets more difficult to determine what
your common factor is. When that happens, ask yourself if they all have a certain number
or variable that can be pulled out of each equation.
5b) Factoring trinomial squares can be simple yet complicated. Here are the three
conditions that make factoring them possible. Be sure to verify to make sure.
1. Two of the terms must be A2 and B2. That is usually the first and last terms
2. There must be no negative sign before A2 and B2.
3. If we multiply the square roots of A2 and B2 and double the result, we get the
remaining two term which is 2 ∙ A ∙ B, or its additive inverse -2 ∙ A ∙ B
A 2  2AB  B 2  (A  B) 2
x 2  8x  16  (x  4) 2
x 2  x and
16  4
A 2  2AB  B 2  (A  B) 2
9y 2  30y  25  (3y  5) 2
9y 2  3y and
25  5
2(4)(x)  8x
2(3y)(5)  30y Notice that we used -5 instead of 5
8x  8x
30y  30y
Solution verified
Solution verified
5c) Factoring the differences of squares is basically taking the square roots of the ‘A’ and
‘B’ terms and plugging it back into the formula.
A 2  B 2  (A  B)(A  B)
x 2  25  (x  5)(x  5)
x 2  x and
25  5
A  x and B  5
(x 2  6x  9)  25y 2
(x  3) 2  (5y) 2
(x  3  5y)(x  3  5y)
A  x  3 and B  5y
Notice that the second example in 5c is a bit complex. The equation has four terms but
you can use trinomial factoring the factor the first three terms. You can then proceed to
using the differences of squares formula. Since x + 3 is treated as one whole term, the
sign in that expression does not change.
5d) Factor by grouping is a method that is commonly used when you have more than
three terms and can’t use trinomial factoring.
y 2  3y  4y  12  (y 2  3y)  (4y  12)
x 2  5x  4x  20  (x 2  5x)  (4x  20)
y(y  3)  4(y  3)
x(x  5)  4(x  5)
(y  4)(y  3)
(x  4)(x  5)
The best way to do factor by grouping is by separating the first two terms and the last two
terms.