Algebra II: Polynomials and Factoring

On this page we hope to clear up problems you might have with polynomials and factoring. All the different methods of factoring and different things such as the difference of cubes are covered. Read on or follow any of the links below to start gaining a better understanding of polynomials.

Combining like terms.
Multiplication of polynomials.
Factoring
Factoring by grouping
Sums and differences of cubes
Quiz on polynomials and factoring

Combining Like Terms

When terms of a polynomial have the same variables raised to the same powers, the terms are called similar, or like terms. Like terms can be combined to make the polynomial easier to deal with.

Example:

1.  Problem: Combine like terms in the following equation:
             3x^2 - 4y + 2x^2.
   Solution: Rearrange the terms so it is easier to deal with.

             3x^2 + 2x^2 - 4y

             Combine the like terms.

             5x^2 - 4y

Multiplication of Polynomilas

Probably the most important kind of polynomial multiplication that you can learn is the multiplication of binomials (polynomials with two terms). An easy way to remember how to multiply binomials is the FOIL method, which stands for first, outside, inside, last.

Example:

1.  Problem: Multiply (3xy + 2x)(x^2 + 2xy^2).
             Simplify the answer.
   Solution: Multiply the first terms of each binomial.  (F)

             3xy * x^2 = 3(x^3)y

             Multiply the outside terms of each binomail.  (O)

             3xy * 2xy^2 = 6(x^2)y^3

             Multiply the inside terms of each binomial.  (I)

             2x * x^2 = 2x^3

             Multiply the last terms of each binomail.  (L)

             2x * 2xy^2 = 4(x^2)y^2

             You now have a polynomial with four terms.  Combine
             like terms if you can to get a
             simplified answer.  There are no like terms, so you have
             your final answer.

             3(x^3)y + 6x(x^2)y^3 + 2x^3 + 4(x^2)y^2

Although it would be nice if all you ever had to do was multiply binomials by other binomials, that isn't even close to reality. A perfect example of this is when you have to cube a binomial.

Example:

1.  Problem: Multiply (A + B)^3 out.
   Solution: Rewrite so you have something you can actually
             multiply out.

             (A + B)(A + B)(A + B)

             Multiply the first two binomials together.

             (A + B)(A + B)
             A^2 + AB + AB + B^2

             After combining like terms, you have
             A^2 + 2AB + B^2.

             You now have a binomial and a trinomial to multiply
             together.

             (A^2 + 2AB + B^2)(A + B)

             This is a slightly more complicated situation than
             multiplying a binomial by another binomial.  Multiply
             the first term of the binomial by each of the terms in 
             the trinomial and then multiply the last term of the
             binomial by each term in the trinomial.

             A^3 + 2(A^2)B + AB^2 + BA^2 + 2AB^2 + B^3

             Combine like terms if possible to simplify the answer.

             A^3 + 3(A^2)B + 3AB^2 + B^3

Factoring

Factoring is the reverse of multiplication. When factoring, look for common factors.

Example:

1.  Problem: Factor out of a common factor of 4y^2 - 8.
   Solution: 4 is a common factor of both terms, so pull it out and
             write each term as the product of factors.

             4y^2 - (4)2

             Rewrite using the distribution law of multiplication,
             which says that a(b + c) = ab + ac.

             4(y^2 - 2)

Sometimes, you will come across a special situation where both terms of a binomial are squares of another number, such as (x^2 + 9). (x^2 is the square of x and 9 is the square of 3.)

There is a special formula for this situation, so you don't have to factor the binomial. The difference of squares formula is listed below.

A^2 - B^2 = (A + B)(A - B)

Example:

1.  Problem: Factor y^2 - 4.
   Solution: Since y^2 is the square of y, and 4
             is the square of 2, this binomial fits the 
             difference of squares formula.

             y^2 - 4 = (y + 2)(y - 2)

Since trinomials are the most common polynomial you will be asked to factor, we will try to help you better understand how to factor qaudratic trinomials, or trinomials whose highest power is two. Also, we assume you know how to multiply binomials (we use the "FOIL" method).

Using a multiplication problem consisting of two binomials, we will show some important things to remember when factoring trinomials, which is the reverse of multiplying binomials.

Example:

             (x - 6)(x + 3) = x^2 - 6x + 3x - 18 = x^2 - 3x - 18

             1. The first term of the trinomial is the product of the
                first terms of the binomials.

             2. The last term of the trinomial is the product of the
                last terms of the binomials.

             3. The coefficient of the middle term of the tri-
                nomial is the sum of the last terms of the
                binomials.

             4. If all the signs in the trinomial are positive, all
                signs in both binomials are positive.

Keeping those important things in mind, you can factor trinomials.

Example:

1.  Problem: Factor x^2 - 14x - 15.
   Solution: First, write down two sets of parentheses to indicate
             the product.

             (     )(     )

             Since the first term in the trinomial is the product of
             the first terms of the binomials, you enter x as
             the first term of each binomial.

             (x    )(x    )

             The product of the last terms of the binomials must
             equal -15, and their sum must equal -14, and
             one of the binomials' terms has to be negative.  Four
             different pairs of factors have a product that equals
             -15.

             (3)(-5) = -15     (-15)(1) = -15
             (-3)(5) = -15     (15)(-1) = -15

             However, only one of those pairs has a sum of -14.

             (-15) + (1) = -14

             Therefore, the second terms in the binomial are -15
             and 1 because these are the only two factors whose
             product is -15 (the last term of the trinomial) and
             whose sum is -14 (the coefficient of the
             middle term in the trinomial).

             (x - 15)(x + 1) is the answer.

Trinomials and binomials are the most common polynomials, but you will sometimes see polynomials with more than three terms. Someitmes, when you are dealing with polynomials with four or more terms, you can group the terms in such a way that common factors can be found.

Example:

4. Problem: Factor 4x^2 - 3x + 20x - 15.
   Solution: Rearrange the terms so common factors can easily be
             found.

             4x^2 + 20x - 3x - 15

             The first two terms have a common factor in 4x.
             The last two terms have a common factor in 3.
             Factor those terms out.

             4x(x + 5) - 3(x + 5)
 
             Now you have a binomial.  Each term has a factor of
             (x + 5).  Factor that out for the final answer.

             (x + 5)(4x - 3)

Sums and Differences of Cubes

When you have a binomial where each term is the cube of something, such as x^3 - 27, there are formulas that you can apply to the binomial so you don't have to factor them. The two formulas are listed below.

1. Sum of Cubes
A^3 + B^3 = (A + B)(A^2 - AB + B^2)
2. Difference of Cubes
A^3 - B^3 = (A - B)(A^2 + AB + B^2)

Use the formulas whenever you can!

Example:

1.  Problem: Factor 125x^3 + y^3.
   Solution: Write the sum of the cube roots.

             (5x + y)(              )

             Take 5x + y to get the next factor.

             Think of the Sum of Cubes formula 
             (A + B)(A^2 - AB + B^2).  5x would be A and
             y would be B.

             (5x + y)(25x^2 - 5xy + y^2)

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