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# Algebra and Probability Study Guide 2 for McGraw-Hill's ASVAB

By McGraw-Hill Professional
Updated on Jun 26, 2011

Practice questions for this study guide can be found at:

Algebra and Probability Practice Problems for McGraw-Hill's ASVAB

### Monomials, Binomials, and Polynomials

You can guess by the prefix mono- that a monomial has something to do with "one." A monomial is a mathematical expression consisting of only one term. Examples include 12x, 3a2, and 9abc.

• A binomial (the prefix bi- means "two") has exactly two terms: 12z + j.
• A polynomial, as indicated by the prefix poly-, meaning "many," has two or more terms.

Examples include x + y, x + y + z, and y2 – 2z + 12.

Multiplying Monomials When multiplying monomials, multiply any numbers, then multiply unknowns. Add any exponents. Keep in mind that in a term like x or 2x, the x is understood to have the exponent 1 even though the 1 is not shown.

Dividing Monomials To divide monomials, divide the numbers and subtract any exponents (the exponent of the divisor from the exponent of the number being divided).

Adding and Subtracting Polynomials Arrange the expressions in columns with like terms in the same column. Add or subtract like terms.

Multiplying Polynomials To multiply polynomials, multiply each term in the first polynomial by each term in the second polynomial. The process is just like regular multiplication. For example, if you multiply 43 times 12, the problem looks like this:

Dividing a Polynomial by a Monomial Just divide the monomial into each term of the polynomial.

Dividing a Polynomial by a Polynomial To divide a polynomial by another polynomial, first make sure the terms in each polynomial are in descending order (i.e., cube → square → first power).

For example, 6c + 3c2 + 9 should be written 3c2 + 6c + 9.

10 + 2c + 5c2 should be written 5c2 + 2c + 10.

Then use long division to solve the problem.

Factoring a Polynomial A factor is a number that is multiplied to get a product. Factoring a mathematical expression is the process of finding out which numbers, when multiplied together, produce the expression.

To factor a polynomial, follow these two steps:

• Find the largest common monomial in the polynomial. This is the first factor.
• Divide the polynomial by that monomial. The result will be the second factor.

Special Case: Factoring the Difference between Two Squares

Examples

Factor the following expression:

y2 – 100

In this expression, each term is a perfect square; that is, each one has a real-number square root. The square root of y2 is y, and the square root of 100 is 10.

When an expression is the difference between two squares, its factors are the sum of the squares (y +10) and the difference of the squares (y – 10). Multiplying the plus sign and the minus sign in the factors gives the minus sign in the original expression.

Factoring Polynomials in the Form ax2 + em>bx + c, where a, b, and c are numbers Remember that you want to find two factors that when multiplied together produce the original expression.

Examples

Factor the expression:

x2 + 5x + 6

First, you know that x times x will give x2, so it is likely that each factor is going to start with x.

(x   ) (x   )

Now you need to find two factors of 6 that when added together give the middle term of 5. Some options are 1 and 6 and 2 and 3. 2 and 3 add to 5, so add those numbers to your factors. Now you have

(x 2) (x 3)

Finally, deal with the sign. Since the original expression is all positive, both signs in the factors must be positive. So the two factors must be

(x + 2)(x + 3)

Check your work by multiplying the two factors to see if you come up with the original expression.

(x + 2)(x + 3) = x2 + 5x + 6

Factor the expression:

6x2 + 8x = 8

6x2 can be factored into either (6x)(x) or (2x)(3x). Using the latter, the first terms in our factors are as follows:

(2x ) (3x )

Now let's consider the –8. Factors of 8 can be (8)(1) or (2)(4). Let's try 2 and 4, so our factors are now:

(2x   2)(3x   4)

Now for the signs. In order to get a minus 8 in the original expression, one of the numbers must be a negative and the other a positive. Let's try making the 4 negative, making the factors:

(2x + 2)(3x – 4) = 6x2 – 2x – 8

That's close, but the original expression was 6x2 + 8x – 8, not 6x2 – 2x – 8. What if we switched the numbers 2 and 4?

(2x + 4)(3x – 2) = 6x2 + 8x – 8

Now the factors give the original expression when multiplied together, so this is the correct answer.