Algebra

Algebra Problems for the CPT

Looking for free practice for the CPT in algebra? You will see a sample algebra test below.

The college placement math test in algebra is one part of the CPT exam.

The algebra part of the test contains twelve questions.

Here is a free sample from the algebra section of our CPT math download.

Example 1:

Polynomials are algebraic expressions that contain more than one term.

You will definitely have questions on the CPT Math Test about multiplying polynomials.

(2a − 3b)2 = ?

Example 2:

This is a question on inequalities. Inequalities contain the less than or greater than signs.

Algebra questions on the CPT will include inequalities problems like this one.

35 − 1/4X > 31, then X < ?

Example 3:

You will also see exam problems on the division of polynomials.

In order to solve this type of problem, you must do long division of the polynomial:

(a2 + a − 12) ÷ (a − 3) = ?


Algebra Problems - Solutions

Solution 1:

When multiplying polynomials, you should use the F-O-I-L method.

This means that you multiply the variables two at a time from each of the two parts of the equation in the parentheses in the order First - Outside - Inside - Last

(2a − 3b)2 = (2a − 3b)(2a − 3b)

So, the first variables in each set of parentheses are 2a and 2a.

FIRST: 2a × 2a = 4a2

The variables on the outside of each set of parentheses are 2a and 3b.

OUTSIDE: 2a × −3b = −6ab

The variables on the inside of each set of parentheses are −3b and 2a.

INSIDE: −3b × 2a = −6ab

The last variables in each set of parentheses are −3b and −3b.

LAST: −3b × −3b = 9b2

Then we add all of the above parts together to get:

4a2 − 6ab − 6ab + 9b2 =

4a2 − 12ab + 9b2

Solution 2:

First, deal with the whole numbers on each side of the equation first:

35 − 1/4X > 31

(35 − 35) − 1/4X > (31 − 35)

1/4X > −4

Then deal with the fraction:

1/4X > −4

4 × −1/4X > −4 × 4

−X > −16

Then, deal with the negative number:

−X > −16

−X + 16 > −16 + 16

−X + 16 > 0

Finally, isolate the unknown variable as a positive number:

−X + 16 > 0

−X + X + 16 > 0 + X

16 > X

X < 16

Solution 3:

Remember, you must to long division of the polynomial until you have no remainder.

                  a + 4
a − 3)a2 + a − 12

      −(a2 − 3a)
                4a − 12

             −(4a − 12)
                          0