  # Intensive Precalculus Diagnostic Pretest

This test is designed to give you and your instructors an idea of your level of skill in doing high school
algebra as you begin Math 1155. You should have been able to do these problems correctly and with
little difficulty. If you are really lost doing these problems you should talk to Jessica or Michelle.

Go through the solutions to the problems and be sure you understand how to do them.

1. Simplify: Follow the order of operations for simplifying arithmetic expressions:

Step 1 Simplify expressions inside grouping symbols, which include parentheses ( ), brackets [ ],
the fraction bar -> a/b absolute value bars | |, and radicals Step 2 Simplify exponents, square roots, and absolute values.
Step 3 Simplify multiplication and division, working left to right.
Step 4 Simplify addition and subtraction, working left to right.

Practice problems: Simplify each of the following. Check the answers at the end of this document. 2. Simplify: Step 1 Find the Least Common Denominator (LCD) by prime factoring each denominator and then
multiplying the largest group of each prime factor.

10 = 2 • 5
12 = 2 • 2 • 3
18 = 2 • 3 • 3

Largest group of 2's is 2 • 2 (in 12)
Largest group of 3's is 3 • 3 (in 18)
Largest group of 5's is 5 (in 10)

So, the LCD = 2 • 2 • 3 • 3 • 5 = 180

Step 2 Convert each given fraction into an equivalent fraction with the LCD as the denominator. Step 3 Now that the fractions have the same denominator, add or subtract the numerators. Step 4 Reduce by canceling factors common to the numerator and denominator. Since there are no factors common to the numerator and denominator the fraction cannot be
reduced
. The final answer is Practice problems: Simplify each of the following. Check the answers at the end of this document. 3. Simplify: Factor the radicand. Each pair of identical factors is a perfect square and so that factor will come
out from under the radical. Any left over single factors will remain under the radical. Rather than writing out all the factors of x and y, we could simply divide their exponents by the
index of the radical, which is 2 since we have a square root . The resulting quotient is the exponent
of the variable outside the radical and the resulting remainder is the exponent of the variable under Practice problems: Simplify each of the following. Check the answers at the end of this document. 4. Factor: 2x^2 + 5x − 12xxx

We are going to do a LOT of factoring in this course. Here is a general procedure to follow:
Step 1 Factor out the Greatest Common Factor (GCF).

For example, factor:  The GCF is 2xy^3 . Notice the third term, 1, must be there for the factorization to be correct.

Step 2 Count the number of terms and look for factoring patterns.

Two terms: Try factoring using one of these patterns (memorize these!)

Difference of perfect squares: • Difference or sum of perfect cubes: Notice the only difference between the above two formulas are the signs .
Three terms:

• Try factoring using the patterns for perfect square trinomials: Notice the only difference between the above two formulas are the signs.

• If the trinomial has the form x^2 + bx + c , find two integers whose product is c and whose sum
is b. Let's say the integers are m and n. Then x^2 + bx + c = (x + m)(x + n).

For example, factor: x^2 + x − 12
We need two integers whose product is –12 and whose sum is 1. Integer factors of 12 are
1•12, 2•6, and 3•4. Since the product is negative, the two factors must have different signs.
Since the sum must be 1, we choose –3•4.
Thus,
x^2 + x − 12 = (x − 3)(x + 4).

• If the trinomial has the form ax^2 + bx + c , find two integers whose product is a•c and whose
sum is b, replace the bx term using these integers, and factor by grouping. For an example,
see below.

Four terms: Try factoring by grouping.
For example, factor: 5x^2 + 15x − 2xy − 6y
Group the first pair of terms and factor out the GCF; do the same for the second pair of terms. Step 3 Factor completely. In some cases, it may be necessary to factor more than once.
Remember that multiplication can be used to check the factorization.

Now, let's factor 2x^2 + 5x − 12:
Step 1 GCF: There is no factor common to all the terms so there is no GCF other than 1 or –1.

Step 2 Number of terms: There are three terms and they match the pattern ax^2 + bx + c , where a = 2,
b = 5, and c = –12. So, we find two integers whose product is a•c = 2•(–12) = –24 and whose sum
is b = 5. Here are the possibilities for 24 (ignore the sign for the moment):
24 = 1•24, 2•12, 3•8, 4•6

The pair that can have a sum of 5 is 3•8 if we attach a negative sign to the 3. The integers we seek
are –3 and 8 since (−3)(8)= −24 and −3 + 8 = 5 .

Replace the middle term, 5x, with its equivalent –3x + 8x and factor by grouping: Step 3 Factor completely: Since the expression cannot be further factored the answer is (2x − 3)(x + 4).

Practice problems: Factor each of the following. Check the answers at the end of this document. 5. Simplify: Be sure you have memorized the laws of exponents : It usually is easiest to simplify the numerator and denominator individually and then simplify the
quotient. Practice problems: Simplify each of the following. Check the answers at the end of this document. 6. Solve: When solving an equation that contains fractions, it usually is easiest if you clear the fractions first
by multiplying each term by the LCD of all the terms. The LCD of 3, 6, and 12 is 12, so multiply
each term by 12 and then simplify. Practice problems: Solve each of the following. Check the answers at the end of this document. 7. Graph the line : y = - 2/3 x + 5

The slope of the line is - 2/3 and the y-intercept is 5. To graph the line, start at the point (0,5) and
then go down 2 and to the right 3 repeatedly, plotting points as you go along. Prev Next

Start solving your Algebra Problems in next 5 minutes!      2Checkout.com is an authorized reseller
of goods provided by Sofmath

Attention: We are currently running a special promotional offer for Algebra-Answer.com visitors -- if you order Algebra Helper by midnight of September 20th you will pay only \$39.99 instead of our regular price of \$74.99 -- this is \$35 in savings ! In order to take advantage of this offer, you need to order by clicking on one of the buttons on the left, not through our regular order page.

If you order now you will also receive 30 minute live session from tutor.com for a 1\$!

You Will Learn Algebra Better - Guaranteed!

Just take a look how incredibly simple Algebra Helper is:

Step 1 : Enter your homework problem in an easy WYSIWYG (What you see is what you get) algebra editor: Step 2 : Let Algebra Helper solve it: Step 3 : Ask for an explanation for the steps you don't understand: Algebra Helper can solve problems in all the following areas:

• simplification of algebraic expressions (operations with polynomials (simplifying, degree, synthetic division...), exponential expressions, fractions and roots (radicals), absolute values)
• factoring and expanding expressions
• finding LCM and GCF
• (simplifying, rationalizing complex denominators...)
• solving linear, quadratic and many other equations and inequalities (including basic logarithmic and exponential equations)
• solving a system of two and three linear equations (including Cramer's rule)
• graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions)
• graphing general functions
• operations with functions (composition, inverse, range, domain...)
• simplifying logarithms
• basic geometry and trigonometry (similarity, calculating trig functions, right triangle...)
• arithmetic and other pre-algebra topics (ratios, proportions, measurements...)

ORDER NOW!         