High School Algebra I Study Guide for the STAAR test
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General Information
If you are a student in the state of Texas, this is one of the tests you’ll need to pass to graduate from high school. Your performance will be rated at one of these three levels and your score must be II or III to be considered “passing.”
 III—Advanced Academic Performance
 II—Satisfactory Academic Performance
 I—Unsatisfactory Academic Performance
You should take the STAAR Algebra I test as soon as possible after completing the Algebra I course in high school so the material will be fresh in your mind. This study guide will help you review all of the tested concepts and let you know what you may need to work on before the test. Be sure to access our practice questions and flashcards to try out your skills.
Here are some basic facts about the test:

There are a total of 54 questions—5 “gridin” (you write in the answer) and the rest multiplechoice.

You will have a reference sheet that looks like this and may refer to it during the test:

You will also be given at least 2 sheets of graph paper for use during the test.

Throughout the test, you will also be assessed on your ability to use “mathematical processes” as you answer the questions. Your performance in this area will not show up as a separate score but is a necessary procedure to answer the questions correctly.
Ready to go? Here’s the content in which you will need to be fluent. We have tried to provide basic explanations and examples, but if there are things with which you still don’t feel comfortable, be sure to find help locally or online before attempting the test.
Number and Algebraic Methods
It will be necessary to demonstrate an understanding of how to use algebraic methods to manipulate numbers, expressions, and equations on the test. Eleven of the 54 questions on the test assess these skills.
Polynomial Expressions
While you know a polynomial is an algebraic expression that is the sum or difference of monomial(single) terms containing numbers or variables, you will also need to know how to rewrite them and perform operations with them.
Also, the way you perform operations is important. Remember to follow the order of operations, or (PEMDAS):
 Do everything in parentheses (P), left to right.
 Evaluate any exponents (E), left to right.
 Do all multiplication and division (MD), in order, left to right.
 Then do all addition and subtraction (AS), in order, from left to right.
A good way to remember: Please Excuse My Dear Aunt Sally
Adding and Subtracting
Remember, when combining polynomial expressions, you can only combine like terms.
Let \(3x+2\) and \(2x^2+3x1\), and \(x+10\) be three polynomial expressions. Find the sum of the first and second polynomials and then find the difference between the third and the second polynomials.
Now, taking the sum of the first and the second, we get
\[(3x+2)+(2x^2+3x1)=2x^2+6x+1\]The difference between the third and the second is:
\[h(x)g(x)=(x+10)(2x^2+3x1)= 2x^22x+11\]If the three polynomial expressions represent the sides of a triangle, we can write a polynomial expression that represents the perimeter, \(P\), of the triangle by adding the three terms as follows:
\[P = (3x+2)+(2x^2+3x1)+(x+10)=2x^2+7x+11\]Multiplying
You will also have to know how to multiply a term such as a number, monomial or binomial by a polynomial expression before combining like terms.
Whenever multiplying two binomial expressions, remember to use the FOIL technique (First Outer Inner Last) and then combine like terms.
Both of these methods require the distributive property.
\[2(x^2+3x4)+2(x^2+x3)\]Simplify this polynomial expression:
Distributing the \(2\) and \(2\) to the respective expressions, we get:
\[2x^26x+8+2x^2+2x6\]Now, combining like terms, we have \(2x^26x+8+2x^2+2x6= 4x+2\)
A rectangle, whose base and height are originally \(5\) and \(7\) units, respectively, are increased by \(x\) units. Write a polynomial expression that represents the area, \(A\) of the new rectangle.
The new base is \(x + 5\), and the new height is \(x + 7\). Since area is the product of the base and the height, we use the FOIL method to get:
\[A = (x+5)(x+7)=x^2+7x+5x+35=x^2+12x+35\]Dividing
To determine if one polynomial is a factor of another polynomial we can use long division. Determine if \(x+2\) is a factor of \(3x^26x+5\). To be a factor, there must be no remainder.
Dividing \(x+2\) into \(3x^26x+5\) we get:
\[\require{enclose} \begin{array}{r} 3x12 \\[3pt] x+2 \enclose{longdiv}{3x^26x+5} \\[3pt] \underline{(3x^2+6x)}\phantom{+5} \\[3pt] {12x+5} \\[3pt] \underline{(12x24)} \\[3pt] 29 \\[3pt] \end{array}\]Since the remainder is \(29\), \(x+2\) is not a factor.
A rectangular table has an area of \(2x^2+11x+12\). The length of the base is \(x+4\). What is the height?
Since \(A=b \cdot h\), \(h=\frac{A}{b}\).
Dividing the base into the area using long division, we get:
\[\require{enclose} \begin{array}{r} 2x\; +\;3 \\[3pt] x+4 \enclose{longdiv}{2x^2+11x+12} \\[3pt] \underline{(2x^2+8x)}\phantom{+12} \\[3pt] 3x+12 \\[3pt] \underline{(3x+12)} \\[3pt] 0 \end{array}\]Notice there is no remainder, so the height of the table is \(2x+3\). As we will see next, the factors of \(2x^2+11x+12\) are \(x+4\) and \(2x+3\).
Using the Distributive Property
You will have to know how to rewrite polynomials by using the distributive method of multiplying.
For example, an equivalent expression for \(3x(x5)\) is \(3x \cdot x3x\cdot 5=3x^215x\).
Also, recall that when multiplying two binomials to distribute, use the FOIL method mentioned in the “multiplying polynomials” section. The first term in the polynomial is produced with F, the middle term is a combination of O and I, and the last term is produced with L.
Write an equivalent expression for \((x+2)(x4)\).
Using the FOIL method to distribute, we get:
\[(x+2)(x4)\] \[x \cdot x+4x+2x+2 \cdot 4\] \[x^22x8\]Next, we’ll look at special polynomials and techniques to determine their factors.
Factoring
Steps for factoring a trinomial:
 Factor out the GCF from each term in the trinomial if possible.
 Identify any special trinomials, such as a perfect square trinomial, \(a^2+2ba+b^2=(a+b)^2\), where the first and last terms are perfect squares.
 If \(ax^2+bx+c\) is not a special trinomial and \(a\ne1\), factor into two binomials using the AC method described in the 2nd example below.
Example I:
Write an equivalent expression for \(x^210x+25\) and \(2x^2+16x+32\).
\(x^210x+25=x^22 \cdot 5x+25\) or \((x5)^2\)
is a perfect square and
\(2x^2+16x+32=2(x^2+8x+16)=2(x^2+2 \cdot 4x+16)\) or \((x+4)^2\)
is also a perfect square, after factoring out a GCF of \(2\).
Example 2:
What are the factors of \(3x^24x4\)?
 Multiply the absolute values \(a\) and \(c\): \(3 \cdot 4 = 12\)
 List all the factors whose product is \(12: 1 \cdot 12, \;2\cdot 6, \;3\cdot 4\)
 Find which combination of factors, either positive or negative, now gives you a sum for \(b\): \(26=4\)
 Replace \(4x\) with \(2x6x\) in the trinomial: \(3x^2+2x6x4\)
 Factor by grouping to get: \(x(3x+2)2(3x+2)\)
 Factor the GCF out of your result to determine the factors of the trinomial:\((3x+2)(x2)\)
 You can always check your answer using the FOIL method.
Difference of Two Squares
Another type of special quadratic expression that can be factored with a formula is the difference of two perfect squares.
The form for a difference of perfect squares is \(a^2b^2\) and factors as \((a+b)(ab)\), where \(a^2\) and \(b^2\) are perfect squares.
For example, we can factor \(4x^281\) to \((2x+9)(2x9)\), where \(a=\sqrt{\mathstrut{4x^2}}=2x\) and \(b=\sqrt{81}=9\).
Factoring \(2x^272\), we first factor out the GCF of \(2\) to get:
\(2(x^236)=2(x+6)(x6)\) where \(a=x^2=x\) and \(b=\sqrt{36}=6\)
Note: A sum of perfect squares is not factorable. So:
\(16x^2+49 \ne (4x+7)(4x7)\) or \((4x+7)(4x+7)\)
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