Accuplacer Review Workshop. Intermediate Algebra. Week Four. Includes internet links to instructional videos for additional resources:

Transcription

1 Accuplacer Review Workshop Intermediate Algebra Week Four Includes internet links to instructional videos for additional resources: (Arithmetic Video Library) Preparing for the Algebra Test: To Place into College Level Math It is important to know that the accuplacer assesses three areas of Algebra:! Numerical Skills/Pre-Algebra! Elementary Algebra! Intermediate Algebra These areas are required prerequisites for College Algebra. The following Workshop will hopefully aide you in your review of Intermediate Algebra. NOTE: Although some questions are basic skills many accuplacer questions require you to put several math concepts together, or work multiple step problems to determine the final answer. 1

2 Intermediate Algebra: Questions in Intermediate Algebra cover a wide range of algebra concepts and skills necessary to enter a College Algebra course. It is important to understand this workshop will only highlight few fundamental concepts from the following categories: Factorization of Polynomials 1. Greatest Common Factor 2. Factor by Grouping 3. Factoring Trinomials 4. Factoring by Special Products Solving Quadratics Equations 1. By factoring 2. Using the Quadratic Formula FACTORING: Factoring in Algebra is the process of finding the factors. For example, we can write the expression 10 as (5)(2) where 5 and 2 are called factors of 10. We can also do this with polynomial expressions. 2

3 Greatest Common Factor (GCF) The GCF for a polynomial is the largest monomial that divides (is a factor of) each term of the polynomial. Example 1: Find the GCF of the list of monomials: x!, x!, x! The exponents on the X s are 8, 7, and 6. If we try using the exponent 8 we cannot divide x! or x! by x!, we don t have enough X s to do that. But, if we use x!, we would have a monomial that we could divide out of ALL the terms. Hence our GCF is x! Note that if all terms have the same variable in common the GCF for the variable part is the variable raised to the lowest exponent that is listed. Example 2: Find the GCF of the list of monomials: 3xy!, 9x! y!, 18x! y We need to figure out what the largest monomial that we can divide out of each of these terms would be. First look at the numerical part. We have 3, 9, and 18. The largest number that can be divided out of those numbers is 3. So the numerical GCF is 3. Now on the variable part. It looks like each term has an X and a Y. In both cases the lowest exponent is 1. So the GCF of our variable part is XY. Putting this together the GCF if 3XY Factoring out the GCF: Step 1. Identify the GCF of the polynomial. Step 2. Divide the GCF out of every term of the polynomial. `Example 3: Factor out the GCF: 8x! + 4x! + 2x Step 1: The largest monomial that can be factored out of each term is 2x Step 2: Divide the GCF out of every term of the polynomial. 8x! + 4x! + 2x = 2x(4x! + 2x + 1) Be careful. If a term of a polynomial is exactly the same as the GCF, when you divide it by the GCF you are left with 1 not 0. As shown above when 2x is divided by 2x we get 1 as the third term inside the ( ). 3

4 Note that if the answered is multiplied out, we should get the original polynomial. Because, factoring gives you another way to write the expression so it will be equivalent to the original problem. Example 4: Factor out the GCF: x! (x + 5) 7(x + 5) This problem looks a little different; because now the GCF is a binomial not a monomial; however we treat it in the same manner. Step 1: Identify the GCF of the polynomial. This time it isn t a monomial but a binomial that we have in common. Our GCF is (x + 5) Step 2: Divide the GCF out of every term of the polynomial. x! (x + 5) 7(x + 5) = (x + 5) (x! 7) Factoring a Polynomial with Four Terms by Grouping In some cases there is not a GCF for ALL terms in a polynomial. If you have four terms with no GCF, then try factoring by grouping. Factoring by Grouping: Step 1. Group the first two terms together and then the last two terms together. Step 2. Factor out a GCF from each separate binomial. Step 3. Factor out the common binomial. Example 5: Factor by Grouping: x! + 7x! + 2x + 14 Note how there is not a GCF for ALL the terms so try factoring by grouping. Step 1: Group the first two terms together and then the last two terms together. x! + 7x! + 2x + 14 = (x! + 7x! ) + (2x + 14) Step 2: Factor out a GCF from each separate binomial. (x! + 7x! ) + (2x + 14) = x! (x + 7) + 2(x + 7) 4

5 Step 3: Factor out the common binomial x! (x + 7) + 2(x + 7) = (x + 7) (x! + 2) Note that if you multiply out the answer, you get the original polynomial. Example 6: Factor by grouping 3x! + xy 12x 4y Step 1: Group the first two terms together 3x! + xy 12x 4y = (3x! + xy) (12x 4y) Step 2: Factor out a GCF from each separate binomial. (3x! + xy) (12x 4y) = x(3x + y) -4(3x + y) Step 3: Factor out the common binomial. x(3x + y) -4(3x + y) = (3x + y) (x 4) Factoring Trinomials of the Form x² + bx + c (Where the number in front of X² is 1) Basically, we are looking for two binomials that when you multiply them you get back the given trinomial. Step 1: Set up a product of two ( ) where each will hold two terms. I will look like this: ( ) ( ) Step 2: Find the factors that go in the first positions of each parenthesis So it would look like this: (x ) (x ) 5

6 Step 3: Find the factors that go in the last positions. The factors for the last positions will have to be two expressions such that their product equals c (the constant term) and at the same time their sum equals b (the number in front of the x term). If c is positive your factors are going to both have the same sign depending on b s sign. If c is negative, your factors are going to have opposite signs depending on b s sign. Example 7: Factor the trinomial y² - 5y + 6 (Note this trinomial does not have a GCF so we go right to factoring) Step 1: Set up a product of two ( ) where each will hold two terms. ( ) ( ) Step 2: Find the factors that go in the first positions. Since we have y² as our first term, we will need the following: (y ) (y ) Step 3: Find factors that go in the last positions. We need two numbers whose product is 6 and whose sum is -5. That would have to be -2 and -3. y² -5y + 6 = (y 2) (y 3) Example 8: Factor the trinomial 2x²y + 2xy 40y Since this trinomial has a GCF of 2y Factor out the GCF: 2x²y + 2xy 40y = 2y(x² + x 20) We are not finished, we can still factor the trinomial, Step 1: 2y( ) ( ) Step 2: 2y (x ) (x ) Step 3: Find factors that go in the last positions We need two numbers whose product is -20 and sum is 1. That would have be 5 and -4. 2x²y + 2xy 40y = 2y (x + 5) (x 4) 6

7 Factoring Trinomials of the Form ax²+ bx + c (where a does not equal 1) The difference between this trinomial and the one discussed above, is there is a number other than 1 in front of the x² term. This means, that not only do you need to find factors of c, but also a. Step 1: Set up a product of two ( ) where each will hold two terms. ( ) ( ) Step 2: Use trial and error to find the factors needed. The factors of a will go in the first terms of the binomials and the factors of c will go in the last terms of the binomials. Your goal is to find the right combination of factors. If the product does not come out to be the trinomial you started with, then you need to try a different combination of factors of a and c. Example 9: Factor the trinomial 3x² - 5x + 2 This trinomial does not have a GCF, so we start factoring. Step 1: Set up a product of two ( ) where each will hold two terms. ( ) ( ) Step 2: Use trial an error to find the factors needed. In the first terms of the binomials we need factors of 3x². This would have to be 3x and x. In the second terms of the binomials, we need factors of 2. This would have to be -2 and -1. Negatives numbers were used here because the middle term is negative. 3x² - 5x + 2 = (3x -2) (x- 1) 7

8 NOTE: Other Factoring Techniques can be explored on the math web sites previously provided. Please use them to continue learning ways to factor the following techniques of Factoring by Special Products: Factoring a Perfect Square Trinomial Factoring a Difference of Two Squares Factoring a Sum of Two Cubes Factoring a Difference of Two Cubes Solving Quadratic Equations: Solving equations in general is a very essential part of Algebra. We will be looking at solving polynomial equations which include quadratic equations by factoring. Polynomial Equation A polynomial equation is one polynomial set equal to another polynomial. The following is an example: 3x² + 5x = 5x³ - 2 Standard Form of a Polynomial Equation When a polynomial is set equal to 0, it is in standard form. The following is an example of a polynomial equation in standard form: 2x² + 5x + 3 = 0 8

9 Quadratic Equation in Standard Form ax² + bx + c = 0 (where a does not equal 0) Note: The degree of a quadratic equation is 2. This is a common type of equation Zero Factor Property 0 is our magic number because the only way a product can become 0 is if at least one of its factors is 0. You cannot guarantee what the factors would have to be if the product was set equal to any other number. For example if ab=1, then a=4 and b=!. But with the product! set equal to 0, we can guarantee finding the solution by setting each factor equal to 0. Solving a Polynomial Equation by Factoring Step 1: Simplify if needed. Clear any fractions or ( ) that are not in standard form in the problem. Step 2: Write in standard form if needed. Step 3: Factor the Polynomial. 9

10 Step 4: Use the Zero Factor Property. Set every factor equal to zero. Step 5: Solve for the equations set up in step 4. Example 9: Solve the equation 2x(x 1)(x + 3) = 0 Step 1: Simplify if needed Step 2: Write in standard form if needed. This problem is already written in standard form. Step 3: Factor the Polynomial. This problem is already factored. Step 4: Use the Zero Factor Property AND Step 5: Solve for the equations set up in step 4. 2x(x 1)(x + 3) = 0 2x = 0 (x + 3) = 0 (x 1) = 0!! =! X = 0-3 x = 0 + 1!! X = 0 x = -3 x = -3 The solutions to this equation are 0, 1, abd -3 Quadratic Formula Often, the simplest way to solve "ax 2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. While factoring may not always be successful, the Quadratic Formula can always find the solution. 10

11 The Quadratic Formula uses the "a", "b", and "c" from "ax 2 + bx + c", where "a", "b", and "c" are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve. For ax 2 + bx + c = 0, the value of x is given by: For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root. And it's a "2a" under there, not just a plain "2". Make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself up. Remember that "b 2 " means "the square of ALL of b, including its sign", so don't leave b 2 being negative, even if b is negative, because the square of a negative is a positive. Example: x² + 3x -4 = 0 a= 1 b= 3 c= -4 The solutions are x = -4, x= 1 11