1. Coming up in Math 110: Today: Review for Quiz 4. Tomorrow: Quiz 4 on sections 5.5 - 5.8 & 8.2 Thursday: Review for Test 2 Next Monday: Test 2 on all of Ch. 5 + sec. 8.2 Reminder: There will be an optional review session for Test 2 on Sunday, March 30 th. That session will be held from 7:00-9:00 p.m. in this classroom (214) and it is open to all Math 110 students from all sections.

2. Any questions on the Section 8.2 homework?

3. Reviewing for Quiz 4

4. Quiz 4 covers Sections 8.2 and 5.5 - 5.8

5. Preparing for the Quiz: 1. Review the homework from all homework assignments using the “Check My Grades” function: 2. Take the Practice Quiz (under Assignments, Review for Quiz 4). REMEMBER: When you review a submitted practice quiz, you will have access to the online help buttons for each problem, just like in homework assignments. 3. Review your notes and/or the online lecture slides (under each Assignment).

6. IMPORTANT NOTE: The formula sheet will be especially important for this quiz because of all the factoring formulas and the quadratic formula. Make sure you have a copy handy while you take the practice quiz.

7. REVIEW of major concepts in Factoring of Polynomials Note: You can view the following review slides from ASSIGNMENTS button, in the folder Review for Quiz 4. Click on the “PowerPoint Slides” link at the top, and then click on the “Daily Lecture Slides” link and scroll down to “Review for Quiz 4”.

8. IMPORTANT!!! Always check your answers! Factoring problems: Multiply out the factors in your final answer to see if you come up with the same polynomial as in the original question. “Solve equation” problems: Plug the number/s you get for answers back into the original equation in the problem and make sure both sides of the equation come out to the same number.

9. REVIEW: Strategy for Factoring a Polynomial: 1.Always look first to see if there is a common factor. If so, factor out the GCF. 2.Determine the number of terms in the polynomial and try factoring as follows: a)If there are two terms, can the binomial be factored by one of the special formulas including difference of two squares, sum of two cubes, or difference of two cubes? b)If there are three terms, is the trinomial a perfect square trinomial? If the trinomial is not a perfect square trinomial, try factoring by the British (factoring by grouping) method. c)If there are four or more terms, try factoring by grouping. 3.Check to see if any factors with more than one term in the factored polynomial can be factored further. If so, factor completely.

10. Example: In this problem, first factor out the greatest common factor, then factor the remaining 4-term polynomial using the grouping method.

11. EXAMPLE: Factor x 2 + 3x – 18 Solution We begin with x 2 + 3x – 18 = (x )( x ). To find the second term of each factor, we must find two numbers whose product is –18 and whose sum is 3. From the table above, we see that 6 and –3 are the required integers. Thus, x 2 + 3x – 18 = (x + 6)(x – 3) or (x – 3)(x + 6). -33-77-1717Sum of Factors -6, 36, -3-9, 29, -2-18, 118, -1Factors of -8 This is the desired sum. Factoring trinomials with a leading coefficient of 1:

12. An alternate way to factor trinomials (This is sometimes referred to as the British Method; and it is especially useful when leading coefficient is NOT 1.) This method is systematic approach that uses factoring by grouping to factor Step 1: Form the product ac Step 2: Find a pair of numbers whose product is ac and whose sum is b Step 3: Rewrite the polynomial so the middle term bx is written as a sum of two terms whose coefficients are the numbers found in step 2 Step 4: Factor by grouping.

13. Example using this method on a 3-term polynomial with a leading coefficient that’s not 1.

14. Factoring Perfect Square Trinomials: Example: Factor 25x 2 – 60x + 36. If we try the British Method on this trinomial, we get a pretty huge number at the first step (25 x 36 = 9000). But wait! Notice that the first and last terms are perfect squares, the squares of 5x and 6. This means we might be able to see if it fits the pattern of the formula ax 2 – 2abx + b 2 = (ax - b) 2 Possible solution: (5x – 6) 2 Now check to see if (5x – 6)(5x – 6) = 25x 2 – 60x + 36

15. Example: Two terms Factor: 81x 2 – 49 Solution: This is a difference of squares, so we can use the formula a 2 – b 2 = (a + b)(a - b) 81x 2 – 49 = (9x) 2 – (7) 2 = (9x + 7)(9x – 7).

16. 64x 3 – 125 = (4x) 3 – 5 3 = (4x – 5)(4x) 2 + (4x)(5) + 5 2 ) = (4x – 5)(16x 2 + 20x + 25) A 3 – B 3 = (A – B)(A 2 + AB + B 2 ) x 3 + 8 = x 3 + 2 3 = (x + 2)( x 2 – x·2 + 2 2 ) = (x + 2)( x 2 – 2x + 4) A 3 + B 3 = (A + B)(A 2 – AB + B 2 ) ExampleType More examples with two terms, when both terms are cubes:

17. Solving Polynomial Equations Make sure you are clear about the difference between factoring a polynomial expression and solving a polynomial equation. When a problem asks you to factor a polynomial, your final answer will contain factors including variables and numbers. [ e.g. (x + 3)(x – 9) ] When a problem asks you to solve a polynomial equation, your answers will be actual numbers. [ e.g. instead of (x + 3)(x – 9), your answers would be x = -3 and x = 9.)

18. Example: Solve 2x 2  7x  4 by factoring and then using the zero  product principle. Step 1 Move all terms to one side and obtain zero on the other side. (Subtract 4 from both sides and write the equation in standard form.) 2x 2  7x  4  4  4 2x 2  7x  4  Step 2 Factor. 2x 2  7x  4  (2x  1)(x  4)  0

19. (Solution continued) Steps 3 and 4 : Set each factor equal to zero and solve each resulting equation. 2 x  1  2 x  1 x = 1/2 x  4  x  4 Steps 5: check your solution

20. EXAMPLE: Solve 10x 2 +17x+3 = 0 Whenever the leading coefficient is not 1, the factoring method becomes more complicated. In this case, it will probably be quicker to solve the equation by using the quadratic formula rather than by using the “British Method” to factor the polynomial.

21. Quadratic Formula Recall: a is the coefficient of x 2, b is the coefficient of x, and c is the constant term of ax 2 + bx + c = 0

22. a = 10, b = 17 and c = 3 so the quadratic formula gives Notice that we came up with the same two answers that we would have gotten by the longer factoring method. To solve 10x 2 +17x+3 = 0:

23. No x-intercepts No real solution; two complex imaginary solutions b 2 – 4ac < 0 One x-intercept One real solution (a repeated solution) b 2 – 4ac = 0 Two x-intercepts Two unequal real solutionsb 2 – 4ac > 0 Graph of y = ax 2 + bx + c Kinds of solutions to ax 2 + bx + c = 0 Discriminant b 2 – 4ac Review: The Discriminant and the Kinds of Solutions to ax 2 + bx +c = 0

24. Do any of you who have already started the practice quiz have any questions you’d like to have explained?

25. If there’s time left, go ahead and start the practice quiz, and we’ll come around to help if you have questions. Remember, the open lab next door will be staffed from 8:00 AM till 6:30 PM Mondays through Thursdays.