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Any questions on the Section 8.2 homework?

Any questions on the Section 8.2 homework?. Reviewing for Quiz 4. Quiz 4 covers Sections 8.2 and 5.5 - 5.8. Preparing for the Quiz :. 1. Review the homework from all homework assignments using the “Check My Grades” function: 2. Take the Practice Quiz

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Any questions on the Section 8.2 homework?

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  1. Any questions on the Section 8.2 homework?

  2. Reviewing for Quiz 4

  3. Quiz 4 covers Sections 8.2 and 5.5 - 5.8

  4. 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).

  5. 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.

  6. REVIEW of major concepts in Factoring of Polynomials View review slides from ASSIGNMENTS button, in the area Review for Quiz 4 . Click on the “PowerPoint Slides” link at the top, and then click on the link to “Review for Quiz 4”.

  7. 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.

  8. A Strategy for Factoring a Polynomial: • Always look first to see if there is a common factor. If so, factor out the GCF. • Determine the number of terms in the polynomial and try factoring as follows: • 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? • 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. • If there are four or more terms, try factoring by grouping. • Check to see if any factors with more than one term in the factored polynomial can be factored further. If so, factor completely.

  9. Factoring by Grouping Factor by grouping is a technique used to factor polynomials with three or more terms when no greatest common factor exists for all the terms. For example,

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

  11. Factors of -8 18, -1 -18, 1 9, -2 -9, 2 6, -3 -6, 3 Sum of Factors 17 -17 7 -7 3 -3 This is the desired sum. Factoring trinomials with a leading coefficient of 1: EXAMPLE: Factor x2 + 3x – 18 Solution We begin with x2 + 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, x2 + 3x – 18 = (x + 6)(x – 3) or (x – 3)(x + 6).

  12. An alternate way to factor trinomials(sometimes referred to as the British Method; 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: • Factor: 25x2 – 60x + 36 . Note: The first and last terms are perfect squares (5x squared and 6 squared). The other term is -2∙5∙6 ∙x =-60x. Remember, if you don’t recognize this as a perfect square trinomial, you can factor it using the techniques to factor trinomials. Solution: 25x2 – 60x + 36 = (5x)2 – 2 · 5x · 6 + 62 = (5x – 6)2.

  15. Example: Two terms • Factor: 81x2 – 49 Solution: This is a difference of squares, so we can use the formula a2 – b2 = (a + b)(a - b) 81x2 – 49 = (9x)2 – 72 = (9x+ 7)(9x – 7).

  16. Type Example A3 + B3 = (A + B)(A2 – AB + B2) x3 + 8 = x3 + 23 = (x + 2)( x2 – x·2 + 22) = (x + 2)( x2 – 2x + 4) A3 – B3 = (A – B)(A2 + AB + B2) 64x3 – 125 = (4x)3 – 53 = (4x – 5)(4x)2 + (4x)(5) + 52) = (4x – 5)(16x2 + 20x + 25) More examples with two terms, when both terms are cubes:

  17. REVIEW: Strategy for Factoring a Polynomial: • Always look first to see if there is a common factor. If so, factor out the GCF. • Determine the number of terms in the polynomial and try factoring as follows: • 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? • 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. • If there are four or more terms, try factoring by grouping. • Check to see if any factors with more than one term in the factored polynomial can be factored further. If so, factor completely.

  18. Example Factor: x3 – 5x2 – 4x + 20 HOW MANY TERMS?What does that tell you? CHECK FIRST: Is there a common factor in all terms? ANSWER: No, but you should ALWAYS check for this first. Solution x3 – 5x2 – 4x + 20 = (x3 – 5x2) + (-4x + 20) Group the terms with common factors. = x2(x – 5) – 4(x – 5) Factor from each group. = (x – 5)(x2 – 4) Factor out the common binomial factor, (x – 5). = (x – 5)(x + 2)(x – 2) Factor completely by factoring x2 – 4 as the difference of two squares.

  19. 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.)

  20. Example: Solve 2x2+ 7x= 4 by factoring and then using the zero-product principle. Step 1Move all terms to one side and obtain zero on the other side. (Subtract 4 from both sides and write the equation in standard form.) 2x2+ 7x- 4= 4 - 4 2x2+ 7x- 4 = 0 Step 2Factor. 2x2+ 7x- 4 = 0 (2x- 1)(x+ 4) = 0

  21. (Solution continued) Steps 3 and 4 : Set each factor equal to zero and solve each resulting equation. 2 x- 1 = 0 2 x= 1 x = 1/2 x+ 4 = 0 x= -4 Steps 5: check your solution

  22. EXAMPLE:Solve 10x2+17x+3 = 0 by factoring • Whenever the leading coefficient is not 1, the factoring method becomes more complicated. (Remember, the leading coefficient is the number in front of the highest degree term, which will be the x2 term in a quadratic polynomial. So the leading coefficient of this polynomial is 10). • To factor this polynomial, we have to either “guess and check” or use the factoring by grouping method.

  23. An overview of solving 10x2+17x+3 = 0 by factoring: • By the grouping method, we multiply 10x3=30, then look for two factors of 30 that add up to 17. How about 15 and 2? • Now we split the middle 17x term up into 15x + 2x: • This gives us: 10x2+15x+2x+3 = 0 • Now factor by grouping: 5x(2x+3)+1(2x+3) = (5x+1)(2x+3) • So the answers will be given by 2x+3 = 0, which gives x = -3/2 and • 5x+1 = 0, which gives x = -1/5 Whew! There must be an easier way… (There is: the quadratic formula)

  24. Quadratic Formula Recall: a is the coefficient of x2, b is the coefficient of x, and c is the constant term of ax2 + bx + c= 0

  25. 10x2+17x+3 = 0 revisited: a = 10, b = 17 and c = 3 so the formula gives Notice that we came up with the same two answers that we got by the factoring method.

  26. Discriminant b2 – 4ac Kinds of solutions to ax2+ bx+ c = 0 Graph of y= ax2+ bx+ c b2 – 4ac > 0 Two unequal real solutions Two x-intercepts b2 – 4ac = 0 One real solution (a repeated solution) One x-intercept b2 – 4ac < 0 No real solution; two complex imaginary solutions No x-intercepts Review: The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0

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

  28. 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 7:30 PM Mondays through Thursdays.

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