Any questions on the Section 5.7 homework?Slide 2Reminder:Section 5.8What possible use is there for factoring polynomials????Slide 6Take 2: What possible use is there for factoring polynomials????Take 3: What possible use is there for factoring polynomials????Take 4: What possible use is there for factoring polynomials????Solving Problems Like This by Factoring:Slide 11Slide 12Slide 13Slide 14Slide 15Example from today’s homework:Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Please Note:Slide 32Any questions on the Section 5.7 homework?Now pleaseCLOSE YOUR LAPTOPSand turn off and put away your cell phones.Sample Problems Page Link(Dr. Bruce Johnston)Reminder:This homework assignment on section 5.8 is due at the start of next class period.Make sure you turn in the worksheet showing all your work for the last 9 problems (#20 – 28) of this assignment. If you don’t turn this in, or if you don’t completely show your work on any of these 9 problems out of the 28 total problems in the online assignment, your online score will be reduced accordingly.Section 5.8Solving Equations by FactoringProblem SolvingWhat possible use is there for factoring polynomials????Take 2: What possible use is there for factoring polynomials????Take 3: What possible use is there for factoring polynomials????Take 4: What possible use is there for factoring polynomials????Solving Problems Like This by Factoring:Polynomial equations•Equations that set 2 polynomials equal to each other.•Standard form has a 0 on one side of the equation.•The maximum number of solutions to a polynomial equation is equal to the degree of the polynomial.Quadratic equations•Polynomial equations of degree 2. (So how many possible solutions?)Zero factor theorem•If a and b are real numbers and ab = 0, then a = 0 or b = 0.•This property is true for three or more factors, as well.Steps for solving a polynomial equation by factoring:1) Write the equation in standard form.2) Clear any fractions.3) Factor the polynomial completely.4) Set each factor containing a variable equal to 0.5) Solve the resulting equations.6) Check each solution in the original equation.Solve x2 – 5x = 24.•First write the polynomial equation in standard form. x2 – 5x – 24 = 0•Now we factor the polynomial using techniques from the previous sections. x2 – 5x – 24 = (x – 8)(x + 3) = 0•We set each factor equal to 0. x – 8 = 0, which will simplify to x = 8 x + 3 = 0 which will simplify to x = -3•Check both possible answers in the original equation, x2 – 5x = 24. x = 8: 82 – 5(8) = 64 – 40 = 24 true x = -3: (-3)2 – 5(-3) = 9 – (-15) = 24 true•So our solutions for x are 8 or –3. ALWAYS REMEMBER TO CHECK YOUR ANSWERS!!! (Especially on quizzes/tests, when there’s no “check answer” button...)•First, simplify the left side using the distributive property: 32x2 + 36x = 5•Then write the polynomial equation in standard form: 32x2 + 36x – 5 = 0•Now we factor the polynomial using techniques from the previous sections: 32x2 + 36x – 5 = (8x – 1)(4x + 5) = 0•Finally, we set each factor equal to 0. 8x – 1 = 0 8x = 1 x = 1/8 4x + 5 = 0 4x = -5 x = -5/4Note: This equation can also be solved (and probably more quickly) using the quadratic formula, which is the topic of tomorrow’s lecture.Solve 4x(8x + 9) = 5•Now check both possible answers (x = 1/8 and x = -5/4) in the original equation. (This can be done fairly quickly if you use your calculator.) 5)10(21)10(814918149818814 true 5)1)(5()1(4549104549458454 true• So our solutions for x are or .8145Solve 4x(8x + 9) = 5Example from today’s homework:Example from today’s homework:General strategy for solving applied (word) problems:1) Understand the problem:•Read and reread the problem.•Choose a variable to represent the unknown.•Construct a drawing, whenever possible.2) Translate the problem into an equation.3) Solve the equation.4) Check your answers in the original equation.5) Interpret the result:•Determine if any or all of the proposed solutions make sense in terms of the applied problem.•Convert your answer(s) into the appropriate form to answer the specific question(s) asked in the word problem.UnderstandThe product of two consecutive positive integers is 132. Find the two integers.Read and reread the problem and choose a variable to represent the unknown quanitity. If we let x = one of the unknown positive integers, then x + 1 = the next consecutive positive integer.Now translate this into an equation:132132Solvex(x + 1) = 132x2 + x = 132 (distributive property)x2 + x – 132 = 0 (write quadratic in standard form)(x + 12)(x – 11) = 0 (factor quadratic polynomial)x + 12 = 0 or x – 11 = 0 (set factors equal to 0)x = -12 or x = 11 (solve each factor for x)InterpretRemember that x is supposed to represent a positive integer. So, although x = -12 satisfies our equation, it cannot be a solution for the problem we were presented.If we let x = 11 (the first integer), then the next consecutive integer is x + 1 = 12. Check: The product of the two numbers is 11 · 12 = 132, our desired result.State Solution: The two positive integers are 11 and 12.Possible solutions: x = -12 or x = 11Pythagorean Theorem (used in homework problem # 21 in today’s assignment)In a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse.(leg a)2 + (leg b)2 = (hypotenuse)2leg ahypotenuseleg bFind the length of the shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotenuse is 10 miles less than twice the shorter leg.(Similar to problem # 21 in today’s assignment.)UnderstandRead and reread the problem. If we let x = the length of the shorter leg, then x + 10 = the length of the longer leg and 2x – 10 = the length of the hypotenuse. Now draw a diagram:x + 102 - 10xxTranslateBy the Pythagorean Theorem,(leg a)2 + (leg b)2 = (hypotenuse)2 x2 + (x + 10)2 = (2x – 10)2x + 102 - 10xxSolve