Any questions on the Section 5 7 homework Now please CLOSE YOUR LAPTOPS and 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 8 Solving Equations by Factoring Problem Solving What 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 2 3 4 5 6 Write the equation in standard form Clear any fractions Factor the polynomial completely Set each factor containing a variable equal to 0 Solve the resulting equations 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 Solve 4x 8x 9 5 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 4 Note 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 1 1 1 1 1 4 8 9 4 1 9 4 10 10 5 2 8 8 8 8 true 5 5 5 5 4 8 9 4 10 9 4 1 5 1 5 4 4 4 4 true So our solutions for x are 1 8 or 5 4 Example from today s homework Example from today s homework General strategy for solving applied word problems 1 Understand the problem 2 3 4 5 Read and reread the problem Choose a variable to represent the unknown Construct a drawing whenever possible Translate the problem into an equation Solve the equation Check your answers in the original equation 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 The product of two consecutive positive integers is 132 Find the two integers Understand 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 132 132 Solve x x 1 132 x2 x 132 x2 x 132 0 x 12 x 11 0 distributive property write quadratic in standard form 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 Interpret Possible solutions x 12 or x 11 Remember 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 Pythagorean 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 2 leg a hypotenuse leg b Similar to problem 21 in today s assignment Find 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 Understand Read 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 2 x 10 x 10 2 x 10 x Translate x 10 By the Pythagorean Theorem leg a 2 leg b 2 hypotenuse 2 x2 x 10 2 2x 10 2 Solve x2 x 10 2 2x 10 2 x2 x2 20x 100 4x2 40x 100 multiply the binomials simplify left side 2x2 20x 100 4x2 40x 100 subtract 2x2 20x 100 from both sides 0 2x2 60x factor right side 0 2x x 30 x 0 or x 30 set each factor 0 and solve x Interpret 2 x 10 x 10 Check Remember that x is supposed to represent the length of the shorter side So although x 0 satisfies our equation it cannot be a solution for the problem we were presented If we let x 30 then x 10 40 and 2x 10 50 Since 302 402 900 1600 2500 502 the Pythagorean Theorem checks out State Solution The length of the shorter leg is 30 miles Remember that is all we were asked for in this problem To solve this problem we set h 0 since 0 ground level Then we solve the equation 16t2 144t 0 Note There are several problems like this in today s homework New question How long will it …