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§ 5.7 Polynomial Equations and Their Applications Solving Polynomial Equations We have spent much time on learning how to factor polynomials. Now we will look at one important use of factoring. In this section, we will use factoring to solve equations of degree 2 and higher. Up to this point, we have only looked at solving equations of degree one. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 5.7 Solving Polynomial Equations, pp 368-369 Definition of a Quadratic Equation A quadratic equation in x is an equation that can be written in the standard form ax2 bx c 0 where a, b, and c are real numbers, with a 0 . A quadratic equation in x is also called a second-degree polynomial equation in x. The Zero-Product Rule If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. If AB = 0, then A = 0 or B = 0. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 5.7 pp 368-369 Solving Polynomial Equations, p 369 Solving a Quadratic Equation by Factoring 1)If necessary, rewrite the equation in the standard form ax2 bx c 0 , moving all terms to one side, thereby obtaining zero on the other side. 2) Factor completely. 3) Apply the zero-product principle, setting each factor containing a variable equal to zero. 4) Solve the equations in step 3. 5) Check the solutions in the original equation. pp 369 Blitzer, Intermediate Algebra, 5e – Slide #4 Section 5.7 Solving Polynomial Equations EXAMPLE (similar to example 1 on page 369) Solve: x 2 4 x 45. SOLUTION 1) Move all terms to one side and obtain zero on the other side. Subtract 45 from both sides and write the equation in standard form. x 2 4 x 45 45 45 x 2 4 x 45 0 Subtract 45 from both sides Simplify 2) Factor. x 9x 5 0 Factor Blitzer, Intermediate Algebra, 5e – Slide #5 Section 5.7 Solving Polynomial Equations CONTINUED 3) & 4) Set each factor equal to zero and solve the resulting equations. x9 0 x9 or x5 0 x 5 5) Check the solutions in the original equation. Check 9: Check -5: x 2 4 x 45 x 2 4 x 45 92 49 ? 45 81 49 ? 45 52 4 5 ? 45 ? 25 4 5 45 Blitzer, Intermediate Algebra, 5e – Slide #6 Section 5.7 Solving Polynomial Equations CONTINUED Check 9: Check -5: 81 36 ? 45 25 20 ? 45 45 45 true 45 45 true The solutions are 9 and -5. The solution set is {9,-5}. 80 The graph of y x 2 4 x 45 60 40 lies to the right. 20 0 -10 -5 0 -20 -40 -60 Blitzer, Intermediate Algebra, 5e – Slide #7 Section 5.7 5 10 15 Solving Polynomial Equations Check Point 1 2x 9x 5 Solve 2 2x 9x 5 0 2 2x 1x 5 0 2x 1 0 x 5 0 1 x 2 x5 1 ,5 2 P 370 Blitzer, Intermediate Algebra, 5e – Slide #8 Section 5.2 Solving Polynomial Equations EXAMPLE (similar to example 2a on page 371 Solve: x 2 4 x. SOLUTION 1) Move all terms to one side and obtain zero on the other side. Subtract 4x from both sides and write the equation in standard form. NOTE: DO NOT DIVIDE BOTH SIDES BY x. WE WOULD LOSE A POTENTIAL SOLUTION!!! x2 4x 4x 4x Subtract 4x from both sides x2 4x 0 Simplify 2) Factor. xx 4 0 Factor Blitzer, Intermediate Algebra, 5e – Slide #9 Section 5.7 Solving Polynomial Equations CONTINUED 3) & 4) Set each factor equal to zero and solve the resulting equations. x0 or x40 x4 5) Check the solutions in the original equation. Check 0: Check 4: x2 4x x2 4x 02 ? 40 42 ? 44 00 16 16 true Blitzer, Intermediate Algebra, 5e – Slide #10 Section 5.7 true Solving Polynomial Equations CONTINUED The solutions are 0 and 4. The solution set is {0,4}. 80 The graph of y x 2 4x 70 60 50 lies to the right. 40 30 20 10 0 -10 Blitzer, Intermediate Algebra, 5e – Slide #11 Section 5.7 -5 -10 0 5 10 15 Solving Polynomial Equations EXAMPLE (similar to 2c on page 371 Solve: x 1x 4 14. SOLUTION Be careful! Although the left side of the original equation is factored, we cannot use the zero-product principle since the right side of the equation is NOT ZERO!! 1) Move all terms to one side and obtain zero on the other side. Subtract 14 from both sides and write the equation in standard form. x 1x 4 14 14 14 x 1x 4 14 0 Subtract 14 from both sides Simplify Blitzer, Intermediate Algebra, 5e – Slide #12 Section 5.7 Solving Polynomial Equations CONTINUED 2) Factor. Before we can factor the equation, we must simplify it first. x 1x 4 14 0 x 2 4 x x 4 14 0 x 2 3x 18 0 FOIL Simplify Now we can factor the polynomial equation. x 3x 6 0 Blitzer, Intermediate Algebra, 5e – Slide #13 Section 5.7 Solving Polynomial Equations CONTINUED 3) & 4) Set each factor equal to zero and solve the resulting equations. x3 0 x3 or x60 x 6 5) Check the solutions in the original equation. Check 3: x 1x 4 14 Check -6: x 1x 4 14 3 13 4 ? 14 6 1 6 4 ? 14 27 ? 14 7 2 ? 14 Blitzer, Intermediate Algebra, 5e – Slide #14 Section 5.7 Solving Polynomial Equations CONTINUED Check 3: 14 14 true Check -6: 14 14 true The solutions are 3 and -6. The solution set is {3,-6}. 120 100 80 The graph of y x 2 3x 18 60 40 lies to the right. 20 0 -15 -10 -5 -20 -40 Blitzer, Intermediate Algebra, 5e – Slide #15 Section 5.7 0 5 10 15 Solving Polynomial Equations EXAMPLE (similar to 2b on page 371) Solve by factoring: x3 2 x 2 x 2 0. SOLUTION 1) Move all terms to one side and obtain zero on the other side. This is already done. 2) Factor. Use factoring by grouping. Group terms that have a common factor. x 3 2x 2 + x 2 0 Common 2 factor is x Common factor is -1. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 5.7 Solving Polynomial Equations CONTINUED x 2 x 2 x 2 0 x 2x2 1 0 x 2x 1x 1 0 Factor x 2 from the first two terms and -1 from the last two terms Factor out the common binomial, x – 2, from each term Factor completely by factoring x 2 1 as the difference of two squares Blitzer, Intermediate Algebra, 5e – Slide #17 Section 5.7 Solving Polynomial Equations CONTINUED 3) & 4) Set each factor equal to zero and solve the resulting equations. or or x20 x 1 0 x 1 0 x2 x 1 x 1 5) Check the solutions in the original equation. Check the three solutions 2, -1, and 1, by substituting them into the original equation. Can you verify that the solutions are 2, -1, and 1? 150 100 The graph of y x 2 x x 2 3 2 50 0 lies to the right. -6 -4 -2 0 -50 -100 -150 Blitzer, Intermediate Algebra, 5e – Slide #18 Section 5.7 2 4 6 Solving Polynomial Equations Check Point 2a Solve 3x 2 2 x 2 0, 3 Check Point 2b Solve x 2 7 10x 18 Check Point 2c Solve x 2x 3 6 5 4,3 P 372 Blitzer, Intermediate Algebra, 5e – Slide #19 Section 5.7 Solving Polynomial Equations Check Point 3 2 x 3x 8x 12 3 Solve 2x 3 2 3x 2 8x 12 0 x 2x 3 42 x 3 0 2 2x 3x2 4 0 2x 3x 2x 2 0 3 2, ,2 2 Blitzer, Intermediate Algebra, 5e – Slide #20 Section 5.7 P 370 Modeling Motion Application EXAMPLE 4 See book p 373-4 (figure on p 374) finds when the ball hits the ground building height is 384 similar to problem 65 on p. 378 similar to 98 on p. 388 Solve st 16t 2 32t 384 0 16t 2 32t 384 Time to this height, substitute 0 for s(t) thrust Beginning height 0 16(t 2 2t 24) 0 16t 4t 6 Factor out -16 factor t 4 0 t 6 0 t 4 4,6 t 6 The ball hits the ground after 6 seconds. Blitzer, Intermediate Algebra, 5e – Slide #21 Section 5.7 Modeling Motion Application Check Point 4 on page 374 finds when the ball height is at 336 feet Solve st 16t 2 32t 384 336 16t 2 32t 384 Time to this height, substitute 336 for s(t) 0 16t 2 32t 48 Subtract 336 0 16(t 2 2t 3) 0 16t 1t 3 Factor out -16 factor t 1 0 t 3 0 t 1 1,3 t3 The ball passes 336 after 3 seconds. (3,336) is the point on the graph Blitzer, Intermediate Algebra, 5e – Slide #22 Section 5.7 Modeling Motion Application Flight of a baseball The height in feet of a baseball after t seconds is given by the following. At what values of t is the height of the baseball 75 feet? s (t ) 16t 2 88t 3 Beginning height Time to this height, substitute 75 for s(t) thrust 75 16t 2 88t 3 Time to this height, substitute 75 for h Modeling Motion Application Flight of a baseball, continued 75 16t 2 88t 3 Multiply both sides by -1 75 16t 2 88t 3 0 16t 2 88t 3 75 16t 2 88t 72 0 Add 75 from both sides Combine like terms 8(2t 2 11t 9) 0 8t 12t 9 0 Answer is 1 second going up and 4.5 seconds going down t 1 0 2t 9 0 t 1 t 9 2 4.5 Both valid Modeling Motion Application Model Rocket #98 on page 388 A model rocket is launched from the top of a cliff 144 feet above sea level. The rocket misses the cliff and falls in the ocean. How long will it take for the rocket to hit the water? s (t ) 16t 2 128t 144 Beginning height thrust 0 16t 2 128t 144 Time to this height, substitute 0 for s(t) Modeling Motion Application 0 16t 2 128t 144 Model rocket #98 on page 388, continued 0 16(t 2 8t 9) 16t 9 t 1 0 t 9 0 t 1 0 t 9 t 1 Answer is 9 seconds Factor out -16 Modeling Motion Application Model Rocket A model rocket is launched from the ground. s (t ) 16t 2 48t How long will it take for the rocket to hit the ground? 0 16t 2 48t Beginning height is zero 0 16t (t 3) Time to this height, substitute 0 for s(t) The rocket strikes the ground after 3 seconds. -16t = 0 t= 0 t-3 = 0 t= 3 Modeling Motion Application Car breaking, #99 on 388 (from extra credit) How much distance do you need to bring your car to a complete stop? d(x) is the stopping distance, in feet, for a care traveling at x miles per hour. It takes 40 feet to come to a complete stop, substitute 40 for d(x) X+40=0 X-20=0 t = -40 t = 20 x2 d ( x) x 20 x2 40 x 20 800 x 2 20 x 0 x 2 20 x 800 0 ( x 40)( x 20) The car was traveling 20 miles per hour. Solving Polynomial Equations-Area Do Check Point 5 on page 375 – see word problem diagram (also see Example 5 on page 374.) Solve 12 2x16 2x 320 192 24x 32x 4 x 2 320 4 x 2 56x 192 320 4 x 2 56x 128 0 4x 16x 2 0 factor 16,2 The path should be 2 feet wide. P 375 Blitzer, Intermediate Algebra, 5e – Slide #29 Section 5.7 The Pythagorean Theorem, p 375 The Pythagorean Theorem The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse. If the legs have lengths a and b, and the hypotenuse has length c, then a 2 b 2 c 2 . B Hypotenuse has length c The two legs have lengths of a and b c A b a C Blitzer, Intermediate Algebra, 5e – Slide #30 Section 5.7 The Pythagorean Theorem EXAMPLE A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire. SOLUTION Since the wire is 1 foot longer than the height that it reaches on the tree, if we call the length of the wire (the quantity we wish to determine) x, then the height of the tree would be x -1. Tree x This is the equation arising from the Pythagorean Theorem Tree x-1 a 2 b2 c2 . x 12 52 x2 5 feet 5 feet Blitzer, Intermediate Algebra, 5e – Slide #31 Section 5.7 The Pythagorean Theorem CONTINUED We can now use the Pythagorean Theorem to solve for x, the length of the wire. x 12 52 x2 x 2 2 x 1 25 x 2 x 2 2 x 26 x 2 2 x 26 0 26 2 x 13 x This is the equation arising from the Pythagorean Theorem Square x – 1 and 5 Add 1 and 25 Subtract x 2 from both sides Add 2x to both sides Divide both sides by 2 Therefore, the solution is x = 13 feet. Blitzer, Intermediate Algebra, 5e – Slide #32 Section 5.7 The Pythagorean Theorem #82 on page 380 See picture on page 380 to solve for x, how far up the pole. x 42 x2 202 x 2 8x 16 x 2 400 2 x 2 8x 16 400 2 x 2 8x 384 0 2( x2 4x 192) 0 2( x 12)(x 16) x 12 0 x 12 x 16 0 x 16 This is the equation arising from the Pythagorean Theorem Square x + 4 and 20 Combine terms Subtract 400 from both sides GCF Factor Therefore, the solution is x = 12 feet. Blitzer, Intermediate Algebra, 5e – Slide #33 Section 5.7 Solving Polynomial Equations Important to Note: A polynomial equation is the result of setting two polynomials equal to each other. The equation is in standard form if one side is 0 and the polynomial on the other side is in standard form, that is in descending powers of the variable. A polynomial equation of degree one is a linear equation and of degree two is a quadratic equation. Some polynomial equations can be solved by writing the equation in standard form, factoring, and then using the zero-product principle: If a product is 0, then at least one of its factors is equal to 0. Blitzer, Intermediate Algebra, 5e – Slide #34 Section 5.7 DONE Modeling Motion Application h 16t 2 88t 3 Time to this height, substitute 75 for h thrust Beginning height 75 16t 2 88t 3 Time to this height, substitute 75 for h h 16t 2 132t 0 thrust Time to this height, substitute 0 for h Beginning height (height of building) Polynomial Equations in Application EXAMPLE A gymnast dismounts the uneven parallel bars at a height of 8 feet with an initial upward velocity of 8 feet per second. The function st 16t 2 8t 8 describes the height of the gymnast’s feet above the ground, s (t), in feet, t seconds after dismounting. The graph of the function is shown below. 10 Height (feet) 8 6 4 2 0 0 0.5 1 1.5 Time (seconds) Blitzer, Intermediate Algebra, 5e – Slide #37 Section 5.7 2 Polynomial Equations in Application CONTINUED When will the gymnast be 8 feet above the ground? Identify the solution(s) as one or more points on the graph. SOLUTION We note that the graph of the equation passes through the line y = 8 twice. Once when x = 0 and once when x = 0.5. This can be verified by determining when y = s (t) = 8. That is, st 16t 2 8t 8 8 16t 2 8t 8 0 16t 2 8t 0 8t 2t 1 Original equation Replace s (t) with 8 Subtract 8 from both sides Factor Blitzer, Intermediate Algebra, 5e – Slide #38 Section 5.7 Polynomial Equations in Application CONTINUED Now we set each factor equal to zero. 8t 0 t0 or 2t 1 0 2t 1 1 t 2 We have just verified the information we deduced from the graph. That is, the gymnast will indeed be 8 feet off the ground at t = 0 seconds and at t = 0.5 seconds. These solutions are identified by the dots on the graph on the next page. Blitzer, Intermediate Algebra, 5e – Slide #39 Section 5.7 Polynomial Equations in Application CONTINUED 10 Height (feet) 8 0 0.5 6 4 2 0 1 1.5 Time (seconds) Blitzer, Intermediate Algebra, 5e – Slide #40 Section 5.7 2