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§ 5.7 Polynomial Equations and Their Applications.

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Presentation on theme: "§ 5.7 Polynomial Equations and Their Applications."— Presentation transcript:

1 § 5.7 Polynomial Equations and Their Applications

2 Blitzer, Algebra for College Students, 6e – Slide #2 Section 5.7 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.

3 Blitzer, Algebra for College Students, 6e – Slide #3 Section 5.7 Solving Polynomial Equations Definition of a Quadratic Equation A quadratic equation in x is an equation that can be written in the standard form where a, b, and c are real numbers, with. 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.

4 Blitzer, Algebra for College Students, 6e – Slide #4 Section 5.7 Solving Polynomial Equations Solving a Quadratic Equation by Factoring 1)If necessary, rewrite the equation in the standard form, 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.

5 Blitzer, Algebra for College Students, 6e – Slide #5 Section 5.7 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: 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. Subtract 45 from both sides Simplify 2) Factor. Factor

6 Blitzer, Algebra for College Students, 6e – Slide #6 Section 5.7 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. or CONTINUED 5) Check the solutions in the original equation. Check 9: Check -5: ? ? ? ?

7 Blitzer, Algebra for College Students, 6e – Slide #7 Section 5.7 Solving Polynomial EquationsCONTINUED Check 9: Check -5: The solutions are 9 and -5. The solution set is {9,-5}. ?? The graph of lies to the right. true

8 Blitzer, Algebra for College Students, 6e – Slide #8 Section 5.7 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: 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! Subtract 4x from both sides Simplify 2) Factor. Factor

9 Blitzer, Algebra for College Students, 6e – Slide #9 Section 5.7 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. or CONTINUED 5) Check the solutions in the original equation. Check 0: Check 4: ?? true

10 Blitzer, Algebra for College Students, 6e – Slide #10 Section 5.7 Solving Polynomial EquationsCONTINUED The solutions are 0 and 4. The solution set is {0,4}. The graph of lies to the right.

11 Blitzer, Algebra for College Students, 6e – Slide #11 Section 5.7 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: 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. Simplify Subtract 14 from both sides

12 Blitzer, Algebra for College Students, 6e – Slide #12 Section 5.7 Solving Polynomial EquationsCONTINUED 2) Factor. Before we can factor the equation, we must simplify it first. FOIL Simplify Now we can factor the polynomial equation.

13 Blitzer, Algebra for College Students, 6e – Slide #13 Section 5.7 Solving Polynomial EquationsCONTINUED 3) & 4) Set each factor equal to zero and solve the resulting equations. or 5) Check the solutions in the original equation. Check 3: Check -6: ?? ??

14 Blitzer, Algebra for College Students, 6e – Slide #14 Section 5.7 Solving Polynomial EquationsCONTINUED Check 3: Check -6: true The solutions are 3 and -6. The solution set is {3,-6}. The graph of lies to the right.

15 Blitzer, Algebra for College Students, 6e – Slide #15 Section 5.7 Solving Polynomial EquationsEXAMPLE SOLUTION Solve by factoring: 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. Common factor is Common factor is -1.

16 Blitzer, Algebra for College Students, 6e – Slide #16 Section 5.7 Solving Polynomial EquationsCONTINUED Factor 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 as the difference of two squares

17 Blitzer, Algebra for College Students, 6e – Slide #17 Section 5.7 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. CONTINUED or 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? The graph of lies to the right.

18 Blitzer, Algebra for College Students, 6e – Slide #18 Section 5.7 Polynomial Equations in ApplicationEXAMPLE 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 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.

19 Blitzer, Algebra for College Students, 6e – Slide #19 Section 5.7 Polynomial Equations in ApplicationSOLUTION When will the gymnast be 8 feet above the ground? Identify the solution(s) as one or more points on the graph. 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, CONTINUED Original equation Replace s (t) with 8 Subtract 8 from both sides Factor

20 Blitzer, Algebra for College Students, 6e – Slide #20 Section 5.7 Polynomial Equations in Application Now we set each factor equal to zero. CONTINUED 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. or

21 Blitzer, Algebra for College Students, 6e – Slide #21 Section 5.7 Polynomial Equations in ApplicationCONTINUED

22 Blitzer, Algebra for College Students, 6e – Slide #22 Section 5.7 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 C B c a b Hypotenuse has length c The two legs have lengths of a and b

23 Blitzer, Algebra for College Students, 6e – Slide #23 Section 5.7 The Pythagorean TheoremEXAMPLE SOLUTION 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. Let’s begin with a picture. Tree 5 feet

24 Blitzer, Algebra for College Students, 6e – Slide #24 Section 5.7 The Pythagorean Theorem 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 5 feet CONTINUED x-1 x

25 Blitzer, Algebra for College Students, 6e – Slide #25 Section 5.7 The Pythagorean Theorem We can now use the Pythagorean Theorem to solve for x, the length of the wire. CONTINUED This is the equation arising from the Pythagorean Theorem Square x – 1 and 5 Add 1 and 25 Subtract from both sides Add 2x to both sides Divide both sides by 2 Therefore, the solution is x = 13 feet.

26 Blitzer, Algebra for College Students, 6e – Slide #26 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.


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