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6.4 Solving Polynomial Equations. One of the topics in this section is finding the cube or cube root of a number. A cubed number is the solution when.

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Presentation on theme: "6.4 Solving Polynomial Equations. One of the topics in this section is finding the cube or cube root of a number. A cubed number is the solution when."— Presentation transcript:

1 6.4 Solving Polynomial Equations

2 One of the topics in this section is finding the cube or cube root of a number. A cubed number is the solution when a number is multiplied by itself three times. A cube root “undos” the cubing operation just like a square root would.

3 Calculator Function – How to take the cube root of a number To take the cube root of a number, press MATH, then select option 4. Example: What is ? 24

4 Solving Polynomials by Graphing We start getting into more interesting equations now... Ex: x 3 + 3x 2 = x + 3 Problem: Solve the equation above, using a graphing calculator

5 Solving Polynomials by Graphing What about something like this??? Ex: x 3 + 3x 2 + x = 10 Use the same principal; plug the first part of the equation in for Y1; the solution (10) for Y2; then find the intersection of the two graphs.

6 FACTORING AND ROOTS CUBIC FACTORING a³ + b³ = (a + b)(a² - ab + b²) a³ - b³ = (a - b)(a² + ab + b²) Difference of Cubes Sum of Cubes Question: if we are solving for x, how many possible answers can we expect? 3 because it is a cubic!

7 CUBIC FACTORING EX- factor and solve 8x³ - 27 = 0 8x³ - 27 = (2x - 3)((2x)² + (2x)3 + 3²) (2x - 3)(4x² + 6x + 9)=0 Quadratic FormulaX= 3/2 a³ - b³ = (a - b)(a² + ab + b²)

8 CUBIC FACTORING EX- factor and solve x³ = 0 a³ + b³ = (a + b)(a² - ab + b²) x³ = (x + 7)(x² - 7x + 7²) (x + 7)(x² - 7x + 49)=0 Quadratic FormulaX= -7

9 Let’s try one Factor a)x 3 -8b)x

10 Let’s try one Factor a)x 3 -8b)x

11 Let’s Try One 81x =0 Hint: IS there a GCF???

12 Let’s Try One 81x =0

13 Factor by Using a Quadratic Form Ex: x 4 -2x 2 -8 Since this equation has the form of a quadratic expression, we can factor it like one. We will make temporary substitutions for the variables = (x 2 ) 2 – 2(x 2 ) – 8 Substitute a in for x 2 = a 2 – 2a – 8 This is something that we can factor (a-4)(a+2) Now, substitute x 2 back in for a (x 2 -4)(x 2 +2) (x 2 -4) can factor, so we rewrite it as (x-2)(x+2) So, x 4 -2x 2 -8 will factor to (x-2)(x+2)(x 2 +2)

14 Let’s Try One Factor x 4 +7x 2 +6

15 Let’s Try One Factor x 4 +7x 2 +6

16 Let’s Try One Where we SOLVE a Higher Degree Polynomial x 4 -x 2 = 12

17 Let’s Try One Where we SOLVE a Higher Degree Polynomial x 4 -x 2 = 12


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