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Solving Polynomial Equations by Factoring

Solving Polynomial Equations by Factoring. Quick Review. FACTORING a polynomial means to break it apart into its prime factors. For example: x 2 – 4 = ( x + 2)( x – 2) x 2 + 6 x + 5 = ( x + 1)( x + 5) 3 y 2 + 10 y – 8 = (3 y – 2)( y + 4).

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Solving Polynomial Equations by Factoring

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  1. Solving Polynomial Equations by Factoring

  2. Quick Review • FACTORING a polynomial means to break it apart into its prime factors. • For example: • x2 – 4 = (x + 2)(x – 2) • x2 + 6x + 5 = (x + 1)(x + 5) • 3y2 + 10y – 8 = (3y – 2)(y + 4)

  3. In this lesson, we want to factor polynomials that have four terms. To do this, we will use a method called grouping.

  4. For example, let’s factor ax + ay + bx + by. • Let’s begin by grouping the first two terms and the last two terms: • ax + ay + bx+ by • Next, factor out the GCF of each group: • a(x + y) + b(x + y) • Notice that (x + y) is a common factor of each term. This means: • ax + ay + bx + by = (x + y)(a + b)

  5. Factor: x3 + 7x2 + 2x + 14 • First, group each set of terms: • x3+ 7x2+ 2x+ 14 • Then, take out GCFs: • x2(x + 7) + 2(x + 7) • Write final answer: • (x + 7)(x2 + 2)

  6. Factor: 3x2 + xy – 12x – 4y • First, group each set of terms: • 3x2+ xy– 12x+ 4y • Notice that because there was a minus here, we needed to use a negative sign in our second group! • Then, take out GCFs: • x(3x + y) – 4(3x + y) • Write final answer: • (3x + y)(x – 4)

  7. You try these. When you are ready to check your answers, move to the next slide. • a3 – 2a2 + 5a – 10 • 15y3 + 24y2 – 35y – 56

  8. How well did you do? 1. a3 – 2a2 + 5a – 10 Group: a3– 2a2+ 5a– 10 GCFs: a2(a – 2) + 5(a – 2) 2.15y3 + 24y2 – 35y – 56 Group: 15y3+ 24y2– 35y+ 56 **Notice the sign change! GCFs: 3y2(5y + 8) – 7(5y + 8) Answer: (a – 2)(a2 + 5) Answer: (5y + 8)(3y2 – 7)

  9. Earlier in the course, you learned how to solve polynomial equations by factoring.

  10. For example, let’s solve the equation 3x2 + 2x = 5. • First, remember that your equation must be set equal to 0 before factoring. • 3x2 + 2x – 5 = 0 • There isn’t a GCF to factor out. So, factor. • (3x + 5)(x – 1) = 0 • Next, using the Zero Product Property, set each factor equal to 0 and solve for x. • 3x + 5 = 0 or x – 1 = 0 • x = -5/3 or x = 1 • The solution set is: Note the graph shows us 2 zeros, so our solution makes sense!

  11. Now, let’s solve some more challenging polynomial equations!

  12. Solve: 4x5 – 6x4 – 4x3 = 0 • These three terms have a GCF that we will need to factor out first: • 2x3(2x2 – 3x – 2) = 0 • Then, factor: • 2x3(2x + 1)(x – 2) = 0 • Set each factor equal to 0: • 2x3 = 0 or 2x + 1 = 0 or x – 2 = 0 • Solve for x: Note the graph shows us 3 zeros, so our solution makes sense!

  13. Solve: 3x4 – 4x2 = -1 • First, we need to set our equation equal to 0: • 3x4 – 4x2 + 1 = 0 • Then, factor: • (3x2 – 1)(x2 – 1) = 0 • Set each factor equal to 0: • 3x2 – 1 = 0 or x2 – 1 = 0 • Solve for x: Note the graph shows us 4 zeros, so our solution makes sense!

  14. Solve: x4 – 1 = 0 • First, factor: • (x2 + 1)(x2 – 1) = 0 • (x2 + 1)(x + 1)(x – 1) = 0 • Set each factor equal to 0: • x2 + 1 = 0 or x + 1 = 0 or x – 1 = 0 • Solve for x: Note the graph shows us 2 zeros, because our calculator only shows REAL zeros.

  15. One of the most widely used applications of solving polynomial equations is modeling expenses and profit. Often times, the sales of a particular item can be modeled using a polynomial equation. Companies can use these equations to predict what prices will result in the largest profit margins.

  16. For example: • A company sells an item for x dollars. Their revenue y (in dollars) is given by the polynomial equation y = -10x4 + 1000x2. At what price will the company stop making money?

  17. Example continued • Look at the graph of the function first: • We want to find the zeros, because then we will know when the company isn’t making any money (revenue of $0). • We can factor the polynomial equation to find these zeros!!!

  18. Example continued • Let’s factor 0 = -10x2(x2 – 100) • 0 = -10x2(x + 10)(x – 10) • Now set each factor equal to 0 and solve for x: • -10x2 = 0 or x + 10 = 0 or x – 10 = 0 • -10 doesn’t make sense in the situation because the company wouldn’t charge -$10 • 0 doesn’t make sense in the situation because the company wouldn’t charge $0 • 10 does make sense! If the company charges $10, they won’t make any money…this would most likely be because people wouldn’t spend that much money for the item.

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