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EQ : How is knowing the zero product property helpful when solving equations by factoring?

EQ : How is knowing the zero product property helpful when solving equations by factoring?. Factor. 1. 16x 4 – 36x 2 y 2 2. 20a 2 – 7a – 6 3. 6t 3 – 21t 2 – 45t 4. 4n 3 – 20n 2 – 9n + 45 5. The area of a rectangle is: 20a 2 – 13a – 15. Find the dimensions.

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EQ : How is knowing the zero product property helpful when solving equations by factoring?

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  1. EQ: How is knowing the zero product property helpful when solving equations by factoring? Factor. 1. 16x4 – 36x2y2 2. 20a2 – 7a – 6 3. 6t3 – 21t2 – 45t 4. 4n3 – 20n2 – 9n + 45 5. The area of a rectangle is: 20a2 – 13a – 15. Find the dimensions.

  2. I’ve Got the Answers: 1. 4x2 (2x – 3y) (2x + 3y) 2. (4a – 3) (5a + 2) 3. 3t (t – 5) (2t + 3) 4. (n – 5) (2n – 3) (2n + 3) 5. (5a + 3) by (4a – 5)

  3. Lesson 10-6 Solving Equations by Factoring

  4. Now that we know how to factor, we can use factoring to solve polynomial equations with one variable such as: k2 - 9k = -18 You can use the zero product property to solve equations by factoring.

  5. Zero Product Property: For all numbers a and b, if a*b = 0, then a = 0, b = 0, or both a and b equal 0. If a product equals zero, at least one of the factors has to equal zero.

  6. Solve: k2 - 9k = -18 So, we try to get the polynomial to equal zero. In our example, that means adding 18 to both sides, so that there is nothing left (or zero) on the other side. Now, we have: k2 – 9k + 18 = 0

  7. Find the Factors… Factor pairs are: 18 and 1, 9 and 2, 6 and 3 Let’s try -3 and -6 Sum-3 + -6 = -9 (fits with middle term) Product-3 * -6 = 18 (fits with last term)

  8. The equation factors to: (k – 3) (k – 6) We will set each of the factors equal to zero to find the solution. (k – 3) = 0 or k – 3 = 0 After solving, k = 3 The other factor (k – 6) also needs to be set equal to zero and solved. k – 6 = 0 So, k =6. There are 2 solutions (which should be written in a solution set) {3, 6}

  9. Let’s try one more: 2x3 – 11x2 = 6x Step 1: Get everything on one side. 2x3 – 11x2 – 6x = 0 Step 2: Remove any common factors. x (2x2 – 11x – 6) = 0 Step 3: Continue factoring if possible.

  10. Factors of 6: 6 and 1 3 and 2 We are going to be multiplying one of the factors by two in order to get 2x2 for our first term. Last term is negative, so we need one positive and one negative factor. - 6 * 2 = - 12 + 1 - 11

  11. Our final factoring: x (2x + 1) (x – 6) = 0 Don’t forget the GCF you factored out So, x = 0, 2x + 1 = 0, x – 6 = 0 Solution set {-½, 0, 6}

  12. 1. x2 + 64 = 16x x2 – 16x + 64 = 0 (x – 8)(x – 8) = 0 x = { 8 } 2. 7n2 = 35n 7n2 – 35n = 0 7n (n – 5) = 0 n = { 0, 5 } 3. 5g + 6 = -g2 g2 + 5g + 6 = 0 (g + 3) (g + 2) = 0 g = { -2, -3 } 4. x2 - 4x - 21 = 0 (x – 7) (x + 3) = 0 x = { -3, 7 } Practice, Practice, Practice!!

  13. Any Questions?? This is the Last Lesson in the Chapter… Start Studying!!!

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