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Lesson 4.3 , For use with pages 252-258

Lesson 4.3 , For use with pages 252-258. Find the product. 1. ( m – 8) ( m – 9) . m 2 – 17 m + 72. ANSWER. 2. ( z + 6 ) ( z – 10 ). z 2 – 4 z – 60. ANSWER. Lesson 4.3 , For use with pages 252-258. Find the product. 3. ( y + 20) ( y – 20). y 2 – 400. ANSWER.

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Lesson 4.3 , For use with pages 252-258

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  1. Lesson 4.3, For use with pages 252-258 Find the product 1. (m – 8) (m – 9) m2 – 17m + 72 ANSWER 2. (z + 6) (z – 10) z2 – 4z – 60 ANSWER

  2. Lesson 4.3, For use with pages 252-258 Find the product 3. (y +20) (y – 20) y2 – 400 ANSWER 4. (d +9)2 d2 + 18d +81 ANSWER

  3. Lesson 4.3, For use with pages 252-258 Find the product 5. (x – 14)2 ANSWER x2– 28x +196 6. A car travels at an average speed of (m+17) miles per hour for (m + 2) hours. What distance does it travel? (m2 + 9m + 14) mi ANSWER

  4. 4.3 Factoring Trinomials

  5. Squares 1-20, and 25 • Write the squares 1-20, 25 • = 4

  6. Factoring Quadratic Rules Given Factored a + bx + c a - bx + c a - bx - c a +bx - c ( + ) ( + ) ( - ) ( - ) ( - Big) ( + Small ) ( + Big) ( - Small )

  7. Factoring Procedures for a Quadratic Equation y = a “No Fuse Method” The trinomial should be set = to y or 0, before beginning to factor. Examine each term to determine if there is a GCF that will factor from each term (Note: the GCF must factor from all three terms). If there is a GCF that will factor from all three terms, factor each term. Look at the Given equations and determine the signs for the factored binomial. If the first term in the trinomial is split the to x and x and put each x in the first term in each binomial. EX: ( x + ) ( x + ). Next multiply the outside term a and c together ( 1 x 3 = 3) and determine the factors of 3. The factors of 3 are 1 and 3, determine if you should add or subtract them to obtain the middle term of 4. EX: 1+3 = 4. Put the factors in the last terms of each binomial, the order does not matter because the signs on the factored binomials are both positive (x + 3) ( x + 1). If the factored binomial had been a big and small factor, then the 3 would have been the big factor and the 1 would have been the small factor. Foil the factored binomial to determine if the quadratic equation was factored properly

  8. ANSWER Notice thatm = – 4andn = – 5. So, x2–9x + 20 = (x –4)(x –5). EXAMPLE 1 Factor trinomials of the form x2+ bx + c Factor the expression. a. x2 – 9x + 20 b. x2 + 3x – 12 SOLUTION a.You wantx2 – 9x + 20 = (x + m) (x + n)where mn= 20 andm + n = –9.

  9. ANSWER Notice that there are no factors mand nsuch that m + n = 3. So, x2 + 3x – 12 cannot be factored. EXAMPLE 1 Factor trinomials of the form x2+ bx + c b.You wantx2 + 3x – 12 = (x + m) (x + n)where mn= – 12 andm + n = 3.

  10. for Example 1 GUIDED PRACTICE Factor the expression. If the expression cannot be factored, say so. 1. x2 – 3x – 18 SOLUTION You wantx2 – 3x – 18 = (x + m) (x + n)where mn= – 18andm + n = –3.

  11. ANSWER Notice m = – 6 and n = 3 so x2 – 3x – 18 = (x – 6) (x + 3) for Example 1 GUIDED PRACTICE

  12. for Example 1 GUIDED PRACTICE 2. n2 – 3n + 9 SOLUTION You wantn2 – 3n + 9 = (x + m) (x + n)where mn= 9 andm + n = –3.

  13. ANSWER Notice that there are no factors mand nsuch that m + n = – 3 . So, n2 – 3x + 9 cannot be factored. for Example 1 GUIDED PRACTICE

  14. for Example 1 GUIDED PRACTICE 3. r2 + 2r – 63 SOLUTION You wantr2 + 2r – 63 = (x + m) (x + n)where mn= –63 andm + n = 2.

  15. ANSWER Notice that m = 9 and n = – 7 . So, r2 + 2r – 63 = (r + 9)(r –7) for Example 1 GUIDED PRACTICE

  16. EXAMPLE 2 Factor with special patterns Factor the expression. a. x2 – 49 = x2 – 72 Difference of two squares = (x + 7) (x –7) b. d 2 + 12d + 36 = d 2 + 2(d)(6) + 62 Perfect square trinomial = (d + 6)2 c. z2 – 26z + 169 Perfect square trinomial = z2 – 2(z) (13) + 132 = (z –13)2

  17. for Example 2 GUIDED PRACTICE Factor the expression. 4. x2 – 9 = x2 – 32 Difference of two squares = (x –3) (x +3) 5. q2 – 100 = q2 – 102 Difference of two squares = (q – 10) (q + 10) 6. y2 + 16y + 64 = y2 + 2(y) 8 + 82 Perfect square trinomial = (y +8)2

  18. for Example 2 GUIDED PRACTICE 7. w2 – 18w + 81 = w2 – 2(w) + 92 Perfect square trinomial = (w –9)2

  19. x – 9 = 0 or x + 4 = 0 x = 9 or x = – 4 ANSWER The correct answer is C. EXAMPLE 3 Standardized Test Practice SOLUTION x2 – 5x – 36 = 0 Write original equation. (x – 9)(x + 4) = 0 Factor. Zero product property Solve for x.

  20. A town has a nature preserve with a rectangular field that measures 600 meters by 400 meters. The town wants to double the area of the field by adding land as shown. Find the new dimensions of the field. EXAMPLE 4 Use a quadratic equation as a model Nature Preserve

  21. or x –200 = 0 x + 1200 = 0 or x = 200 x = –1200 EXAMPLE 4 Use a quadratic equation as a model SOLUTION 480,000 = 240,000 + 1000x + x2 Multiply using FOIL. 0 = x2+ 1000x –240,000 Write in standard form. 0 = (x –200) (x + 1200) Factor. Zero product property Solve for x.

  22. EXAMPLE 4 Use a quadratic equation as a model ANSWER Reject the negative value, – 1200. The field’s length and width should each be increased by 200 meters. The new dimensions are 800 meters by 600 meters.

  23. x + 6 = 0 or x – 7 = 0 x = – 6 or x = 7 for Examples 3 and 4 GUIDED PRACTICE 8. Solve the equationx2 – x – 42 = 0. SOLUTION x2 – x – 42 = 0 Write original equation. (x + 6)(x – 7) = 0 Factor. Zero product property Solve for x.

  24. New width (meters) New Area New Length (meters) = 2(1000)(300) (1000 + x) (300 + x) = or x –200 = 0 x + 1500 = 0 for Examples 3 and 4 GUIDED PRACTICE 9. What If ? In Example 4, suppose the field initially measures 1000 meters by 300 meters. Find the new dimensions of the field. SOLUTION Multiply using FOIL. 600000 = 300000 + 1000x + 300x + x2 0 = x2+ 1300x –300000 Write in standard form. 0 = (x –200) (x + 1500) Factor. Zero product property

  25. ANSWER Reject the negative value, – 1200. The field’s length and width should each be increased by 200 meters. The new dimensions are 1200 meters by 500 meters. for Examples 3 and 4 GUIDED PRACTICE or x = 200 x = –1500 Solve for x.

  26. EXAMPLE 5 Find the zeros of quadratic functions. Find the zeros of the function by rewriting the function in intercept form. a. y = x2 – x – 12 b. y = x2 + 12x + 36 SOLUTION a. y = x2 – x – 12 Write original function. = (x + 3) (x – 4) Factor. The zeros of the function are –3 and 4. CheckGraph y = x2 – x – 12. The graph passes through (–3, 0) and (4, 0).

  27. EXAMPLE 5 Find the zeros of quadratic functions. Find the zeros of the function by rewriting the function in intercept form. a. y = x2 – x – 12 b. y = x2 + 12x + 36 SOLUTION b. y = x2 + 12x + 36 Write original function. = (x + 6) (x + 6) Factor. The zeros of the function is – 6 CheckGraph y = x2 + 12x + 36. The graph passes through ( – 6, 0).

  28. for Example for Example 5 GUIDED PRACTICE GUIDED PRACTICE Find the zeros of the function by rewriting the function in intercept form. 10. y = x2 + 5x – 14 SOLUTION y = x2 + 5x – 14 Write original function. = (x + 7) (x – 2) Factor. The zeros of the function is – 7 and 2 CheckGraph y = x2 + 5x – 14. The graph passes through ( – 7, 0) and (2, 0).

  29. for Example for Example 5 GUIDED PRACTICE GUIDED PRACTICE 11. y = x2 – 7x – 30 SOLUTION y = x2 – 7x – 30 Write original function. = (x + 3) (x – 10) Factor. The zeros of the function is – 3 and 10 CheckGraph y = x2 – 7x – 30. The graph passes through ( – 3, 0) and (10, 0).

  30. for Example for Example 5 GUIDED PRACTICE GUIDED PRACTICE 12. f(x)= x2 – 10x + 25 SOLUTION f(x)= x2 – 10x + 25 Write original function. = (x – 5) (x – 5) Factor. The zeros of the function is 5 CheckGraph f(x)= x2 – 10x + 25. The graph passes through ( 5, 0).

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