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Factoring Polynomials

Factoring Polynomials. Finding GCF. If you remember multiplication with real numbers, then you should remember the following facts:. If 6 • 2 = 12, then we know 12 is the product of 6 and 2 6 and 2 are factors , or divisors , of 12. the quotient of 12 divided by 6 is 2.

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Factoring Polynomials

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  1. Factoring Polynomials Finding GCF

  2. If you remember multiplication with real numbers, then you should remember the following facts: If 6 • 2 = 12, then we know • 12 is the product of 6 and 2 • 6 and 2 are factors, or divisors, of 12. • the quotient of 12 divided by 6 is 2. • the quotient of 12 divided 2 is 6. So, to factor a number is to write it as the product of two or more numbers, usually natural numbers. Factoring and division are closely related.

  3. Review: a factor - is a number that is multiplied by another number to produce a product. A prime number - is any natural number greater than 1 whose only factors are 1 and itself. A compositenumber - is a number greater than 1 that has more than two factors.The prime factorization - is the factorization of a natural number that contains only prime numbers or powers of prime numbers.

  4. The figure below shows 24 square tiles arranged to form a rectangle. (4 tiles wide, 6 tiles long)Sketch other ways that the 24 tiles can be arranged to form a rectangle. ANSWER

  5. If you notice each rectangle has an area of 24, which includes (4 x 6, 3 x 8, 2 x 12, 1 x 24).Each of the numbers involved in these multiplications is a factor of 24. There are no other natural number pairs that have a product of 24, so 1, 2, 3, 4, 6, 8, 12, and 24 are the only factors of 24. Here are two examples

  6. Write 3 different factorizations of 16. Use natural numbers. 2 x 8 4 x 4 8 x 2 1 x 16

  7. Finding Greatest Common Factor GCF’s

  8. Let’s make a list of factors of 36, written in order from least to greatest.36: 1, 2, 3, 4, 6, 9, 12, 18, 36Make a list of factors for the number 54.54: 1, 2, 3, 6, 9, 18, 27, 54Examine the two lists for factors that appear in both list.36: 1, 2, 3, 6, 9, 12, 18, 3654: 1, 2, 3, 6, 9, 18, 27, 54common factors: 1, 2, 3, 6, 9, 18GCF: 18

  9. What if the numbers in our previous example were expressions 36c3 and 54c2? 36c3 2 ∙ 2 ∙ 3 ∙ 3 ∙ c ∙ c ∙ c 21 ∙ 32 ∙ c2 18 ∙ c2 54c2 2 ∙ 3 ∙ 3 ∙ 3 ∙ c ∙ c 21 ∙32 ∙c2 18 ∙ c2 GCF 18c2 Try again

  10. Find the GCF: a) 18d5 and 108db) 18d and 5c) 3m3n3 and 9m2n2d) 4mn3, 4m2n3, and 16m2n2

  11. Factoring a monomial from a polynomial Using GCF

  12. Factoring a polynomial reverses the multiplication process. To factor a monomial from a polynomial, first find the greatest commonfactor (GCF) of its terms. Find the GCF of the terms of: 4x3 + 12x2 – 8x List the prime factors of each term. 4x3 = 2 · 2 · x · x x 12x2 = 2 · 2 · 3 · x · x 8x = 2 · 2 · 2 · x The GCF is 2 · 2 · x or 4x. Factoring a Monomial from a Polynomial

  13. Find the GCF of the terms of each polynomial.a) 5v5 + 10v3b) 3t2 – 18c) 4b3 – 2b2 – 6bd) 2x4 + 10x2 – 6x

  14. Use the GCF to factor each polynomial.a) 8x2 – 12xb) 5d3 + 10dc) 6m3 – 12m2 – 24md) 4x3 – 8x2 + 12x Try to factor mentally by scanning the coefficients of each term to find the GCF. Next, scan for the least power of the variable.

  15. Factor 3x3 – 12x2 + 15x Step 1 Find the GCF 3x3 = 3 · x · x · x 12x2 = 2 · 2 · 3 · x · x 15x = 3 · 5 · x The GCF is 3 · x or 3x Step 2 Factor out the GCF 3x3 – 12x2 + 15x = 3x(x2) + 3x(-4x) + 3x(5) = 3x(x2 – 4x + 5) Factoring Out a Monomial To factor a polynomial completely, you must factor until there are no common factors other than 1.

  16. Factoring x2 + bx + c Factoring x2 + bx + cwhen c is positive

  17. Observe the two columns below, the multiplication of binomials on the left and the products on the right.What do you notice about the two list? • (x + 5)(x + 6) • (x + 3)(x + 10) • (x + 2)(x + 15) 4. (x + 1)(x + 30) [x2 + 11x + 30] [x2 + 13x + 30] [x2 + 17x + 30] [x2 + 31x + 30]

  18. The product of the constants = 30The sum of the constants = the coefficient of the x-terms • (x + 5)(x + 6) • (x + 3)(x + 10) • (x + 2)(x + 15) • (x + 1)(x + 30) [x2 + 11x + 30] [x2 + 13x + 30] [x2 + 17x + 30] [x2 + 31x + 30] TRY THIS

  19. Remember the patterns you say earlier.x2 + 17x + 30Write the binomial multiplication that gives this product.( )( ) The sum of what two of those constants give you 17? What two constants multiplied together gives you 30?

  20. In other words, to factor x2 + bx + c, look for the factor pairs (the two numbers) whose product is c. Then choose the pair whose sum is b.Factor x2 + 5x + 61. c = 6: write all (+ and -) the factor pairs of 6. 1 • 6 2 • 3 -1 • -6 -2 • -32. b = 5: choose the pair whose sum is 5.1 + 6 = 7 2 + 3 = 5 -1 - 6 = -7 -2 – 3 = -53. Write the product using 2 and 3. Thus(x + 2)(x + 3) = x2 + 5x + 6 TRY THIS a2 + 9a + 20 NEXT

  21. Factor y2 – 10y + 241. c = 24 Write all (+ and -) factor pairs of 24.1 • 24 -1 • -24 4 • 6 -4 • -62 • 12 -2 • -12 3 • 8 -3 • -82. b = -10: choose the pair whose sum is -10.1 + 24 = 25 -1 +(-24) = -25 4 + 6 = 10 -4 + (-6) = -103. Write the product using -4 and -6.(y – 4)(y – 6) = y2 -10y + 24 TRY THIS n2 -13n + 36

  22. Practice and Problem Solving. • t2 + 7t + 10 = (t + 2)(t + □) • x2 – 8x + 7 = (x – 1)(x - □) • x2 + 9x + 18 = (x + 3)(x + □) Factor each expression • r2 +4r + 3 • n2 – 3n + 2 • k2 + 5k + 6 • x2 – 2x + 1 • y2 + 6y + 8

  23. Factoring x2 + bx + c Factoring x2 + bx + c, when c is negative

  24. Find each product.What do you notice about the c term in each? (x – 5)(x + 6) (x – 3)(x + 10) (x – 2)(x + 15) (x – 1)( x + 30) (x + 5)(x - 6) (x + 3)(x - 10) (x + 2)(x – 15) (x + 1)(x – 30) x2 + x – 30 x2 + 7x – 30 x2 + 13x – 30 x2 + 29x – 30 x2 – x – 30 x2 – 7x – 30 x2 – 13x – 30 x2 – 29x - 30 What do you notice about c in each expression? What do you notice about c and the coefficient of the x-term?

  25. You can also factor x2 + bx – cFactor x2 + x – 201. c = -20: write all (+ and -) factors of -20.(-1) • 20 1 • (-20) (-2) • 10 2 • (-10) (-4) • 5 4 • (-5)2. b = 1: find the pair whose sum is 1.(-1) • 20 1 • (-20) (-2) • 10 2 • (-10) (-4) • 5 4 • (-5)3. Write the product using -4 and 5 = 1(x – 4)(x + 5) = x2 + x - 20 TRY THIS n2 + 3n - 40

  26. Factor z2 – 4z – 121. c = -12: write all (+ and -)factors of -12.(-1) • 12 1 • (-12) (-2) • 6 2 • (-6) (-3) • 4 3 • (-4)2. b = -4: find the factor pair of -4.(-1) • 12 1 • (-12) (-2) • 6 2 • (-6) (-3) • 4 3 • (-4)3. write the product using 2 and -6.(x + 2)(x – 6) = x2 – 4z - 12 TRY THIS n2 – 3n - 40

  27. Not every polynomial of the form x2 + bx + cis factorable.Factor x2 + 3x – 1c = -1: the only factors of -1, are 1 and -1.b = 3: because -1 + 1 ≠ 3,x2 + 3x – 1 cannot be factored TRY THIS t2 + 5t - 8

  28. Factor:1. y2 + 10y – 112. x2 – x – 423. b2 – 17b – 384. s2 + 4s – 55. y2 + 2y – 63

  29. Factoring ax2 + bx + c Factoring ax2 + bx + c, when c is positive

  30. Before we tackle this factoring let’s go back and review the F O I L method of how the product of two binomials works.(2x + 3)(5x + 4) =10x2 + (8x +15x) + 12 =10x2 + 23x + 12 Last Terms First Terms Inner Terms Outer Terms Notice what happens when the multiply the Outer Terms and Inner Terms. NEXT

  31. 10x2 + 23x + 12To factor it, think of 23x as 8x + 15x.8x + 15x = 23x10x2 + 23x + 12 = 10x2 + 8x + 15x + 12Where did we get 8x and 15x?Notice that multiplying (a) 10 and (c) 12 gives you 120, which is the product of the x2-coefficient (10) and the constant term (12).8 and 15 are factors of 1201•120 2•60 3•40 4•30 5•24 6•20 8•15 10•12also 8x + 15x = 23This example suggest that, to factor a trinomial, you should look for factors of the product ac that have a sum of b. In the form of ax2 + bx + c Let’s see if it works.

  32. Consider the trinomial 6x2 + 23x + 7. To factor it, think of 23x as 2x + 21x.Where did we get 2 and 21?If we multiply 6 and 7 we get 42, which is the product of the x2-coefficient (6) and the constant (7).2 and 21 are factor of 421•42 2•21 3•14 6•7and2x + 21x = 23xYes it does work!So we must find the product of ac that have the sum b. NEXT

  33. Now we must rewrite the trinomial using the factors you found for b. (2 and 21)6x2 + 2x + 21x + 7Now we are going to find the GCF by grouping terms. (Remember the Associative Property)(6x2 + 2x) + (21x + 7)2x(3x + 1) + 7(3x + 1)Now lets use the Distributive Property to write the two binomials.(2x + 7)(3x + 1) FACTOR FACTOR What do you notice about the terms in the parenthesis? Now we have factored 6x2 + 23x + 7 to (2x + 7)(3x + 1)

  34. Factor 5x2 + 11x + 25x2 + 11x + 2 = 5x2 + 1x + 10x + 2Rewrite bx: 11x = 1x + 10x. = (5x2 + 1x)(10x + 2)Group terms, Associative Property. = x(5x + 1) + 2(5x + 1)Factor GCF of each pair of terms. = (x + 2)(5x + 1)Use Distributive Property to write factored terms. Step 1: Find factors of ac that have a sum b. Since ac = 10 and b = 11, find the positive factors of 10 that have a sum 11. Step2: To factor the trinomial, use the factors you found (1 + 10) to rewrite bx. TRY ANOTHER 6x2 + 13x + 5

  35. What is the factored form of 6x2 + 13x + 5?6x2 + 13x + 5 = 6x2 + 3x + 10x + 5Rewrite bx: 13x = 3x + 10x.(6x2 + 3x)(10x + 5)Group terms together to factor.3x(2x + 1) + 5(2x + 1)Factor GCF of each pair of terms.(3x + 5)(2x + 1)Use Distributive Property to write binomials.

  36. Factor:1) 2n2 + 11n + 52) 5x2 + 34x + 243) 2y2 – 23y + 604) 4y2 + 62y + 305) 8t2 + 26t + 15

  37. Factoring ax2 + bx +c Factoring when ac is negative

  38. Can we apply the same steps we have learned to factor trinomials that contain negative numbers?Yes. Your goal is still to find factors of ac that have sum b. Because ac< 0 (less than), the factors must have different signs.We need to use all combinations of factors.Ex: factors of -15.1•(-15) (-1)•15 3•(-5) (-3)•51+(-15)=-14 (-1)+15=14 3+(-5)=-2 (-3)+5 =2 The sums of positive and negative numbers gives us our b.

  39. Factor 3x2 + 4x – 15Find factors of ac with the sum b.Since ac = -45 and b = 4, find factors of -45.3x2 -5x +9x – 15 Rewrite bx: 4x = -5x + 9x.(3x2 -5x) + (9x – 15) Group terms together to factor. x(3x – 5) + 3(3x– 5) Factor GCF of each pair of terms.(x + 3)(3x – 5) Use Distributive to rewrite binomials.

  40. Factor:1) 3k2 + 4k – 42) 5x2 + 4x – 13) 10y2 – 11y – 64) 6q2 – 7q – 495) 2y2 + 11y – 90 Not all expressions of the form ax2 + bx – c can be factored. This is especially common when the polynomial contains subtraction. Try this. -10x2 + 21x - 5 TRY THIS

  41. Geometry The area of a rectangle is 2y2 – 13y – 7.What are the possible dimensions of the rectangle? Use factoring.2y2 – 13y – 7 =

  42. Simplifying before factoring Some polynomials can be factored repeatedly.

  43. Some polynomials can be factored repeatedly. This means you can continue the process of factoring until there are no common factors other than 1. If a trinomial has a common monomial factor, factor it out before trying to find binomial factors.Take for example:(6h + 2)(h + 5) and (3h + 1)(2h + 10)Find each product.6h2 + 32h + 10 and 6h+ 32h + 10 What do you notice about these two polynomials? Can you factor a monomial before factoring a binomial? NEXT

  44. Factor 20x2 + 80x + 35 completely.20x2 + 80x + 35 =Factor out GCF monomial.5(4x2 + 16x + 7) = Rewrite bx: 16x = 2x + 14x.5[4x2 + 2x + 14x + 7)] = Factor GCF of each pair of terms.5[2x(2x + 1) + 7(2x + 1)] = 5(2x + 7)(2x + 1)Rewrite using the Distributive Property. Step 1: Find factors of ac with sum b. Step 2: Rewrite bx using factors.

  45. Solving Polynomial Equations by Factoring Finding x

  46. If you remember graphing linear equations, many times the line crossed the x-axis. You could find the x- and y-intercepts of these lines.Polynomials have the same characteristics, but quadratics can have no x-intercept, one x-intercept or two x-intercepts. x-intercept x-intercept x-intercept y-intercept y-intercept

  47. Make a table of the polynomial shown below. y = x2 + 2x – 3Identify the numbers that appear to be the x-intercepts.Rewrite the equation by factoring the right side.[y = (x + 3)(x – 1)]If you notice the x-intercept are solutions to the equation (x + 3)(x – 1) = 0 x-intercept = -3 x-intercept = 1

  48. Remember the Standard Form of a Quadratic Equationax2 + bx + c, where a ≠ 0. The value of the variable in a standard form equation is called the solution, or the root, of the equation.Let’s discuss the Multiplication Property of Zero which states that if a = 0 or b = 0, then ab = 0. We can use the Zero-Product Property to solve quadratic equations once the quadratic expression has been factored into a product of two linear factors.Let’s see how we can use this property to solve quadratic expressions.

  49. Solve (4x + 5)(3x – 2) = 0 (4x + 5)(3x – 2) = 0 The quadratic expression has already been factored.4x + 5 = 0 or 3x – 2 = 0 If (4x + 5)(x – 2) = 0, then (4x + 5) = 0 or 3x – 2) = 0. 4x = -5 3x = 2 x = -5/4 x = 2/3 Solve each equation for x. Apply the Zero-Product Property. x-intercept x-intercept TRY THIS

  50. Solve (x + 5)(2x – 6) = 0(x + 5)(2x – 6) = 0 x + 5 = 0 or 2x – 6 = 0Use the Zero-Product Property. 2x = 6Solve for x. x = -5 or x = 3Substitute – 5 for x. Substitute 3 for x.(x + 5)(2x – 6) = 0 (x + 5)(2x – 6) = 0(-5 + 5)[2(-5) – 6] = 0 (3 + 5)[2(3) – 6) = 0 (0)(-16) = 0 (8)(0) = 0

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