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## Factoring Polynomials

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**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. • 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.**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.**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**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**Write 3 different factorizations of 16. Use natural**numbers. 2 x 8 4 x 4 8 x 2 1 x 16**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**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**Find the GCF: a) 18d5 and 108db) 18d and 5c) 3m3n3 and**9m2n2d) 4mn3, 4m2n3, and 16m2n2**Factoring a monomial from a polynomial**Using GCF**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**Find the GCF of the terms of each polynomial.a) 5v5 +**10v3b) 3t2 – 18c) 4b3 – 2b2 – 6bd) 2x4 + 10x2 – 6x**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.**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.**Factoring x2 + bx + c**Factoring x2 + bx + cwhen c is positive**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]**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**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?**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**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**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**Factoring x2 + bx + c**Factoring x2 + bx + c, when c is negative**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?**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**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**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**Factor:1. y2 + 10y – 112. x2 – x – 423. b2 – 17b**– 384. s2 + 4s – 55. y2 + 2y – 63**Factoring ax2 + bx + c**Factoring ax2 + bx + c, when c is positive**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**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.**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**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)**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**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.**Factor:1) 2n2 + 11n + 52) 5x2 + 34x + 243) 2y2 – 23y +**604) 4y2 + 62y + 305) 8t2 + 26t + 15**Factoring ax2 + bx +c**Factoring when ac is negative**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.**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.**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**Geometry The area of a rectangle is 2y2 – 13y – 7.What**are the possible dimensions of the rectangle? Use factoring.2y2 – 13y – 7 =**Simplifying before factoring**Some polynomials can be factored repeatedly.**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**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.**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**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**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.**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**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