Today we will Finish chapter 5 Math 083 Bianco Today we will Finish chapter 5

Factoring by Special Products § 5.7 Factoring by Special Products

Perfect Square Trinomials (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 So if the first and last terms of our polynomial to be factored can be written as expressions squared, and the middle term of our polynomial is twice the product of those two expressions, then we can use these two previous equations to easily factor the polynomial. a2 + 2ab + b2 = (a + b)2 a2 – 2ab + b2 = (a – b)2

Difference of Two Squares Perfect Square Trinomials a2 – b2 = (a + b)(a – b) A binomial is the difference of two square if both terms are squares and the signs of the terms are different. 9x2 – 25y2 – c4 + d4

Difference of Two Squares Example: Factor the polynomial x2 – 9. The first term is a square and the last term, 9, can be written as 32. The signs of each term are different, so we have the difference of two squares Therefore x2 – 9 = (x – 3)(x + 3). Note: You can use FOIL method to verify that the factorization for the polynomial is accurate.

Difference of Two Squares Example: Factor x2 – 16. Since this polynomial can be written as x2 – 42, x2 – 16 = (x – 4)(x + 4). Factor 9x2 – 4. Since this polynomial can be written as (3x)2 – 22, 9x2 – 4 = (3x – 2)(3x + 2). Factor 16x2 – 9y2. Since this polynomial can be written as (4x)2 – (3y)2, 16x2 – 9y2 = (4x – 3y)(4x + 3y).

Difference of Two Squares Example: Factor x8 – y6. Since this polynomial can be written as (x4)2 – (y3)2, x8 – y6 = (x4 – y3)(x4 + y3). Factor x2 + 4. Oops, this is the sum of squares, not the difference of squares, so it can’t be factored. This polynomial is a prime polynomial.

Sum and Difference of Two Cubes There are two additional types of binomials that can be factored easily by remembering a formula. Sum and Difference of Two Cubes a3 + b3 = (a + b)(a2 – ab + b2) a3 – b3 = (a – b)(a2 + ab + b2)

Sum and Difference of Two Cubes Example: Factor x3 + 1. Since this polynomial can be written as x3 + 13, x3 + 1 = (x + 1)(x2 – x + 1). Factor y3 – 64. Since this polynomial can be written as y3 – 43, y3 – 64 = (y – 4)(y2 + 4y + 16).

Sum and Difference of Two Cubes Example: Factor 8t3 + s6. Since this polynomial can be written as (2t)3 + (s2)3, 8t3 + s6 = (2t + s2)((2t)2 – (2t)(s2) + (s2)2) = (2t + s2)(4t2 – 2s2t + s4). Factor x3y6 – 27z3. Since this polynomial can be written as (xy2)3 – (3z)3, x3y6 – 27z3 = (xy2 – 3z)((xy2)2 + (3z)(xy2) + (3z)2) = (xy2 – 3z)(x2y4 + 3xy2z + 9z2).

Choosing a Factoring Strategy Steps for Factoring a Polynomial Factor out any common factors. Look at number of terms in polynomial If 2 terms, look for difference of squares, difference of cubes or sum of cubes. If 3 terms, use techniques for factoring into 2 binomials. If 4 or more terms, try factoring by grouping. See if any factors can be further factored. Check by multiplying.

Solving Equations by Factoring and Problem Solving § 5.8 Solving Equations by Factoring and Problem Solving

Polynomial Equations Polynomial Equations Quadratic Equations Equations that set two polynomials equal to each other. Standard form has a 0 on one side of the equation. Quadratic Equations Polynomial equations of degree 2. Zero Factor Property If a and b are real numbers and ab = 0, then a = 0 or b = 0. This property is true for three or more factors, as well.

Solving Polynomial Equations Solving Polynomial Equations by Factoring Write the equation in standard form so that one side of the equation is 0. Factor the polynomial completely. Set each factor containing a variable equal to 0. Solve the resulting equations. Check each solution in the original equation.

Solving Polynomial Equations Example: Solve x2 – 5x = 24. First write the polynomial equation in standard form. x2 – 5x – 24 = 0 Now we factor the polynomial using techniques from the previous sections. x2 – 5x – 24 = (x – 8)(x + 3) = 0 We set each factor equal to 0. x – 8 = 0 or x + 3 = 0, which will simplify to x = 8 or x = -3 Continued

Solving Polynomial Equations Example continued: Check both possible answers in the original equation. 82 – 5(8) = 64 – 40 = 24 true (–3)2 – 5(–3) = 9 – (–15) = 24 true So our solutions for x are 8 or –3.

Solving Polynomial Equations Example: Solve 4x(8x + 9) = 5 First write the polynomial equation in standard form. 32x2 + 36x = 5 32x2 + 36x – 5 = 0 Now we factor the polynomial using techniques from the previous sections. 32x2 + 36x – 5 = (8x – 1)(4x + 5) = 0 We set each factor equal to 0. 8x – 1 = 0 or 4x + 5 = 0 8x = 1 or 4x = –5, which simplifies to x = or Continued

Strategy for Problem Solving General Strategy for Problem Solving Understand the problem Read and reread the problem Choose a variable to represent the unknown Construct a drawing, whenever possible Propose a solution and check Translate the problem into an equation Solve the equation Interpret the result Check proposed solution in problem State your conclusion

Finding an Unknown Number Example: The product of two consecutive positive integers is 132. Find the two integers. 1.) Understand Read and reread the problem. If we let x = one of the unknown positive integers, then x + 1 = the next consecutive positive integer. Continued

Finding an Unknown Number Example continued: 3.) Solve x(x + 1) = 132 x2 + x = 132 Distributive property x2 + x – 132 = 0 Write quadratic in standard form. (x + 12)(x – 11) = 0 Factor the quadratic polynomial. x + 12 = 0 or x – 11 = 0 Set factors equal to 0. x = -12 or x = 11 Solve each factor for x. Continued

Finding an Unknown Number Example continued: 4.) Interpret Check: Remember that x is suppose to represent a positive integer. So, although x = 12 satisfies our equation, it cannot be a solution for the problem we were presented. If we let x = 11, then x + 1 = 12. The product of the two numbers is 11 · 12 = 132, our desired result. State: The two positive integers are 11 and 12.

The Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse. (leg a)2 + (leg b)2 = (hypotenuse)2

The Pythagorean Theorem Example: Find the length of the shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotenuse is 10 miles less than twice the shorter leg. 1.) Understand Read and reread the problem. If we let x = the length of the shorter leg, then x + 10 = the length of the longer leg and 2x – 10 = the length of the hypotenuse. Continued

The Pythagorean Theorem Example continued: 2.) Translate By the Pythagorean Theorem, (leg a)2 + (leg b)2 = (hypotenuse)2 x2 + (x + 10)2 = (2x – 10)2 3.) Solve x2 + (x + 10)2 = (2x – 10)2 x2 + x2 + 20x + 100 = 4x2 – 40x + 100 Multiply the binomials. 2x2 + 20x + 100 = 4x2 – 40x + 100 Simplify the left side. 0 = 2x2 – 60x Subtract 2x2 + 20x + 100 from both sides. 0 = 2x(x – 30) Factor the right side. x = 0 or x = 30 Set each factor = 0 and solve, Continued

The Pythagorean Theorem Example continued: 4.) Interpret Check: Remember that x is suppose to represent the length of the shorter side. So, although x = 0 satisfies our equation, it cannot be a solution for the problem we were presented. If we let x = 30, then x + 10 = 40 and 2x – 10 = 50. Since 302 + 402 = 900 + 1600 = 2500 = 502, the Pythagorean Theorem checks out. State: The length of the shorter leg is 30 miles. (Remember that is all we were asked for in this problem.)

Math 083 Homework 5.7 #’s 1-11 odd, 15-27 odd, 43-57 odd 5.8 #’s 1,5,9,15,17,21,23,29,33,37,41,81,85,87

Math 083 Homework Wednesday is Chapter 5 Test. All HW from Chapter 5 is due on Wednesday. Make sure to staple or clip your papers togeter before you get here!