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

2. Factoring by Special Products
§ 5.7
Factoring by Special Products

3. 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

4. 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

5. 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.

6. 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).

7. 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.

8. 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)

9. 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).

10. 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).

11. 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.

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

13. 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.

14. 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.

15. 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

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

17. 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

18. 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

19. Finding an Unknown Number
Example:
The product of two consecutive positive integers is 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

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

21. 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.

22. 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

23. 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

24. 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 = 4x2 – 40x + 100
Multiply the binomials.
2x2 + 20x = 4x2 – 40x + 100
Simplify the left side.
0 = 2x2 – 60x
Subtract 2x2 + 20x from both sides.
0 = 2x(x – 30)
Factor the right side.
x = 0 or x = 30
Set each factor = 0 and solve,
Continued

25. 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 = = 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.)

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

27. 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!