 # Any questions on the Section 5.7 homework?

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Any questions on the Section 5.7 homework?

and turn off and put away your cell phones, and get out your note-taking materials.

Reminder: This homework assignment on section 5.8 is due at the start of next class period. Make sure you turn in the worksheet showing all your work for the last 9 problems (#20 – 28) of this assignment. If you don’t turn this in, or if you don’t completely show your work on any of these 9 problems out of the 28 total problems in the online assignment, your online score will be reduced accordingly. 3

Solving Equations by Factoring Problem Solving
Section 5.8 Solving Equations by Factoring Problem Solving

What possible use is there for factoring polynomials????

Take 2: What possible use is there for factoring polynomials????

Take 3: What possible use is there for factoring polynomials????

Take 4: What possible use is there for factoring polynomials????

Solving problems like these by factoring:
Polynomial equations Equations that set 2 polynomials equal to each other. Standard form has a 0 on one side of the equation. The maximum number of solutions to a polynomial equation is equal to the degree of the polynomial. Quadratic equations Polynomial equations of degree 2. (So how many possible solutions?) Zero factor theorem 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.

Steps for solving a polynomial equation by factoring:
Write the equation in standard form. Clear any fractions. Factor the polynomial completely. Set each factor containing a variable equal to 0. Solve the resulting equations. Check each solution in the original equation.

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, which will simplify to x = 8 x + 3 = 0 which will simplify to x = -3

Example (cont.) Check both possible answers in the original equation, x2 – 5x = 24. x = 8: – 5(8) = 64 – 40 = true x = -3: (-3)2 – 5(-3) = 9 – (-15) = true So our solutions for x are 8 or –3. ALWAYS REMEMBER TO CHECK YOUR ANSWERS!!! (Especially on quizzes/tests, when there’s no “check answer” button...)

Example Solve 4x(8x + 9) = 5 First, simplify the left side using the distributive property: 32x2 + 36x = 5 Then write the polynomial equation in standard form: 32x2 + 36x – 5 = 0 Now we factor the polynomial using techniques from the previous sections: 32x2 + 36x – 5 = (8x – 1)(4x + 5) = 0 Finally, we set each factor equal to 0. 8x – 1 = x = x = 1/8 4x + 5 = x = x = -5/4 Note: This equation can also be solved (and probably more quickly) using the quadratic formula, which is the topic of tomorrow’s lecture.

Solve 4x(8x + 9) = 5 Example (cont.)
Now check both possible answers (x = 1/8 and x = -5/4) in the original equation. (This can be done fairly quickly if you use your calculator.) true true So our solutions for x are or

Example from today’s homework:

Example from today’s homework:

General strategy for solving applied (word) problems:
Understand the problem: Read and reread the problem. Choose a variable to represent the unknown. Construct a drawing, whenever possible. Translate the problem into an equation. Solve the equation. Check your answers in the original equation. Interpret the result: Determine if any or all of the proposed solutions make sense in terms of the applied problem. Convert your answer(s) into the appropriate form to answer the specific question(s) asked in the word problem.

Example The product of two consecutive positive integers is Find the two integers. Understand Read and reread the problem and choose a variable to represent the unknown quanitity. If we let x = one of the unknown positive integers, then x + 1 = the next consecutive positive integer.

Example (cont.) Now translate this into an equation: The product of •
= 132 two consecutive positive integers x (x + 1)

Example (cont.) Solve x(x + 1) = 132
x2 + x = (distributive property) x2 + x – 132 = (write quadratic in standard form) (x + 12)(x – 11) = (factor quadratic polynomial) x + 12 = 0 or x – 11 = (set factors equal to 0) x = -12 or x = (solve each factor for x)

Example (cont.) Interpret Possible solutions: x = -12 or x = 11
Remember that x is supposed 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 (the first integer), then the next consecutive integer is x + 1 = 12. Check: The product of the two numbers is 11 · 12 = 132, our desired result. State Solution: The two positive integers are 11 and 12.

Pythagorean Theorem (used in homework problem # 21 in today’s assignment) 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

Example (Similar to problem # 21 in today’s assignment.) 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.

x = the length of the shorter leg, then x + 10 = the length of the longer leg and 2x – 10 = the length of the hypotenuse. Now draw a diagram:

Example (cont.) Translate By the Pythagorean Theorem,
(leg a)2 + (leg b)2 = (hypotenuse)2 x2 + (x + 10)2 = (2x – 10)2

Example (cont.) Solve x2 + (x + 10)2 = (2x – 10)2
x2 + x2 + 20x = 4x2 – 40x + 100 (multiply the binomials) 2x2 + 20x = 4x2 – 40x + 100 (simplify left side) 0 = 2x2 – 60x (subtract 2x2 + 20x from both sides) 0 = 2x(x – 30) (factor right side) x = 0 or x = 30 (set each factor = 0 and solve)

Example (cont.) Interpret
Check: Remember that x is supposed 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 Solution: The length of the shorter leg is 30 miles. (Remember that is all we were asked for in this problem.)

To solve this problem, we set h = 0 (since 0 = ground level).
Then we solve the equation -16t t = 0 Factoring this equation gives two answers: t = 0, which means the rocket is at ground level when it is launched, and t = 9, which tells us that the rocket returns to the ground 9 seconds after it is launched. Note: There are several problems like this in today’s homework. New question: How long will it take for the rocket to reach its peak height of 324 feet? ANSWER: Solve 324 = -16t t Factoring gives: 4(2t – 9)(2t - 9)= 0 Answer: t = 4.5 sec.

Reminder: This homework assignment on section 5.8 is due at the start of next class period. Make sure you turn in the worksheet showing all your work for that last 9 problems (#20 – 28) of this assignment. If you don’t turn this in, or if you don’t completely show your work on any of these 9 problems out of the 28 total problems in the online assignment, your online score will be reduced accordingly. 30

Please Note: You DO NOT have to memorize formulas for the word problems. You will have the pink sheet of formulas to use during quizzes and tests on this material. Let us know if you need another copy of the yellow formula handout to keep in your notebook to use as you do your homework and study for the quizzes and tests.

and begin working on the homework assignment.
You may now OPEN your LAPTOPS and begin working on the homework assignment.