Download Solving Quadratic Functions by Factoring Find common factors

4.7
Solving Quadratic Functions by Factoring
Lesson Objectives:
Find common factors
between numbers
Put equations into factored form
Solve quadratic equations in factored form
Understand the meaning of the factors of a quadratic and
how they relate to the graph of a quadratic function
Consider This…
After the semester is over, you discover that the math department has changed textbooks (again)
so the old books are not needed. Your friend Herman and you decide to get creative. You go to the
roof of a twelve-story building and look over the edge to the reflecting pool 160 feet below. You
drop your book over the edge at the same instant that Herman chucks his book straight down at
48 feet per second (the initial velocity is negative).
The quadratic equation h(t)  h0  v0  16t 2 models height as a function of time for these situations
where h0 represents the initial height and v0 represents the initial velocity.
1. What is the equation that represents the path your book takes?
2. Rewrite this equation in factored form.
3. What is the equation Herman’s book takes?
4. Rewrite this equation in factored form.
5. How long does it take for your book to hit the water? How long does it take Herman’s book
to hit the water? (Find the x-intercepts)
6. By how many seconds does his book beat yours into the water?
1
Summarize
When we solve the above equations in factored form, we are identifying the
roots or x-intercepts of the function.
EXAMPLE: Solve (x
– 3)(x – 4) = 0.
If I multiply two things together and the result is zero, what can I say about those two things?
 At least one of them must also be zero. That is, the only way to multiply and get zero is to
multiply by zero.
This is sometimes called "The Zero Factor Property" or "Rule" or "Principle".
 You cannot make this statement about any other number! You can only make the conclusion
about the factors ("one of them must equal zero") if the product itself equals zero.
The Zero Factor Principle tells me that at least one of the factors must be equal to zero. Since at
least one of the factors must be zero, I'll set them each equal to zero:
x – 3 = 0 or x – 4 = 0
This gives me simple linear equations, and they're easy to solve:
x = 3 or x = 4  And this is the solution: x = 3, 4
Practice
1)
(k + 1)(k − 5) = 0
2)
(a + 1)(a + 2) = 0
3)
(4k + 5)(k + 1) = 0
4)
(2m + 3)(4m + 3) = 0
5)
(2k - 5)(-k - 3) = 0
6)
(0.5x - 15)(x + 2) = 0
2
Summarize
REMEMBER from 4.4,
WHEN the “a” value is equal to 1:
Standard Form
y  x 2  bx  c
Factored Form

y  (x  m)(x  n)
To get the FACTORED form:
1. Identify the a, b and c values
2. Find the factors or the numbers that MULTIPLY to “c” AND that ADD to “b”
_______ + _______ = b
_______  _______ = c
3. These are the factors of the equation
4. Plug these factors in for m and n in the factored form of the equation
Now to SOLVE A QUADRATIC EQUATION:
Solve x2
+ 5x + 6 = 0.
x2 + 5x + 6 = (x + 2)(x + 3)
1. Set the equation equal to 0.
(x + 2)(x + 3) = 0
2. Factor the quadratic equation (notes above)
x + 2 = 0 or x + 3 = 0
3. Like before, set each factor equal to 0.
x = –2 or x = – 3
4. Solve for x.
DON’T FORGET TO
CHECK YOUR
SOLUTIONS by plugging
the values back into the
equation for x.
The solution to x2 + 5x + 6=0 is x = –3, –2
3
Summarize
Solve:
Remember: You must have
the quadratic equal to zero
before you can solve.
(x + 2)(x + 3) = 12.
You cannot set each of the factors
above equal to the other side of the
equation and "solve".
Instead, you first have to multiply out and simplify the left-hand side, then subtract the 12
over to the left-hand side, and re-factor. Only then can you solve.
(x + 2)(x + 3) = 12
x + 5x – 6 = 0 (x + 6)(x – 1) = 0 x2 + 5x + 6 = 12
2
x + 6 = 0 or x – 1 = 0
x = –6 or x = 1
Then the solution to (x
+ 2)(x + 3) = 12 is x = –6, 1
4
Practice
Find the value(s) of x for the following equations. Check using FOIL.
1. x 2  8x  15  0
2. x(x  5)  0
3. x 2  4x  32  0
4. (2x  3)(x  2)  0
5. (4x  3)(2x  7)  0
6. (2x  9)(3x  5)  0
7. x(x  5)  2
8. (x  3)(x  8)  30
9. (x  1)(x  4)  14
10. (x  4)(x  5)  52
11. (x  4)(x  8)  35
12. (x  1)(x  6)  10
13. x 2  12x  35  0
14. x 2  22x  120  0
15. 2x 2  22x  24  0
5
Summarize
REMEMBER from 4.4,
What happens when “a” is not equal to 1?
FIRST: CHECK TO SEE IF YOU CAN FACTOR OUT THE “a” value! If not, then do the following:
The Not As Simple Case (a  1)
The Box Method can be used to factor quadratics, including the simple case, but it is very useful
when a  1. Be sure to have the quadratic in its simplest form.
• Factor
4x2 + 4x – 15
5) Determine which sum or difference of these
factors will give you the bx term:
1) Make a box and divide it into 4 squares:
bx = 4x → -6x+10x = 4x
6) Put these terms in the remaining boxes:
4x2
-6x
2) Put the ax2 term in the top left corner:
10x -15
4x2
7) Factor out common terms to the outside of the
box
3) Put the c term in the bottom right corner:
4x
2x
2
5
-15
4) Multiply both terms (ax2 and c) → generate a
list of factors.
4x2 • -15 = -60x2 → ± 2x, ± 30x
3x, 20x
4x, 15x
5x, 12x
6x, 10x
2x
-3
4x2
-6x
10x -15
8) Read factors from the sides of the box:
(2x – 3) (2x + 5), which is the factored
form of the quadratic.
6
BE SURE TO CHECK YOUR
ANSWER by using FOIL.
Practice
Solve the following equations. Check your answer using FOIL.
1. 2x 2  6x  4  0
2. 3x 2  15x  18  0
3. x 2  7x  18  0
4. 2x 2  3x  1  0
5. 5x 2  6x  1  0
6. 2x 2  3x  1  0
7. 11x 2  54x  5  0
8. 3x 2  16x  16  0
9. 3x 2  22x  16  0
10. 16b2− 40b + 25 = 0
11. 3b2− 6b + 3 = 0
12. 100x2− 180x + 81 = 0
7
Summarize
When we have quadratics in the form
Solve x
2
y  ax 2  bx :
– 5x = 0.
Do not divide both sides by x,
or you'll lose one of your
solutions!
This two-term quadratic is easier to factor than were
the previous quadratics.
FIRST, you can factor an x out of both terms, taking the x out
front.
x(x – 5) = 0
THEN, set each factor equal to 0 and solve.
x = 0 or x – 5 = 0
x = 0 or x = 5
The solution to x2 – 5x = 0 is x = 0, 5
______________________________________________________________________
When we have quadratics in the form
Solve x2
y  ax 2  c :
– 4 = 0.
x2 – 4 = 0
With quadratics in this form, you
can isolate the squared variable
term, putting the number over on
the other side, and solving for x.
x2 = 4
When solving an
equation, you can do
whatever you like to that
equation as long as you
do the same thing to
both sides of the
equation. On the lefthand side of this
particular equation, you
have an x2, and you need
a plain x. To turn an x2
into an x, you can take
the square root of each
side of the equation.
I know that:
x=±2
Then the solution is x = ± 2
When you square root both sides, remember, the solution gives both the
POSITIVE and NEGATIVE value since a negative number squared gives
you a positive.
8
Practice
Using any of the methods mentioned in this lesson, factor the following expressions.
2. 4x2 
1. 16n2 − 9n = 0
4. x2  


10. -4x2 
13. -16x x2 
 n2 − 49 = 0
8. 1- x2
7. -36x2 + 400 = 0



 9x2 − 1 = 0
6. a4  

-4x2 x 
14. 2xx2  x2
9


9. -36n2 − 142 = 2

-x2 x 
15. -12x x2 x

Apply
Quadratic functions appear in many different types of word problems. Try these problems by
drawing a diagram first, then solving the quadratic equation that corresponds.
1. The length of a rectangle is 6 inches more than its width. The area of the rectangle is 91
square inches. Find the dimensions of the rectangle.
2. The hypotenuse of a right triangle is 6 more than the shorter leg. The longer leg is three
more than the shorter leg. Find the length of the shorter leg.
3. One leg of a right triangle is one inch shorter than the other leg. If the hypotenuse is 5
inches, find the length of the shorter leg.
4. The length of a rectangle is 2 less than three times the width. Find the dimensions of the
rectangle if the area is 65 square meters.
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