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3 (4 5)x x 212 15x x
4 (6 2)x x 224 8x x
6 ( 1)x x 26 6x x
(2 1)(2 1)x x 24 1x
23 (2 4 5)x x x 3 26 12 15x x x
(6 5)(6 5)x x 236 25x
M3U4D1 Warm-UPDistribute each problem:
NEW SEATS!
M3U4D1 Evaluating and Operating with
Polynomials
OBJ: To review adding, subtracting, multiplying,
and factoring polynomials
How do I evaluate polynomial functions?
You have 3 minutes to complete the top of
handout page 1.
Discuss
How do I operate with polynomial functions?
Let’s review…
The sum f + g
xgxfxgf This just says that to find the sum of two functions, add them together. You should simplify by finding like terms.
1432 32 xxgxxf
1432 32 xxgf
424 23 xx
Combine like terms & put in descending order
The difference f - g
xgxfxgf To find the difference between two functions, subtract the first from the second. CAUTION: Make sure you distribute the – to each term of the second function. You should simplify by combining like terms.
1432 32 xxgxxf
1432 32 xxgf
1432 32 xx
Distribute negative
224 23 xx
The product f • g
xgxfxgf To find the product of two functions, put parenthesis around them and multiply each term from the first function to each term of the second function.
1432 32 xxgxxf
1432 32 xxgf31228 325 xxx
FOIL
Good idea to put in
descending order
32128 235 xxx
The quotient f /g
xgxf
xg
f
To find the quotient of two functions, put the first one over the second.
1432 32 xxgxxf
14
323
2
x
x
g
f Nothing more you could do here. (If you can reduce these you should. More later…)
Operations with
Polynomials
Now you try the four operations on the bottom of page 1 of your handout.
Factoring Review:#1: GCF Method
GCF Method is just
distributing backwards!!
Review: What is the GCF of
25a2 and 15a?5a
Let’s go one step further…1) FACTOR 25a2 + 15a.
Find the GCF and divide each term25a2 + 15a = 5a( ___ + ___ )
Check your answer by distributing.
225
5
a
a
15
5
a
a
5a 3
2) Factor 18x2 - 12x3.Find the GCF
6x2
Divide each term by the GCF18x2 - 12x3 = 6x2( ___ - ___ )
Check your answer by distributing.
2
2
18
6
x
x
3
2
12
6
x
x
3 2x
3) Factor 28a2b + 56abc2.
GCF = 28abDivide each term by the GCF
28a2b + 56abc2 = 28ab ( ___+ ___)
Check your answer by distributing.28ab(a + 2c2)
228
28
a b
ab
256
28
abc
ab
a 2c2
4) Factor 20x2 - 24xy
1. x(20 – 24y)2. 2x(10x – 12y)3. 4(5x2 – 6xy)4. 4x(5x – 6y)
5) Factor 28a2 + 21b - 35b2c2
GCF = 7Divide each term by the GCF
28a2 + 21b - 35b2c2 = 7 ( ___ + ___ - ____ )
Check your answer by distributing.7(4a2 + 3b – 5b2c2)
228
7
a 21
7
b
4a2 5b2c2
2 235
7
b c
3b
Factor 16xy2 - 24y2z + 40y2
1. 2y2(8x – 12z + 20)2. 4y2(4x – 6z + 10)3. 8y2(2x - 3z + 5)4. 8xy2z(2 – 3 + 5)
Factor out the GCF for each polynomial:Factor out means you need the GCF times the
remaining parts.
a) 2x + 4yb) 5a – 5bc) 18x – 6yd) 2m + 6mne) 5x2y – 10xy
2(x + 2y)
6(3x – y)
5(a – b)
5xy(x - 2)
2m(1 + 3n)
Greatest Common Factorsaka GCF’s
How can you check?
Ex 1
•15x2 – 5x•GCF = 5x•5x(3x - 1)
Ex 2
•8x2 – x•GCF = x•x(8x - 1)
Ex 3
•8x2y4+ 2x3y5 - 12x4y3
•GCF = 2x2y3
•2x2y3 (4y + xy2 – 6x2)
#2: X-box Factoringaka Diamond Method
#2: X-box Factoringaka Diamond Method
X- Box
3 -9
Product
Sum
X-box FactoringX-box Factoring• This is a guaranteed method for
factoring quadratic equations—no guessing necessary!
• We will review how to factor quadratic equations using the x-box method
• Background knowledge needed:
– Basic x-solve problems
– General form of a quadratic equation
Factor the x-box wayExample: Factor 3x2 -13x -10
-13
(3)(-10)=
-30
-15 2
-10
-15x
2x
3x2
x -5
3x
+2
3x2 -13x -10 = (x-5)(3x+2)
Factor the x-box way
Middleb=m+
nSum
Product
ac=mnm n
First and Last
Coefficients
y = ax2 + bx + c
Last term
1st Term
Factorn
Factorm
Base 1 Base 2
GCF
Height
ExamplesExamplesFactor using the x-box method.1. x2 + 4x – 12
a) b)
x -12
4 6 -2
x2 6x
-2x -12
x
-2
+6
Solution: x2 + 4x – 12 = (x + 6)(x - 2)
Examples continuedExamples continued
2. x2 - 9x + 20
a) b) 20
-9
x2 -4x
-5x 20
x
x -4
-5
Solution: x2 - 9x + 20 = (x - 4)(x - 5)
-4 -5
Examples continuedExamples continued
3. 2x2 - 5x - 7
a) b) -14
-5
2x2 -7x
2x -7
x
2x -7
+1
Solution: 2x2 - 5x – 7 = (2x - 7)(x + 1)
-7 2
Examples continuedExamples continued
3. 15x2 + 7x - 2
a) b) -30
7
15x2 10x
-3x -2
5x
3x +2
-1
Solution: 15x2 + 7x – 2 = (3x + 2)(5x - 1)
10 -3
#3: Difference of Squares
•a2 – b2 = (a + b)(a - b)
What is a Perfect Square
• Any term you can take the square root evenly (No decimal)
• 25• 36• 1• x2
• y4
5
6
1
x
2y
Difference of Perfect Squares
x2 – 4=
the answer will look like this: ( )( )
take the square root of each part:( x 2)(x 2)
Make 1 a plus and 1 a minus:(x + 2)(x - 2 )
FACTORINGDifference of Perfect
Squares
EX:
x2 – 64
How:
Take the square root of each part. One gets a + and one gets a -.
Check answer by FOIL.
Solution:
(x – 8)(x + 8)
Example 1
•9x2 – 16•(3x + 4)(3x – 4)
Example 2
•x2 – 16•(x + 4)(x –4)
Ex 3
•36x2 – 25•(6x + 5)(6x – 5)
More than ONE Method• It is very possible to use more than one
factoring method in a problem• Remember:
•ALWAYS use GCF first
Example 1
•2b2x – 50x•GCF = 2x•2x(b2 – 25) •2nd term is the diff of 2 squares•2x(b + 5)(b - 5)
ClassworkM3U4D1 Factoring Review
Evens
Distribute Interims!
Due back TOMORROW with parent signature
Distribute Interims!
Due back TOMORROW with parent signature
HomeworkM3U4D1 Factoring Review
Odds
Unit 3 Geometry Test due tomorrow!!!
Show all your work to receive credit– don’t forget to check by
multiplying!