math 31 lessons precalculus 1. simplifying and factoring polynomials
TRANSCRIPT
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MATH 31 LESSONS
PreCalculus
1. Simplifying and Factoring Polynomials
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A. Simplifying Polynomials
When you simplify a polynomial,
you are removing the brackets.
e.g.
(2x - 3) (4x + 1) = 8x2 - 10x - 3
Also, you are reducing a polynomial to the smallest
number of terms.
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1. Adding and Subtracting Polynomials
You can add or subtract monomials
only with like terms.
e.g.
5x + 7x = 12x
11y2 - 7y2 = 4y2
6ab3 + 11ab3 = 17ab3
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If they are not like terms,
then you cannot add them.
e.g.
2x + 3y
5y2 - 8y3
12xy2 + 8x2y
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Ex. 1 Simplify 2x - 11y + 7x + 3y + 5x
Try this example on your own first.Then, check out the solution.
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2x - 11y + 7x + 3y + 5xIdentify the like terms
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2x - 11y + 7x + 3y + 5x
= 2x + 7x + 5x - 11y + 7yCollect the like terms
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2x - 11y + 7x + 3y + 5x
= 2x + 7x + 5x - 11y + 3y
= 14x - 8y
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2. Multiplying Polynomials
Monomial Monomial
Consider
5a2b3 10ab4 =
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5a2b3 10ab4 = (5 10) (a2 a) (b3 b4)
Multiply numbers and like variables separately
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5a2b3 10ab4 = (5 10) (a2 a) (b3 b4)
= 50 a3 b7
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Monomial Polynomial
Consider
5x (6x - 7) =
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5x (6x - 7) = 5x (6x) - 5x (7)
Multiply the monomial to each term of the polynomial
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5x (6x - 7) = 5x (6x) - 5x (7)
= 30x2 - 35x
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Binomial Binomial
Consider
(2x - 3) (4x + 1) =
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(2x - 3) (4x + 1) = 2x (4x)
Use FOIL: First
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(2x - 3) (4x + 1) = 2x (4x) + 2x (1)
Use FOIL: First
Outside
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(2x - 3) (4x + 1) = 2x (4x) + 2x (1) - 3 (4x)
Use FOIL: First
OutsideInside
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(2x - 3) (4x + 1) = 2x (4x) + 2x (1) - 3 (4x) - 3 (1)
Use FOIL: First
OutsideInsideLast
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(2x - 3) (4x + 1) = 2x (4x) + 2x (1) - 3 (4x) - 3 (1)
= 8x2 + 2x - 12x - 3
= 8x2 - 10x - 3
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Polynomial Polynomial
Consider
(x + 2y) (5x - 3y + 6) =
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(x + 2y) (5x - 3y + 6) = x (5x) - x (3y) + x (6)
Multiply the first term to the entire polynomial
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(x + 2y) (5x - 3y + 6) = x (5x) - x (3y) + x (6)
+ 2y (5x) - 2y (3y) + 2y (6)
Then, multiply the second term to the entire polynomial
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(x + 2y) (5x - 3y + 6) = x (5x) - x (3y) + x (6)
+ 2y (5x) - 2y (3y) + 2y (6)
= 5x2 - 3xy + 6x + 10xy - 6y2 + 12y
= 5x2 + 6x + 7xy - 6y2 + 12y
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Ex. 2 Simplify 2 (3a + 4) (5a - 6) - (2a - 7)2
Try this example on your own first.Then, check out the solution.
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2 (3a + 4) (5a - 6) - (2a - 7)2
= 2 (3a + 4) (5a - 6) - (2a - 7) (2a - 7)
If it is a perfect square, then you should write both binomials. Then, you will remember to FOIL.
Notice:
(2a - 7)2 (2a)2 - (7)2
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2 (3a + 4) (5a - 6) - (2a - 7)2
= 2 (3a + 4) (5a - 6) - (2a - 7) (2a - 7)
= 2 (15a2 - 18a + 20a - 24) - (4a2 - 14a - 14a + 49)
Be certain to show the brackets around the entire product
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2 (3a + 4) (5a - 6) - (2a - 7)2
= 2 (3a + 4) (5a - 6) - (2a - 7) (2a - 7)
= 2 (15a2 - 18a + 20a - 24) - (4a2 - 14a - 14a + 49)
= 2 (15a2 + 2a - 24) - (4a2 - 28a + 49)
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2 (3a + 4) (5a - 6) - (2a - 7)2
= 2 (3a + 4) (5a - 6) - (2a - 7) (2a - 7)
= 2 (15a2 - 18a + 20a - 24) - (4a2 - 14a - 14a + 49)
= 2 (15a2 + 2a - 24) - (4a2 - 28a + 49)
= 30a2 + 4a - 48 - 4a2 + 28a - 49
Distribute the negative to all terms
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2 (3a + 4) (5a - 6) - (2a - 7)2
= 2 (3a + 4) (5a - 6) - (2a - 7) (2a - 7)
= 2 (15a2 - 18a + 20a - 24) - (4a2 - 14a - 14a + 49)
= 2 (15a2 + 2a - 24) - (4a2 - 28a + 49)
= 30a2 + 4a - 48 - 4a2 + 28a - 49
= 26a2 + 32a - 97 Add like terms
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B. Factoring Polynomials
When you factor a polynomial,
you are adding brackets.
e.g.
8x2 - 10x - 3 = (2x - 3) (4x + 1)
You are making a polynomial into a product.
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1. Greatest Common Factor (GCF)
The GCF is:
the largest number that divides evenly into
the coefficients
the smallest power of each variable
Taking out the GCF is usually the first step of factoring.
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e.g.
Factor 12 x3 y4 + 18 x8 y2
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12 x3 y4 + 18 x8 y2
= 6 x3 y2 (
The largest number that divides into 12 and 18 evenly
The smallest power of each variable
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12 x3 y4 + 18 x8 y2
= 6 x3 y2 ( 2 x3-3 y4-2 + 3x8-3 y2-2 )
When you factor (divide), you subtract the exponents
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12 x3 y4 + 18 x8 y2
= 6 x3 y2 ( 2 x3-3 y4-2 + 3x8-3 y2-2 )
= 6 x3 y2 ( 2 x0 y2 + 3x5 y0 )
= 6 x3 y2 ( 2 y2 + 3x5 )
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2. Difference of Squares
Formula:
A2 - B2 = (A + B) (A - B)
Note:
There is no formula for A2 + B2.
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e.g.
Factor 81 m8 - 16 y6 z4
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81 m8 - 16 y6 z4
= (9 m4)2 - (4 y3 z2)2
Put into the form A2 - B2.
48 981 mmA 2346 416 zyzyB
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81 m8 - 16 y6 z4
= (9 m4)2 - (4 y3 z2)2
= (9 m4 + 4 y3 z2) (9 m4 - 4 y3 z2)
A2 + B2 = (A + B) (A - B)
where A = 9 m4 and B = 4 y6 x2
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3. Sum / Difference of Cubes
Formulas:
A3 - B3 = (A - B) (A2 + 2AB + B2)
A3 + B3 = (A + B) (A2 - 2AB + B2)
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e.g. 1
Factor x3 - 64y3
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x3 - 64y3
= (x)3 - (4 y)3
Put into the form A3 - B3
xxA 3 3
yyB 4643 3
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x3 - 64y3
= (x)3 - (4 y)3
= (x - 4y) [ x2 + (x) (4y) + (4y)2 ]
A3 - B3 = (A - B) (A2 + AB + B2)
where A = x and B = 4y
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x3 - 64y3
= (x)3 - (4 y)3
= (x - 4y) [ x2 + (x) (4y) + (4y)2 ]
= (x - 4y) (x2 + 4xy + 16y2)
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e.g. 2
Factor 8x3 + 27y6
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8x3 + 27y6
= (2x)3 + (3 y2)3
Put into the form A3 + B3
xxA 283 3
23 6 327 yyB
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8x3 + 27y6
= (2x)3 + (3 y2)3
= (2x + 3y2) [ (2x)2 (2x) (3y2) + (3y2)2 ]
A3 + B3 = (A + B) (A2 - AB + B2)
where A = 2x and B = 3y2
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8x3 + 27y6
= (2x)3 + (3 y2)3
= (2x + 3y2) [ (2x)2 (2x) (3y2) + (3y2)2 ]
= (2x + 3y2) (4x2 6xy2 + 9y4)
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4. Grouping
When there are 4 terms, try grouping:
Group pairs of terms (you may need to rearrange)
Factor each pair
Factor out the common polynomial
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e.g.
Factor ac bd + bc ad
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ac bd + bc ad
No common factors for each pair.
Thus, we need to rearrange.
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ac bd + bc ad
= ac ad + bc bd
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ac bd + bc ad
= ac ad + bc bd
= a (c d) + b (c d)
They must have a common factor.
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ac bd + bc ad
= ac ad + bc bd
= a (c d) + b (c d)
= (a + b) (c d)
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5. Factoring Trinomials
Trinomials are polynomials with 3 terms.
They have the form
Ax2 + Bx + C = 0
We will deal with two cases:
Case 1: A = 1 (By inspection)
Case 2: A ≠ 1 (Decomposition)
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Case 1: A = 1 (By inspection)
To factor x2 + Bx + C,
Find 2 numbers that add to B and multiply to C
Simply substitute the numbers into the two
binomial factors
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e.g.
Factor x2 + 2x - 15
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x2 + 2x - 15
Find two
numbers that ... add to 2
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x2 + 2x - 15
Find two
numbers that ... add to 2 and multiply to -15
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x2 + 2x - 15
2 numbers:
Sum = 2
Product = -155, -3
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x2 + 2x - 15
2 numbers:
Sum = 2
Product = -15
= (x + 5) (x - 3)
Simply sub the numbers in
5, -3
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Case 2: A ≠ 1 (Decomposition)
To factor Ax2 + Bx + C,
Find 2 numbers that add to B and multiply to AC
Replace B with these two numbers
Factor by grouping
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e.g.
Factor 3x2 - 17x + 10
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3x2 - 17x + 10
Find 2 numbers:
Sum = -17
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3x2 - 17x + 10
Find 2 numbers:
Sum = -17
Product = 30
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3x2 - 17x + 10
Find 2 numbers:
Sum = -17
Product = 30-15, -2
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3x2 - 17x + 10
= 3x2 - 15x - 2x + 10Replace B with the two numbers, -2 and -15
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3x2 - 17x + 10
= 3x2 - 15x - 2x + 10
= 3x (x - 5) - 2 (x - 5) Factor by grouping
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3x2 - 17x + 10
= 3x2 - 15x - 2x + 10
= 3x (x - 5) - 2 (x - 5)
= (x - 5) (3x - 2)
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Summary (Factoring methods)
GCF first
Look at the # of terms:
2 terms : - Difference of squares
- Sum / difference of cubes
3 terms: - Inspection (if A = 1)
- Decomposition (if A ≠ 1)
4 terms: - Grouping
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Ex. 3
Factor 80 xy3 + 10xz6 completely.
Try this example on your own first.
Then, check out the solution.
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80 xy3 + 10xz6
= 10x (8y3 + z6) Factor GCF first.
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80 xy3 + 10xz6
= 10x (8y3 + z6)
Don’t stop here.
Do you see what else can be factored?
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80 xy3 + 10xz6
= 10x (8y3 + z6)
= 10x [ (2y)3 + (z2)3 ] Sum of cubes
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80 xy3 + 10xz6
= 10x (8y3 + z6)
= 10x [ (2y)3 + (z2)3 ]
= 10x (2y + z2) [ (2y)2 - (2y) (z2) + (x2)2 ]
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80 xy3 + 10xz6
= 10x (8y3 + z6)
= 10x [ (2y)3 + (z2)3 ]
= 10x (2y + z2) [ (2y)2 - (2y) (z2) + (x2)2 ]
= 10x (2y + z2) (4y2 - 2yz2 + x4)
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Ex. 4
Factor x2y - 54 + 6x2 - 9y completely.
Try this example on your own first.Then, check out the solution.
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x2y - 54 + 6x2 - 9y
We will factor by grouping (4 terms).
However, we must rearrange so that there will be
common factors.
Can you see how?
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x2y - 54 + 6x2 - 9y
= x2y - 9y + 6x2 - 54
This is one way to do so.
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x2y - 54 + 6x2 - 9y
= x2y - 9y + 6x2 - 54
= y (x2 - 9) + 6 (x2 - 9)
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x2y - 54 + 6x2 - 9y
= x2y - 9y + 6x2 - 54
= y (x2 - 9) + 6 (x2 - 9)
= (x2 - 9) (y + 6)
Don’t stop here.
Can you see what else can be factored?
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x2y - 54 + 6x2 - 9y
= x2y - 9y + 6x2 - 54
= y (x2 - 9) + 6 (x2 - 9)
= (x2 - 9) (y + 6)
= (x + 3) (x - 3) (y + 6) Difference of squares
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Ex. 5
Factor 3a4 - 7a2 - 20 completely.
Try this example on your own first.Then, check out the solution.
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Notice that 3a4 - 7a2 - 20 is a trinomial.
To make it easier to factor, let’s do a substitution.
i.e.
Let x = a2
Then,
3 (a2)2 - 7 (a2) - 20 = 3x2 - 7x - 20
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3x2 - 7x - 20
Find 2 numbers:
Sum = -7
Product = -60-12, 5
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3x2 - 7x - 20
Find 2 numbers:
Sum = -7
Product = -60
= 3x2 - 12x + 5x - 20
-12, 5
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3x2 - 7x - 20
Find 2 numbers:
Sum = -7
Product = -60
= 3x2 - 12x + 5x - 20
= 3x (x - 4) + 5 (x - 4)
-12, 5
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3x2 - 7x - 20
Find 2 numbers:
Sum = -7
Product = -60
= 3x2 - 12x + 5x - 20
= 3x (x - 4) + 5 (x - 4)
= (x - 4) (3x + 5)
-12, 5
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= (x - 4) (3x + 5)
Finally, we have to back-substitute x = a2:
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= (x - 4) (3x + 5)
Finally, we have to back-substitute x = a2:
= (a2 - 4) (3a2 + 5)
Don’t stop here.
Do you see what else can be factored?
![Page 92: MATH 31 LESSONS PreCalculus 1. Simplifying and Factoring Polynomials](https://reader036.vdocuments.us/reader036/viewer/2022062408/56649ed45503460f94be558b/html5/thumbnails/92.jpg)
= (x - 4) (3x + 5)
Finally, we have to back-substitute x = a2:
= (a2 - 4) (3a2 + 5)
= (a + 2) (a - 2) (3a2 + 5)