Post on 15-Jul-2015
METHODS
1.Trial and Error (Takes the most time)
2.Factoring by Grouping
3.Box Method (Grouping with a Box)
Lehmann, Elementary and Intermediate Algebra, 1edSection 8.3 Slide 10Copyright © 2011 Pearson Education, Inc.
Method 1: Factoring Trinomials by Trial and ErrorFactoring by Grouping
Factor
Example
23 14 8.x x+ +
Lehmann, Elementary and Intermediate Algebra, 1edSection 8.3 Slide 11Copyright © 2011 Pearson Education, Inc.
Method 1: Factoring Trinomials by Trial and ErrorFactoring by Grouping
Factor 3x2 + 14x + 8, the result will be of the form
(3x +?)(x +?)
Product of the last terms must be 8, so the last terms must be 1 and 8 or 2 and 4
Rule out negative last terms in the factors, because the middle term of 3x2+14x +8 has the positive coefficient 14
Solution
Lehmann, Elementary and Intermediate Algebra, 1edSection 8.3 Slide 12Copyright © 2011 Pearson Education, Inc.
Method 1: Factoring Trinomials by Trial and ErrorFactoring by Grouping
• Decide between the two pairs of possible last terms
by multiplying:
Solution Continued
Lehmann, Elementary and Intermediate Algebra, 1edSection 8.3 Slide 13Copyright © 2011 Pearson Education, Inc.
Method 1: Factoring Trinomials by Trial and ErrorMethod 1: Factoring Trinomials by Trial and ErrorFactoring by Trial and ErrorFactoring by Trial and Error
FactorFactor
ExampleExample22 5 25.x x− −
Lehmann, Elementary and Intermediate Algebra, 1edSection 8.3 Slide 14Copyright © 2011 Pearson Education, Inc.
Method 1: Factoring Trinomials by Trial and ErrorFactoring by Trial and Error
Factor 2x2 – 5 – 25, the result will be of the form
(2x +?)(x +?)
Product of the last terms must be –25, so the last terms must be 1 and –25, 5 and –5, or –1 and 25
Decide amongst the three pairs of possible last
terms by multiplying:
Solution
Lehmann, Elementary and Intermediate Algebra, 1edSection 8.3 Slide 15Copyright © 2011 Pearson Education, Inc.
Method 1: Factoring Trinomials by Trial and ErrorFactoring by Trial and Error
Solution Continued
Factoring Trinomials Review
X2 + 6x + 5 (x + )(x + )
Find factors of 5 that add to 6:
1*5 = 5 1+5 = 6 (x + 1)(x + 5)
Factoring Trinomials where a ≠ 1BY GROUPING! IT’S FASTER! Follow these steps:
1. Find two numbers that multiply to ac and add to b for ax2 + bx + c
2. Replace bx with the sum of the 2 factors found in step 1.
ie: ax2 + bx + c becomes ax2 + mx + nx + c, where m and n are the factors found in step 1.
3. Use grouping to factor this expression into 2 binomials
2x2 + 5x + 2
Step 1: ac = 2*2 = 4 1*4 = 4 1+4 = 5 2*2 = 4 2+2 = 4 m = 1 and n = 4
Step 2: Rewrite our trinomial by expanding bx 2x2 + 1x + 4x + 2
Step 3: Group and Factor (2x2 + 1x) + (4x + 2) x(2x + 1) + 2( 2x + 1) (2x + 1) (x + 2)
2x2 + 5x + 2
Questions for thought:
1. Does it matter which order the new factors are entered into the polynomial?
2. Do the parenthesis still need to be the same?
3. Will signs continue to matter when finding m and n?
4. Does it matter how we group the terms for factoring?
3z2 + z – 2
Step 1: ac = 3*-2 = -6 -1*6 = -6 -1+6 = 5 1* -6 = -6 1+-6 = -5 -2*3 = -6 -2+3 = 1 2* -3 = -6 2+-3 = -1
m = -2 and n = 3Step 2: Rewrite our trinomial by expanding bx
3z2 + 3z – 2z – 2 Step 3: Group and Factor
(3z2 + 3z) + (-2z - 2) 3z(z + 1) - 2( z + 1) (z + 1) (3z - 2)
3z2 + z – 2
Step 1: ac = 3*2 = 6 -1*6 = -6 -1+7 = 6 1* -6 = -6 1+-7 = -6 -2*3 = -6 -2+3 = 1 2* -3 = -6 2+-3 = -1
m = -2 and n = 3Step 2: Rewrite our trinomial by expanding bx
3z2 + 3z – 2z – 2
Notice that I changed the order of m and n between step 1 and step 2. Why do you think I did this? Do you have to change the order to get the correct answer?
3z2 + z – 2
What are the 3 steps for factoring this quadratic equation? Step 1: Multiply a*c. Find the factors that multiply
to ac and add to b Step 2: Expand bx to equal mx + nx Step 3: Group and Factor
4x3 – 22x2 + 30x
Step 0: Factor out the GCF: 2x
2x(2x2 – 11x + 15) Step 1: a*c = 30
-1*-30 = 30 -1+-30 = -31
-2*-15 = 30 -2+-15 = -17
-3*-10 = 30 -3+-10 = -13
-5*-6 = 30 -5+-6 = -11
4x3 – 22x2 + 30x
Step 0: Factor out the GCF: 2x
2x(2x2 – 11x + 15) Step 1: a*c = 30
-1*-30 = 30 -1+-30 = -31
-2*-15 = 30 -2+-15 = -17
-3*-10 = 30 -3+-10 = -13
-5*-6 = 30 -5+-6 = -11
4x3 – 22x2 + 30x
Step 2: Expand bx to equal mx + nx-11x = -5x + -6x
2x(2x2 – 5x – 6x + 15) Step 3: Group and Factor
2x((2x2 – 5x )(– 6x + 15))
2x(x(2x – 5) -3(2x – 5))
2x(2x – 5) (x – 3)
4x3 – 22x2 + 30x
Step 2: Expand bx to equal mx + nx-11x = -5x + -6x
2x(2x2 – 5x – 6x + 15) Step 3: Group and Factor
2x((2x2 – 5x )(– 6x + 15))
2x(x(2x – 5) -3(2x – 5))
Note: The Parenthesis are the Same
2x(2x – 5) (x – 3)
Practice
1. 3x2 + 5x + 2
(3x + 2)(x + 1)
2. 6x2 + 7x – 3
(3x – 1)(2x + 3)
3. 6 + 4y2 – 11y
(4y – 3)(y – 2)
Review
What is Step 0? When do you need to include this step?
When will your factors both be negative? When will you have one negative and one
positive factor? How do you check your answers?
When the leading coefficient is negative, factor out –1 from each term before using other factoring methods.
When you factor out –1 in an early step, you must carry it through the rest of the steps and into the answer.
Caution!
Additional Example 4: Factoring ax2 + bx + c When a is Negative
Factor –2x2 – 5x – 3.
–1(2x2 + 5x + 3) –1( x + )( x+ )
Factor out –1. a = 2 and c = 3;
Outer + Inner = 5Factors of 2 Factors of 3 Outer + Inner
1 and 2 3 and 1 1(1) + 3(2) = 7 1 and 2 1 and 3 1(3) + 1(2) = 5
–1(x + 1)(2x + 3) (x + 1)(2x + 3)
Additional Example 4: Factoring ax2 + bx + c When a is Negative
Factor –2x2 – 5x – 3.
–1(2x2 + 5x + 3)
–1( x + )( x+ )
Factor out –1.
a = 2 and c = 3; Outer + Inner = 5
Factors of 2 Factors of 3 Outer + Inner
1 and 2 3 and 1 1(1) + 3(2) = 7 1 and 2 1 and 3 1(3) + 1(2) = 5
–1(x + 1)(2x + 3)
(x + 1)(2x + 3)
Difference of Squares
Think back to Chapter 5. What happened when we multiplied a sum and difference?
(a – b)(a + b) = a2 – b2
So, the reverse is also true.
a2 – b2 = (a – b)(a + b)
x2 – 25
Notice that we do not have a bx term. This means that we only have the F and L in foil; therefore, none of the procedures from 6.1, 6.2, or 6.3 will work.
We need to use a2 – b2 = (a – b)(a + b)
where a = x and b = 5 X2 – 25 = (x – 5)(x + 5)
Practice 4x2 – 9
a = 2x, b = 3
(2x – 3) (2x + 3) 100 – 16t2
a = 10, b = 4t
(10 – 4t) (10 + 4t) 49y2 – 64z2
a = 7y, b = 8z
(7y – 8z) (7y + 8z)
Perfect Square Trinomials
Think back to Chapter 5. What happened when we squared a binomial?
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
So, the reverse is also true.
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
x2 + 10x + 25
This can be worked 2 different ways The first way is the simplest, but depends on
whether you recognize the equation as a perfect square trinomial.
a2 + 2ab + b2 = (a + b)2
Where a = x and b = 5
x2 + 10x + 25 = (x + 5)2
x2 + 10x + 25
This can be worked 2 different ways The second way is to use the method we learned in
6.2
x2 + 10x + 25
5*5 = 25 and 5+5 = 10
(x + 5) (x + 5) or (x + 5)2
4x2 - 4x + 1
This can be worked 2 different ways The first way is the simplest, but depends on
whether you recognize the equation as a perfect square trinomial.
a2 + 2ab + b2 = (a + b)2
Where a = 2x and b = 1
4x2 - 4x + 1 = (2x – 1)2
4x2 - 4x + 1
This time we need to use the 6.3 method
4*1 = 4
-2 * -2 = 4 and -2 + -2 = -4
(4x2 – 2x) ( – 2x + 1)
2x(2x – 1) – 1(2x – 1)
(2x – 1) (2x – 1) or (2x – 1)2
Practice x2 – 4xy + 4y2
a = x, b = 2y
(x – 2y)2
9a2 – 60a + 100
a = 3a, b = 10
(3a – 10) 25y2 + 20yz + 4z2
a = 5y, b = 2z
(5y + 2z)
Review
What methods can you use to factor a Difference of Squares?
What methods can you use to factor a Perfect Square Trinomial?
What clues should you look for to identify a Difference of Squares?
What clues should you look for to identify a Perfect Square Trinomial?