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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]PowerPoint Presentation

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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5.4

Review §- Any QUESTIONS About
- §5.4 → Factoring TriNomials

- Any QUESTIONS About HomeWork
- §5.4 → HW-14

§5.5 Factoring Special Forms

- Factoring Perfect-Square Trinomials and Differences of Squares
- Recognizing Perfect-Square Trinomials
- Factoring Perfect-Square Trinomials
- Recognizing Differences of Squares
- Factoring Differences of Squares
- Factoring SUM of Two Cubes
- Facting DIFFERENCE of Two Cubes

Recognizing Perfect-Sq Trinoms

- A trinomial that is the square of a binomial is called a perfect-square trinomial
A2 + 2AB + B2 = (A + B)2

A2− 2AB + B2 = (A−B)2

- Reading the right sides first, we see that these equations can be used to factor perfect-square trinomials.
- A2 + 2AB + B2 = (A + B)(A + B)
- A2− 2AB + B2 = (A−B)(A−B)

Recognizing Perfect-Sq Trinoms

- Note that in order for the trinomial to be the square of a binomial, it must have the following:
1. Two terms, A2 and B2, must be squares, such as: 9, x2, 100y2, 25w2

2. Neither A2 or B2 is being SUBTRACTED.

3. The remaining term is either 2 A B or −2 A B

- where A &B are the square roots of A2 & B2

Example Trinom Sqs

- Determine whether each of the following is a perfect-square trinomial.
a) x2 + 8x + 16 b) t2− 9t− 36

c) 25x2 + 4 – 20x

- SOLUTION a) x2 + 8x + 16
- Two terms, x2 and 16, are squares.
- Neither x2 or 16 is being subtracted.
- The remaining term, 8x, is 2x4, where x and 4 are the square roots of x2 and 16

Example Trinom Sqs

- SOLUTION b) t2– 9t– 36
- Two terms, t2 and 36, are squares. But
- But 36 is being subtracted so t2– 9t– 36 is nota perfect-square trinomial.
- SOLUTION c) 25x2 + 4 – 20x
It helps to write it in descending order.

25x2– 20x + 4

Example Trinom Sqs

- SOLUTION c) 25x2− 20x + 4
- Two terms, 25x2 and 4, are squares.
- There is no minus sign before 25x2 or 4.
- Twice the product of the square roots is 2 5x 2, is 20x, the opposite of the remaining term, −20x
- Thus 25x2− 20x + 4 is a perfect-square trinomial.

Factoring a Perfect-Square Trinomial

- The Two Types of Perfect-Squares
A2 + 2AB + B2 = (A + B)2

A2− 2AB + B2 = (A−B)2

Example Factor Perf. Sqs

- Factor: a) x2 + 8x + 16
b) 25x2− 20x + 4

- SOLUTION a)
x2 + 8x + 16 = x2 + 2 x 4 + 42 = (x + 4)2

A2 + 2 A B + B2 = (A + B)2

Example Factor Perf. Sqs

- Factor: a) x2 + 8x + 16
b) 25x2− 20x + 4

- SOLUTION b)
25x2– 20x + 4 = (5x)2–2 5x 2 + 22 = (5x– 2)2

A2– 2 A B + B2 = (A – B)2

Example Factor 16a2– 24ab + 9b2

- SOLUTION
16a2− 24ab + 9b2= (4a)2− 2(4a)(3b) + (3b)2

= (4a− 3b)2 = (4a− 3b)(4a− 3b)

- CHECK:
(4a− 3b)(4a− 3b) = 16a2− 24ab + 9b2

- The factorization is (4a− 3b)2.

Expl Factor 12a3 –108a2 + 243a

- SOLUTION
- Always look for a common factor. This time there is one. Factor out 3a.
12a3− 108a2 + 243a = 3a(4a2− 36a + 81)

= 3a[(2a)2− 2(2a)(9) + 92]

= 3a(2a− 9)2

- The factorization is 3a(2a− 9)2

Recognizing Differences of Squares

- An expression, like 25x2− 36, that can be written in the form A2−B2 is called a difference of squares.
- Note that for a binomial to be a difference of squares, it must have the following.
- There must be two expressions, both squares, such as: 9, x2, 100y2, 36y8
- The terms in the binomial must have different signs.

Difference of 2-Squares

- Diff of 2 Sqs → A2−B2
- Note that in order for a term to be a square, its coefficient must be a perfect square and the power(s)of the variable(s) must be even.
- For Example 25x4− 36
- 25 = 52
- The Power on x is even at 4 → x4 = (x2)2
- Also, in this case 36 = 62

- For Example 25x4− 36

Example Test Diff of 2Sqs

- Determine whether each of the following is a difference of squares.
a) 16x2− 25 b) 36 −y5 c) −x12 + 49

- SOLUTION a) 16x2− 25
- The 1st expression is a sq: 16x2 = (4x)2
The 2nd expression is a sq: 25 = 52

- The terms have different signs.
- Thus, 16x2− 25 is a difference of squares, (4x)2− 52

Example Test Diff of 2Sqs

- SOLUTION b) 36 −y5
- The expression y5 is not a square.
- Thus, 36 −y5 is not a diff of squares
- SOLUTION c) −x12 + 49
- The expressions x12 and 49 are squares:x12 = (x6)2 and 49 = 72
- The terms have different signs.
- Thus, −x12 + 49 is a diff of sqs, 72− (x6)2

Factoring Diff of 2 Squares

- A2−B2 = (A + B)(A−B)

- The Gray Area by Square Subtraction

- The Gray Area by(LENGTH)(WIDTH)

Example Factor Diff of Sqs

- Factor: a) x2− 9 b) y2− 16w2
- SOLUTION a) x2− 9 = x2– 32 = (x + 3)(x− 3)
A2−B2 = (A + B)(A−B)

b) y2− 16w2 = y2− (4w)2 = (y + 4w)(y− 4w)

A2−B2 = (A + B) (A−B)

Example Factor Diff of Sqs

- Factor: c) 25 − 36a12 d) 98x2− 8x8
- SOLUTION
c) 25 − 36a12 = 52− (6a6)2 = (5 + 6a6)(5 − 6a6)

d) 98x2− 8x8

Alwayslook for a common factor. This time there is one, 2x2:

98x2− 8x8 = 2x2(49 − 4x6)

= 2x2[(72− (2x3)2]

= 2x2(7 + 2x3)(7 − 2x3)

Grouping to Expose Diff of Sqs

- Sometimes a Clever Grouping will reveal a Perfect-Sq TriNomial next to another Squared Term
- Example Factor m2−4b4 + 14m + 49
rearranging

m2 + 14m + 49 − 4b4

GROUPING

(m2 + 14m + 49) − 4b4

Grouping to Expose Diff of Sqs

- Example Factor m2− 4b4 + 14m + 49
- Recognize m2 + 14m + 49 as Perfect Square Trinomial → (m+7)2
- Also Recognize 4b4 as a Sq → (2b)2
(m2 + 14m + 49) − 4b4

Perfect Sqs

(m + 7)2− (2b2)2

- In Diff-of-Sqs Formula: A→m+7; B→2b2

Grouping to Expose Diff of Sqs

- Example Factor m2− 4b4 + 14m + 49
(m + 7)2− (2b2)2

Diff-of-Sqs → (A − B)(A + B)

([m+7] − 2b2)([m + 7] + 2b2)

Simplify → ReArrange

(−2b2 + m + 7)(2b2 + m + 7)

- The Check is Left for us to do Later

Factoring Two Cubes

- The principle of patterns applies to the sum and difference of two CUBES. Those patterns
- SUM of Cubes

- DIFFERENCE of Cubes

TwoCubes SIGN Significance

- Carefully note the Sum/Diff of Two-Cubes Sign Pattern

SAME Sign

OPP Sign

SAME Sign

OPP Sign

Example: Factor x3 + 64

- Factor

Recognize Pattern as Sum of CUBES

Determine Values that were CUBED

Map Values to Formula

Substitute into Formula

Simplify and CleanUp

Example: Factor 8w3−27z3

- Factor

Recognize Pattern as Difference of CUBES

Determine CUBED Values

Simplify by Properties of Exponents

Map Values to Formula

Sub into Formula

Simplify & CleanUp

Example: Check 8w3−27z3

- Check

Use Distributive property

Use Comm & Assoc. properties, and Adding-to-Zero

Factoring Completely

- Sometimes, a complete factorization requires two or more steps. Factoring is complete when no factor can be factored further.
- Example: Factor 5x4− 3125
- May have the Difference-of-2sqs TWICE

Factoring Completely

- SOLUTION
5x4− 3125 = 5(x4− 625)

= 5[(x2)2− 252]

= 5(x2− 25)(x2 + 25)

= 5(x− 5)(x + 5)(x2 + 25)

- The factorization: 5(x− 5)(x + 5)(x2 + 25)

Factoring Tips

- Always look first for a common factor. If there is one, factor it out.
- Be alert for perfect-square trinomials and for binomials that are differences of squares.
- Once recognized, they can be factored without trial and error.

- Always factor completely.
- Check by multiplying.

WhiteBoard Work

- Problems From §5.5 Exercise Set
- 14, 22, 48, 74, 94, 110

- The SUM (Σ) & DIFFERENCE (Δ) of Two Cubes

All Done for Today

Sum ofTwoCubes

Graph y = |x|

- Make T-table

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