1

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5.1 Multiplying Polynomials

Recall Adding Polynomials:

(2g+3b) + (1g+2b)=3g+5b

What happens with(2g+3b) * (1g+2b)?

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x

1

x

x x2 x2

x x

A=l*w=2x(x+1)=2x^2+2x

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x

1

x x2

x

­x

­x­1

­x

(2x-2)(x-1) =2x^2-2x-2x+22x^2-4x+2Success!!!

x

x2

11

­1 ­1

­x

­x ­x

x2

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x

1

x

x2

x2

xx

11

x x

11

­x

­1 ­1

­x2 ­x2

­1 ­1

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2

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2x^2-8xy-xy+4y^2F O I L2x2-9xy+4y2

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Test Rewrite options:-redo just one section-redo just one learning outcome-redo whole test?!

Before the retest:-finish review-do corrections-tomorrow lunch tutorial

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5.2 Common Factorsx

1

x

x x2 x2

x x

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3

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Cam Jillian

66 23716

2x3x11 2x2x7x7x11x11

2x2x3x7x7x11x11=71148

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Find the "All-Star" "Dream Team" please-everyone, Mountain High, Super Hero Pizza

1: Crust2: Cheese3: Pepperoni5: Pineapple7: Mushroomsx: "mystery" meaty: "mega mouthwatering mystery" meat

Jasmine: 63x2 Nick: 12xy Gregory: 18x

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p. 220 #(1-7, 12)o.l.

4

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§ 5.3 Factoring Trinomials

x2 x2

x x

x x

11

To factor this expression, set up algebra tiles that give a product of 2x2+4x+2

2x2+4x+2=(x+1)(2x+2)

x

x x 1 1

1

Nov 2­2:37 PM

x

1

x

xx2

xx

1 1

x x

11

Ex 2: set up tiles for x2+7x+10

xxxx

111111

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x

1

x

x x2

xx

1 1

x x

11

Ex 2: set up tiles for x2+7x+10

xxxx

111111

11

11111

x2+7x+10=(x+5)(x+2)

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x

1

x

x x2

x x

1 1

x

x

11

Ex 2: set up tiles for 2x2+3x+1

x

x

xx

111111

1

1 1

1111

x2+7x+10=(x+5)(x+2)

x2

2x2+3x+1=(x+1)(2x+1)

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x 1x

x x2

x x

1 1

x

x

11

Ex 2b: set up tiles for 3x2+5x-2

x

x

xx

111111

11 1

1111

x2+7x+10=(x+5)(x+2)

x2

­x

­1

­1

x2

x

x

x

x x ­1

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5

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Can we do this without algebra tiles?

Ex 1: x2+3x+2=(x+_)(x+_)

Our missing numbers should be integers with a sum of 3 and a product of 2

x2+3x+2=(x+1)(x+2)

Integers Sum Product

­1,­2 ­3 2

1,2 3 2

­1,4 3 ­4

1,­2 ­1 ­2

1,3 4 3

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Start p. 235: #1-4

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Ex 2: x2+3x­4=(x+_)(x­_)

Our missing numbers should be integers with a sum of 3 and a product of -4

x2+3x­4=(x+4)(x­1)

Integers Sum Product

­1,­2 ­3 2

1,2 3 2

­1,4 3 ­4

1,­2 ­1 ­2

1,3 4 3

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Ex 3: 2x2+6x­8

If all three terms have a common factor, we can deal with this first!

x2+3x­4=(x+4)(x­1)

Ex 3: 2x2+6x­8=2(x2+3x­4)

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The difficult ones:

Ex: factor 2x2+11x+12

try splitting up the middle term to get two binomials that factor nicely :)

2x2+8x+3x+122x(x+4)+3(x+4)(2x+3)(x+4)

6

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How do you know?

Ex: factor 2x2+11x+12-look for two integers with a sum of 11 and a product of 24!

2x2+8x+3x+122x(x+4)+3(x+4)(2x+3)(x+4)

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Method 3

We know (x+a)(x+b) "foils" to give us x2+(a+b)x+abSo why not try skipping straight to

(x+_)(x+_) when factoring?

Ex: factor x2+5x+6

try writing (x+ )(x+ )now fill in the blanks with an educated guess!

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7

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Finish p. 235 #1-4, continue #(5-9) o.l.

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"Perfect" Trinomials

ex 1) x2-4=x2+0x-4

perfect squares

A difference of squares always factors like:(x+2)(x-2)

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8

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p. 246 #(1-7) o.l., 14

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Review: p. 252 #1-2, 6-7, 10-11, 13-15

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