Unit 1A: Quadratics Revisited

2018

pebblebrook high schoolALGBRA 2

1A.1 – 1A.8

1A.1 Multiplying Binomials

Example 1Multiply: (x +2)(x +4)

Example 2Multiply: (2x +3)(5x + 8)

You Try……

Think-Write-Discuss

Section 1A.1 Independent PRACTICE

1A.2 Factoring Quadratics: GCF

_______________________ is rewriting an expression the _________________ of its factors.

The ______________ is the common factor with the greatest __________________ and the greatest __________________.

Example 1: Identify the GCF1)4x2, 20x, 12 2) 9n2, 24n

Steps to factor a quadratic expression:

Identify the GCF Divide EVERY term by the GCF Rewrite using the Distributive

Property

Example 2: Factoring using GCF1) 9x2 + 3x – 18 2) 7p2 + 21

3) 4w2 + 2w

You Try….1) 3a2 – 9a 2) 25b2 – 35b + 5

CHALLENGE: Factor: 8x3y2 – 4xy

Section 1A.2 Independent PRACTICEFactor the GCF

1A.3 Factoring Quadratics: a = 1A ____________________ is an expression in the form ax2 + bx + c.

You can factor many quadratic trinomials into _____ __________ _____________.

Example: Factor

1) x2 + 8x + 7

2) x2 -17x + 72

3) x2 – x -12

4) x2 - 9

You Try....1) x2 + 12x + 32

2) x2 – 11x + 24

3) x2 + 3x – 10

4) x2 - 144

Section 1A.3 Independent PRACTICE

1A.4 Factoring Quadratics: a ≠ 1 & GCF

Factoring completely implies factoring out the GCF 1 st , then factor as a product of 2 binomials.

Steps for factoring completely:

Identify the GCF Divide EVERY term by the GCF Rewrite using the Distributive

Property Factor the remaining trinomial using

AC-Method

Examples: Factor completely

1) 2x2 + 6x + 4

2) 3x2 – 39x + 36

3) 2x2 – 10x – 28

4) 3x2 - 12

You Try…..

1) 3x2 + 12x – 15 2) 9x2 - 36

Sometimes you don’t have a GCF. Examples: Factor completely.

1) 2x2 + 7x – 9

2) 3x2 – 16x -12

3) 4x2 + 5x - 6

You Try…..

1) 3x2 + 7x – 20

2) 2x2 – 19x + 24

Section 1A.4 Independent PRACTICE

1.5 Solving Quadratics: Factoring

“Solving” a quadratic implies finding the __________ that make the equation equal to zero.

This is sometimes referred to _________ of the

function.

Finding zeros by factoring… Write equation in standard form; equal to zero.

Factor the trinomial. Set each binomial equal to zero.

Solve each binomial for x.

Examples: Solve by factoring

1) 2x2 – 11x = -15

2) 2x2 + 4x = 6

3) x2 + 6x + 8 = 0

4) 3x2 – 5x – 4 = 0

You Try …..

1)16x2 = 8x

2) x2 + 7x = 18

3) 2x2 – x – 3 = 0

Section 1A.5 Independent PRACTICE

1.6 Solving Quadratics: Square Roots

This method is best used when the linear term is missing.

ax2 + c = 0

Examples:

1) 5x2 – 180 = 0

2) 4x2 – 25 = 0

3) 3x2 = 24

4) x2 = 14

You Try….1) 5x2 = 80

2) x2 = 4

3) 3x2 = 15

Section 1A.6 Independent PRACTICE

1A.7 Solving Quadratics: Quadratic Formula(Real Solutions)

What happens when you can’t factor? Use the Quadratic Formula!

Examples: 1) 2x2 + 6x + 1 = 0

2) 3x2 – 5x = -2

You Try…1) x2 – x – 1 =0

2) x2 + 4x = 41

Section 1A.7 Independent PRACTICE

1A.8 Discriminant

NO SOLUTION MEANS COMPLEX SOLUTION.

Example 1: Tell the type & number of solutions

1) x2 + 6x + 8 = 0

2) x2 – 4x – 5 = 0

3) 2x2 + 7x – 15 = 0

You Try…1) x2 + 6x + 10 = 0

2) 6x2 – 2x + 5 = 0

Example 2: Finding Complex Solutions1) 4x2+ 100 = 0

2) 3x2+ 48 = 0

3) -5x2 – 150 = -200

4) 2x2 = -6x – 7

Section 1A.8 Independent PRACTICESolve Each and follow the directions.