Math - Chapter 5

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MAT.105X A7

Ved David Kedrowski

MTE 280_Spring 2019

Ved Traci Nahsonhoya

MTE 280_Spring 2019

Ved Traci Nahsonhoya

Quadratic Formula

Ved Shane Harvey

Perfect square trinomials

Square root of a and the square root of c all multiplied by 2 is equal to b

4x² + 12x + 9

Square root of 4 -> 2 square root of 9 is -> 3

2(2)(3)

= 12 = b

∴ This is a perfect square trinomial

(2x+3)²

(3x+4)(8x+5)

24x² + 15x + 32x + 20

24x² + 47x + 20

Floating topic

used to simplify equations/expressions

done so by finding the greatest common factor in an expression/equation

Math - Chapter 5

Expanding

squaring binomials

(a+b)²

can be solved in 2 ways

(a+b)(a+b)

Subtopic

FOIL it

a²+ab+ab+b²

using Formula

a² - 2ab+b²

(a-b)²

a²+2ab+b²

Short form

distributive property

multiply a single term by 2 or more terms

used to expand equations/expressions

Multiply binomials

methods to do this

FOIL

F - First O - Outer I - Inner L - Last

its like applying the distributive property twice

multiply each term in the first factor by each in the second

2 factors of 2 terms being multiplied to each other

(x-4)(x-5)

difference of squares

part of factoring as well

(4x+5y)(4x-5y)

simplifies to

(4x)² - (5y)²

16x² - 25y²

(a+b)(a-b)

all term a = to the same thing all term b = to the same thing

a² - b²

Factoring

Factoring Quadratic expressions (a=1)

find 2 integers that when multiplied is equal to c

those 2 integers have to also equal to b when added together

ax² + 5x +6

__ x __ = 6 (c) __ + __ = 5 (b)

Those two integers are 2 and 3

2 x 3 = 6 (c) 2 + 3 = 5 (b)

r = 2 s = 3

(x+2)(x+3)

can be switched around

r ans s can equal to which ever integer

always see if you can common factor something out first

in this case no

where a, b, and c are integers

in this case a is equal to one

Factoring Quadratic expressions (a≠1)

ax² + bx + c

Where a, b, c are integers

Find 2 integers that when multiplied together is equal to a X c

Those same two integers have to add up to b

a(x+r)(x+s)

3x² + 8x +4

__ x __ = 12 ( a X c) __ + __ = 8 (b)

Those two integers are 2 and 6

2 x 6 = 12 (a X c) 2 + 6 = 8 (b)

Break down b into 2x and 6x and add that to the equation

3x² + 2x + 6x +4

Group factor

x(3x+2) + 2(3x+2)

(3x+2)(x+2)

refer to group factoring

Group common factors

4 terms separated into 2 groups

find common factor for both groups

produces a binomial common factor

2wx + 10w +7x + 35 l_________l l______l group one Group two

Group 2 -> 7 is the greatest common factor

7(x+5)

Group one -> 2 and w = greatest common factor

2w(x+5)

2w(x+5) + 7(x+5)

binomial common factor

refer to binomial common factor

(x+5)(2w+7)

Binomial common factors

A binomial factor that is shared by all the terms of a polynomial

common binomial among a polynomial

7(x-3) - 2x (x-3)

(x-3) is the greatest common factor

(x-3) [ 7(x-3) 2x (x-3)] ______ - ______ (x-3) (x-3)

(x-3)(7 - 2x)

When (x-3) is removed from the terms

(7-2x)

divide everything by (x-3)

put it outside the bracket

∴ equation not changed

Binomial = 2 terms

Monomial common factors

factor out a monomial out of a set of terms

6x² - 14x + 8

the biggest number that can divide all of these terms is 2

no biggest common variable

2 is the only thing common among all 3 terms

divide everything by 2

6x² - 14x + 8 ___________ 2

2(3x² - 7x + 4)

you cannot change the equation

that's why the 2 is put in the front of the brackets

if you ere to expand it you would end up with the same equation as before

nothing was changed

only simplified

Monomial = 1 term