Algebra Honors (Period 6 & 7)

Solving Equations by Factoring Section 5-12

a⋅0 = 0
and if a = 0 or b = 0
then we know that ab= 0
This is an if, then statement
conversely
if ab = 0 then either a= 0 or b = 0
THis Zero Products Property helps us solve equations.

(x +2)(x -5) = 0
either x + 2 must equal zero or x - 5 must equal zero
so set each expression equal to zero and solve
x + 2 = 0
x= -2
and x-5 = 0
x = 5

{-2. 5}

5m(m-3)(m-4) = 0
now you have three expressions so set each of them to zero
5m = 0 so m = 0
m-3 = 0 so m=3
m-4 = 0 so m=4
{0,3,4}

What happens with
3x²+ x = 2
It isn't the 2 products property but the ZERO products property so set the expression equal to ZERO
3x²+ x -2 = 0
Now factor
(x+1)(3x -2) = 0
set each of these equal to zero
x + 1 = 0 x = -1
3x -2 = 0 so x = 2/3

10x³ - 15x² = 0
factor
5x²(2x -3) = 0
again set each equal to zero
5x² = 0 so x = 0
and
2x -3 = 0 so x = 3/2

polynomial equation named by the term of highes degree

ax + b = 0 linear equation

ax² + bx + c = 0 quadratic equations

ax³ + bx² + cx + d = 0 cubic equation

2x² + 5x = 12
becomes
2x² +5x - 12 = 0
(x + 4)(2x-3) = 0
so x = -4 and x = 3/2
{-4, 3/2}

18y³ + 8y + 24y² = 0
Rearrange first
18y³+ 24y ² + 8y = 0
Then factor the GCF
2y(9y² +12y +4) = 0
WAIT--> its a trinomial SQ
2y(3y +2)² = 0
2y = 0 so y = 0
and 3y + 2 = 0 so y = -2/3

-2/3 is a double or multiple root but you only list it once in solution set.
That is,
{-2/3, 0}

y = x² + x - 12
solve for the roots means you set this quadratic equal to ZERO
so
x² + x - 12 = 0
(x+4)(x -3) = 0

x = -4 and x = 3

Algebra Honors (Period 6 & 7)

Using Several Methods of Factoring Section 5-11

1) Always factor the GCF first
2) look for he difference of 2 SQ's
3) Look for a Perfect SQ trinomial
4) If trinomial is NOT SQ look for a pair of factors
5) If 4 or more terms-- look for a way to group the terms into pairs or into a group of 3 terms that is a perfect SQ trinomial
6) Make sure each binomial or trinomial factor is PRIME
7) check your work
-4n⁴ + 40 n³ -100n²
GCF
-4n²(n² -10n + 25)
-4n²(n-5)²

What about
5a³b² + 3a⁴b - 2a²b³
Again factor the GCF
a²b(5ab + 3a²-2b²)
reorder this and use either XBOX or factor pairs to solve

a²b(3a²+5ab-2²)


a²b(3a²+6ab-ab-2b²)


a²b[(3a²+6ab)+(-ab-2b²)]

a²b[3a(a+2b)-b(a + 2b)]
a²b(a+2b)(3a-b)


a²bc -4bc + a² -4b
b(a²c-4c+a²-4)
b(a²c+a²-4c-4)
b[(a²c+a²) -(4c+4)]
b[a²(c + 1) -4(c +1)]
b(c+1)(a²-4)
but we aren't finished...
b(c+1)(a+2)(a-2)



6c²+18cd+12d²
6(c²+3cd+2d²)
6(c+2d)(c+d)

3xy²-27x³

3x(y²-9x²)
3x(y+3x)(y-3x)


-n⁴-3n²-2n³

-n²(n²+3+2n)
-n²(n²+2n+3)

Its factored completely!!


16x²+16y -y²-64
16x²-y²+16y-64
16x²-y²+16y-64
16²-(y²-16y+64)
16x²-(y-8)²
becomes the difference of two squares
(4x+y-8)(4x-y+8)

x¹⁶ -1
(x⁸+1)(x⁸-1) =
(x⁸+1)(x⁴+1)(x⁴-1)=
(x⁸+1)(x⁴+1)(x²+1)(x² -1) =
(x⁸+1)(x⁴+1)(x²+1)(x +1)(x-1)


2(a +2)² + 5(a +2) - 3
Think of this as letting a+ 2 = x

2x² +5x -3
Factoring that is easy
(2x-1)(x +3)
so substitute in a + 2 for each x
[2(a+2) -1]{a+2 +3]


2a + 4 -1)(a +5)
(2a +3)(a +5)

Algebra Honors (Period 6 & 7)

Factoring Pattern for ax² + bx + c Section 5-9

When a > 1
We used a different method than what is taught in the book. I showed you X box

2x² + 7x -9
Multiply the 2 and the 9
put eighteen in the box
Your controllers are
2x² and -9
THen using a T chart find the factors of 19 such that the difference is 7x
we found that +9x and -2x worked

so
2x² +9x -2x -9
Then separate them in groups of 2
such that


(2x² +9x) + (-2x -9)

Then realize you can factor a - from the second pair

(2x² +9x) - (2x + 9)
Then wht is the GCF in each of the hugs( )
x(2x +9) -1(2x +9)
look they both have 2x + 9
:)
(2x +9)(x-1)
But what if you said -2x + 9x instead to make the +7x in the middle
Look what happens
(2x² -2x) + (9x -9)
now, factor te GCF of each
2x(x -1) + 9(x -1)
now they both have x -1
(x-1)(2x +9)
SAME RESULTS!!

14x² -17x +5
remember the second sign tells us that the numbers are the same and the first sign tells us that they are BOTH negative

create your X BOX with the product of 14 and 5 in it
70

Place your controllers on either side

14x² and + 5

Now do your T Chart for 70
You will need two numbers whose product is 70 and whose sum is 17
that's 7 and 10

14x² -7x -10x + 5

Now group in pairs

(14x² -7x) + (-10x + 5)
which becomes

(14x² -7x) - (10x - 5)

FACTOR each
7x(2x -1) - 5(2x-1)
(2x-1)(7x-5)

10 + 11x - 6x ²

sometimes its better to arrange by decreasing degree so this becomes

- 6x ² +11x + 10

now factor out the -1 from each terms


- (6x ² - 11x - 10)

Se up your X BOX with the product of your two controllers :)
60 We discover that +4x and -15x are the two factors

-1(6x ² +4x - 15x - 10)

-1[(6x ² +4x) + (- 15x - 10)]
-1[6x ² +4x) - (15x +10)
-1[2x(3x +2) -5(3x+2)]
-(3x+2)(2x-5)


If you had worked it out as
10 + 11x -6x² you would have ended up factoring
(5 -2x)(2 + 3x)
and we all know that
5 -2x = -(2x-5) Right ?


Next, we looked at the book and the example of
5a² -ab - 22b²
We discussed the books instructions to test the possibilities and decided that the X BOX method was much better.... I need to check out hotmath.com... did you????

5a² -ab - 22b² Using X BOX method we have 110 in the box and the controllers are
5a² and - 22b²
What two factors will multiply to 110 but have the difference -1?
Why 10 and 11

5a² +10ab -11ab - 22b²

separate and we get
(5a² +10ab) + (-11ab - 22b²)
( 5a² +10ab) - (11ab + 22b²)

5a(a + 2b) -11b(a + 2b)
(a + 2b)(5a - 11b)



Factoring by Grouping 5-10


5(a -3) - 2a (3 -a)

a-3 and 3-a are OPPOSITES
so we could write 3-a as -(-3 +a) or -(a -3)
sp we have
5(a-3) -2a [-(a-3)]
which is really
5(a-3) + 2a(a-3)
wait... look... OMG they both have a-3
so
(a-3)(5 + 2a)

What about
2ab-6ac + 3b -9c

What can you combine...
some saw the following:

(2ab -6ac) + 3b -9c)
then
2a(b-3c) + 3( b-3c)
(b -3c)(2a + 3)

BUT others look at 2ab-6ac + 3b -9c and saw
2ab +3b -6ac -9c
which lead them to
(2ab + 3b) + (-6ac -9c)
b(2a +3) -3c(2a +3)
(2a +3)(b-3c)
wait that's the same!!
Hooray

What about 4p² -4q² +4qr -r²
First look carefully and you will see

4p² -4q² +4qr -r²
That's a trinomial square OMG

so isn't that
4p² - ( 2q -r)²

BUT WAIT look at

4p² - ( 2q -r)² That's the
Difference of Two Squares
Which becomes
(2p + 2q -r)(2p -2q +r)