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

Math 6 Honors ( Periods 1, 2, & 3)

Dividing Decimals 3-9 cont'd

For word Problems use the 5 step plan found on Page 18 of our textbook

296.06 ÷ (18.7 + 3.9)
Following Aunt Sally ( or PEMDAS... remember our singing...
we do the operation inside the hugs!! ( )using a sidebar
18.7 + 3.9 make sure to stack them lining up the decimals and you will get 22.6

296.06 ÷ 22.6

When dividing by a decimal remember the rule from yesterday, multiply the divisor ( 22.6) by a power of ten which makes it a natural number

then use that same power of ten and multiply the dividend,
WHen you divide you have
2960.6 ÷ 226
Please do that problem and your quotient should be 13.1

(47.1 - 16.9) ÷ (21.9 -6.8)
Again you need to do the operations inside the ( ) first. Using a side bar and lining up the decimals
47.1 - 16.9 = 30.2
and 21.9 - 6.8 = 15.1
Just take a look at those two numbers and you will notice a relationship!!
30.2 ÷ 15.1
BUT... practice your division skills and confirm what you can tell...
30.2 ÷ 15.1 = 2



At an average rate of 55 km/hour how long will it take to drive 225 km to the nearest tenth of an hour?

d = rt
What must we find and what are the clues? Well, how long... is usually time and the fact that we need to round to the nearest tenth of an hour indicates we are finding TIME as well.
So what is the distance? 225 km and what is the rate? 55 km/h
so plug into the formula
225= 55t
Now, how do we solve this one step problem?

divide both sides by 55
225/55 = 55t/55

do the division as a side bar

225/55 ≈ 4.09 so
t ≈ 4.1
and the answer is 4.1 hour

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)

Math 6 Honors ( Periods 1, 2, & 3)

Dividing Decimals 3-9

According to our textbook-
In using the division process to divide a decimal by a counting number, place the decimal point in the quotient directly over the decimal point in the dividend.

Check out our textbook for some examples!!

When a division does not terminate-- or does not come out evenly-- we usually round to a specified number of decimal places. This is done by adding zeros to the end of the dividend, which as you know, does NOT change the value of the decimal. We then divide ONE place beyond the specified number of places.

Divide 2.745 by 8 to the nearest thousandths.
See the set up in our textbook on page 89. Notice that they have added a zero and the end of the dividend ( 2.745 becomes 2.7450) because you want to round to the thousandths and we need to go ONE place additional.
DIVIDE carefully!!

the quotient is 0.3431 which rounds to 0.343


To divide one decimal by another

Multiply the dividend and the divisor by a power of ten that makes the DIVISOR a counting number


Divide the new dividend by the new divisor

Check by multiplying the quotient and the divisor.