Algebra Honors (periods 5 & 6)

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

{-1, 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

{0, 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 PERFECT 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