Algebra Honors ( Periods 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 called an  “ if, then”  statement

 conversely

if ab = 0 then either a= 0 or b = 0

That is the converse.

Not all converse of true statements are true. For example

"If a polygon is a square, then it is a rectangle," is a true statement but its converse

"If a polygon is a rectangle, then it is a square"—is NOT true.

The words “ if and only if” are used to combine a statement and its converse when BOTH are true.

The Zero Products Property  is one such statement

This Zero Products Property helps us solve equations.

For all real numbers a and b,

ab= 0 if and only if a = 0 and b = 0

A product of factors is zero if and only if one or more of the factors is zero.

For example: 

(x +2)(x -5) = 0

either x + 2 must equal zero or x - 5 must equal zer0

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 zer0

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" --> we need to use the ZERO Products Property,  so set the expression equal to zer0

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 zer0

5x² = 0 so x = 0

and

2x -3 = 0 so x = 3/2

{0, 3/2}

Polynomial equations are named by the term of highest degree

(where a≠ 0)

ax + b = 0  is a linear equation

ax² + bx + c = 0 is a quadratic equations

ax³ + bx² + cx + d = 0  is a cubic equation

Many polynomial equations can be solved by factoring and then using the zero-products property. The first step is often to transform the equation into STANDARD FORMà in which one side is zer0. The other side should be a simplified polynomial arranged in order of decreasing degree of the variable.

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}

Find an equation in standard form with integral coefficients that has the following solutions set: {2/3, -4}

This is just working backwards…

Since we know that the solution set must be {2/3, -4}, we know that

( x – 2/3)(x + 4) = 0  because we get the solution set from those two sets of hugs by using the Zero Products Property

( x – 2/3)(x + 4) = 0   is NOT in standard form so we need to use the BOX METHOD   However… since we have a fraction in one of the two sets of parenthesis.. we may want to clear that BEFORE we start using FOIL, or the BOX method. If we multiply BOTH SIDES by 3, we haven’t changed the value so  3( x – 2/3)(x + 4) = 0   Now distribute the 3( x - 2/3) becomes (3x -2)  so now we have (3x – 2)(x + 4) = 0   Now FOIL, or BOX, or FIREWORKS…. and we get 3x²+ 10x – 8 = 0

(x-1)(x+3) = 12

We can’t immediately solve this—there isn’t a “12 products property.” That is, we need to set this to “zer0.”

(x-1)(x+3)  - 12 = 0

Now FOIL, or BOX, or FIREWORKS…. à  x² -2x – 3 -12 = 0

which is x² -2x – 15 = 0

FACTOR

(x -3)(x + 5) = 0

so

x = 3 and x = -5

{-5, 3}

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