World 1-8
Solving by Factoring
Recall: it requires one equation to solve for one unknown.
Traditionally we find solutions to equations by isolating x.
5x + 10 = 25
-10 -10
5x = 15
5
5
x=3
y2 + 25 = 61
-25 -25
y2 = √36
y=±6
x2 + x = 20
Cannot get x
alone, by old
methods.
In the past solving an equation with an x2 and an x
that didn’t cancel would not have been possible.
But now you CAN by factoring.
Consider the equation ab= 0
For this to be true. Either a=0, or b =0
ab= 0
ab=0
So a=0 and/or b=0 is
(0)(b)=0
the solution.
(a)(0)=0
works
works as well
STEPS to use this property
1. Rearrange the equation to let one side = 0
2. Factor the expression
3. If a factor has a variable let it = 0
4. Solve for the variable
Solve by factoring
1. 3x – 6 = 0
3(x – 2)= 0
3=0
nonsense
x–2=0
x=2 
3. 4x + 20 = 0
4(x +5)= 0
x+5=0
x = -5
2. 7x – 7 = 0
7(x – 1)= 0
x–1=0
x=1
4. 16 – 12x = 0
4(4 – 3x)= 0
4 – 3x = 0
-4
-4
- 3x = -4
x = 4/3
Solve by factoring
1. 3x2 – 9x = 0
3x(x – 3)= 0
x= 0
x–3=0
x=3
or
Two solutions
2. 4x – 3 = x + 3
-3
-3
4x – 6 = x
-x
-x
3x – 6 = 0
3(x – 2) = 0
x–2=0
x=2
Factor and Solve
1. x(x+1) – 5(x+1) = 0
(x + 1)(x- 5)= 0
x= -1
3.
2. x(x - 4) + 5(x - 4)
Not an equation!
or x = 5
x2 - 16= 0
(x -4)(x + 4)= 0
4. x2 – 5x + 14= 0
(x - 7 )(x + 2 )= 0
x = 7 or x = - 2
x= 4
or x = - 4
so x = ±4
Tougher One
x3 + 2x2 -3x = 0
x(x2 +2x -3)= 0
x(x - 1)(x + 3)= 0
x= 0, x=1, and x= -3
are solutions
Homework
World 1-8