ACT Math : ACT Math
Study concepts, example questions & explanations for ACT Math
All ACT Math Resources
Example Questions
Example Question #2 : Distributive Property
What is the greatest common factor in the evaluated expression below?
This is essentially a multi-part question that at first may seem confusing, until it's realized that the question only involves basic algebra, or more specifically, using FOIL and greatest common factor concepts.
First, we must use FOIL (first, outside, inside, last), to evaluate the given expression: ⋅
First:
Outside:
Inside:
Last:
Now add all of the terms together:
Which simplifies to:
Now, we must see what is greatest common factor shared between each of these two terms. They are both divisible by 
Therefore, 
Example Question #2 : How To Use Foil With The Distributive Property
Given that i2 = –1, what is the value of (6 + 3i)(6 – 3i)?
36 – 9i
45
25
36 + 9i
36 + 9(i2)
45
We start by foiling out the original equation, giving us 36 – 9(i2). Next, substitute 36 – 9(–1). This equals 36 + 9 = 45.
Example Question #3 : Distributive Property
Expand the following expression:
(B – 2) (B + 4)
B2 + 2B + 8
B2 + 4B – 8
B2 + 2B – 8
B2 – 4B – 8
B2 – 2B – 8
B2 + 2B – 8
Here we use FOIL:
Firsts: B * B = B2
Outer: B * 4 = 4B
Inner: –2 * B = –2B
Lasts: –2 * 4 = –8
All together this yields
B2 + 2B – 8
Example Question #4 : Distributive Property
A rectangle's length L is 3 inches shorter than its width, W. What is an appropriate expression for the area of the rectangle in terms of W?
2W2 – 3W
2W2 – 3
W2 – 3W
2W – 3
W2 – 3
W2 – 3W
The length is equal to W – 3
The area of a rectangle is length x width.
So W * (W – 3) = W2 – 3W
Example Question #5 : Foil
Expand the following expression:
(f + 4) (f – 4)
f2 – 4f – 16
f2 + 4f – 16
f2 + 16
2f – 4f – 16
f2 – 16
f2 – 16
using FOIL:
First: f x f = f2
Outer: f x – 4 = –4f
Intter: 4 x f = 4f
Lasts: 4 x – 4 = –16
Adding it all up:
f2 – 4f +4f – 16
Example Question #6 : Distributive Property
Expand the following expression: (x+3) (x+2)
x2 + 3x + 6
x2 + 5x + 6
x2 + 5x + 3
2x + 5x + 6
x2 + 4x + 6
x2 + 5x + 6
This simply requires us to recall our rules from FOIL
First: X multiplied by X yields x2
Outer: X multplied by 2 yields 2x
Inner: 3 multiplied by x yields 3x
Lasts: 2 multiplied by 3 yields 6
Add it all together and we have x2 + 2x + 3x + 6
Example Question #11 : Foil
FOIL:
By FOIL (First Outer Inner Last), we obtain –8x2 – 8x – 6x – 6.
Simplify: –8x2 – 14x – 6
Example Question #12 : Foil
What is the product of the two real solutions of x2 + 5x = 6?
–1
–6
1
6
–6
x2 + 5x – 6 = 0 factors to (x+6)(x–1) = 0.
Therefore, the two real solutions are x = –6 and x = 1. Their product is simply –6.
Example Question #13 : Foil
What is the value of (5 + 3i)(6 – 2i)?
24 + 8i
36 + 8i
12 + 8i
16 – 2i
20 – 8i
36 + 8i
We FOIL the equation to obtain 30 – 10i + 18i – 6i2, combining like terms and substituting i2 = –1, we obtain 36 + 8i.
Example Question #12 : Foil
Expand the following expression:
FOIL 
Then multiply it by 
All ACT Math Resources
