ACT Math : Algebra
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Example Questions
Example Question #101 : Variables
Factor the following polynomial:
To factor a polynomial of the form 
Since 
When you factor 
Since:
Simply plug them into the parentheses and you have:
Example Question #1 : Factoring
Two consecutive positive multiples of five have a product of 300. What is their sum?
35
20
45
15
25
35
Define the variables as x = 1st number and x + 5 = 2nd number, so the product is given as x(x + 5) = 300, which becomes x2 + 5x – 300 = 0.
Factoring results in (x + 20)(x – 15) = 0, so the positive answer is 15, making the second number 20.
The sum of the two numbers is 35.
Example Question #1 : Factoring
Factor 12x3y4 + 156x2y3
12x2y3
12xy(xy + 13)
x2y3(xy + 13)
12x2y3(xy + 13)
12x2y3(xy + 13)
The common factors are 12, x2, and y3.
So 12x2y3(xy + 13)
Example Question #1 : Factoring
Solve for all solutions of 
First move all of the variables to the left side of the equation. Combine similar terms, and set the equation equal to zero. Then factor the equation to get
Thus the solutions of 
Example Question #2 : Factoring
Simplify:
One 
Example Question #4 : Factoring
Factor: 
In the form of 
Therefore 
To check, multiply the two expressions out and it should equal
Example Question #5 : Factoring
Factor the following expression:
The expression is already simplified as much as possible.
To factor an expression we look for the greatest common factor.
Remember that
Thus:
Example Question #4 : Factoring
Factor the following expression:
To factor, you are looking for two factors of 40 that add to equal 13.
Factors of 40 include: (1, 40), (2, 20), (4, 10), (5, 8). Of these factors which two will add up to 13?
Also, since the first sign (-) and the second sign is (+) this tells us both binomials will be negative. This is because two negatives multiplied together will result in the positive third term, while two negatives added together will result in a larger negative number.
Thus,
Example Question #1 : Exponents
Solve: 5x2 – 3y1 where x = 4, y = 5.
17
95
60
65
65
Substitute the values for x and y within the equation: 5(4)2 - 3(5)1. Proceed according to proper order of operations: 5(16) – 3(5). Therefore: 80-15= 65.
Example Question #1 : Exponents
If 64t+1 = (√2)10t + 4, what is the value of t?
1/3
–1/3
4
–4
1/2
–4
In order to set the exponents equal to each other and solve for t, there must be the same number raised to those exponents.
64 = (√2)n?
(√2)2 = 2 and 26 = 64, so ((√2)2)6= (√2)2*6 = (√2)12.
Thus, we now have (√2)12(t+1) = (√2)10t + 4.
12(t+1) = 10t + 4
12t + 12 = 10t + 4
2t + 12 = 4
2t = –8
t = –4
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