All SAT Math Resources
Example Questions
Example Question #1 : Factoring Equations
Factor the following equation.
x2 – 16
(x2)(4 – 2)
(x + 4)(x – 4)
(x + 4)(x + 4)
(x)(x – 4)
(x – 4)(x – 4)
(x + 4)(x – 4)
The correct answer is (x + 4)(x – 4)
We neen to factor x2 – 16 to solve. We know that each parenthesis will contain an x to make the x2. We know that the root of 16 is 4 and since it is negative and no value of x is present we can tell that one 4 must be positive and the other negative. If we work it from the multiple choice answers we will see that when multiplying it out we get x2 + 4x – 4x – 16. 4x – 4x cancels out and we are left with our answer.
Example Question #195 : Algebra
If x3 – y3 = 30, and x2 + xy + y2 = 6, then what is x2 – 2xy + y2?
25
cannot be determined
24
5
180
25
First, let's factor x3 – y3 using the formula for difference of cubes.
x3 – y3 = (x – y)(x2 + xy + y2)
We are told that x2 + xy + y2 = 6. Thus, we can substitute 6 into the above equation and solve for x – y.
(x - y)(6) = 30.
Divide both sides by 6.
x – y = 5.
The original questions asks us to find x2 – 2xy + y2. Notice that if we factor x2 – 2xy + y2 using the formula for perfect squares, we obtain the following:
x2 – 2xy + y2 = (x – y)2.
Since we know that (x – y) = 5, (x – y)2 must equal 52, or 25.
Thus, x2 – 2xy + y2 = 25.
The answer is 25.
Example Question #4 : Equations / Inequalities
if x – y = 4 and x2 – y = 34, what is x?
9
15
10
12
6
6
This can be solved by substitution and factoring.
x2 – y = 34 can be written as y = x2 – 34 and substituted into the other equation: x – y = 4 which leads to x – x2 + 34 = 4 which can be written as x2 – x – 30 = 0.
x2 – x – 30 = 0 can be factored to (x – 6)(x + 5) = 0 so x = 6 and –5 and because only 6 is a possible answer, it is the correct choice.
Example Question #196 : Algebra
If x2 + 2ax + 81 = 0. When a = 9, what is the value of x?
3
–18
9
–9
0
–9
When a = 9, then x2 + 2ax + 81 = 0 becomes
x2 + 18x + 81 = 0.
This equation can be factored as (x + 9)2 = 0.
Therefore when a = 9, x = –9.
Example Question #91 : Algebra
If f(x) has roots at x = –1, 0 and 2, which of the following could be the equation for f(x)?
f(x) = x4 + x3 – 2x2
f(x) = x3 – x2 – 2x
f(x) = x3 – x2 + 2x
f(x) = x2 – x – 2
f(x) = x2 + x – 2
f(x) = x3 – x2 – 2x
In general, if a function has a root at x = r, then (x – r) must be a factor of f(x). In this problem, we are told that f(x) has roots at –1, 0 and 2. This means that the following are all factors of f(x):
(x – (–1)) = x + 1
(x – 0) = x
and (x – 2).
This means that we must look for an equation for f(x) that has the factors (x + 1), x, and (x – 2).
We can immediately eliminate the function f(x) = x2 + x – 2, because we cannot factor an x out of this polynomial. For the same reason, we can eliminate f(x) = x2 – x – 2.
Let's look at the function f(x) = x3 – x2 + 2x. When we factor this, we are left with x(x2 – x + 2). We cannot factor this polynomial any further. Thus, x + 1 and x – 2 are not factors of this function, so it can't be the answer.
Next, let's examine f(x) = x4 + x3 – 2x2 .
We can factor out x2.
x2 (x2 + x – 2)
When we factor x2 + x – 2, we will get (x + 2)(x – 1). These factors are not the same as x – 2 and x + 1.
The only function with the right factors is f(x) = x3 – x2 – 2x.
When we factor out an x, we get (x2 – x – 2), which then factors into (x – 2)(x + 1). Thus, this function has all of the factors we need.
The answer is f(x) = x3 – x2 – 2x.
Example Question #92 : Algebra
Factor 36x2 – 49y2.
cannot be factored
(6x + 7y)(6x + 7y)
6x2 – 7y2
(6x – 7y)(6x – 7y)
(6x + 7y)(6x – 7y)
(6x + 7y)(6x – 7y)
This is a difference of squares. The difference of squares formula is a2 – b2 = (a + b)(a – b). In this problem, a = 6x and b = 7y.
So 36x2 – 49y2 = (6x + 7y)(6x – 7y).
Example Question #7 : How To Factor An Equation
Solve for x:
Find two numbers that add to and multiply to
Factors of
You can use
Then make each factor equal 0.
and
and
Example Question #93 : Algebra
Find the roots of
Factoring yields giving roots of and .
Example Question #2 : How To Factor An Equation
Find the root of the equation above.
The numerator can be factored into .
Therefore, it can cancel with the denominator. So imples .
Example Question #94 : Algebra
Solve for .
Find all factors of 24
1, 2, 3,4, 6, 8, 12, 24
Now find two factors that add up to and multiply to ; and are the two factors.
By factoring, you can set the equation to be
If you FOIL it out, it gives you .
Set each part of the equation equal to 0, and solve for .
and
and
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