### All SAT Math Resources

## Example Questions

### Example Question #1 : Equations / Solution Sets

Factor the following equation.

x^{2 }– 16

**Possible Answers:**

(x + 4)(x – 4)

(x + 4)(x + 4)

(x^{2})(4 – 2)

(x – 4)(x – 4)

(x)(x – 4)

**Correct answer:**

(x + 4)(x – 4)

The correct answer is (x + 4)(x – 4)

We neen to factor x^{2 }– 16 to solve. We know that each parenthesis will contain an x to make the x^{2}. 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 x^{2 }+ 4x – 4x – 16. 4x – 4x cancels out and we are left with our answer.

### Example Question #2 : Factoring Equations

If x^{3} – y^{3} = 30, and x^{2} + xy + y^{2} = 6, then what is x^{2} – 2xy + y^{2}?

**Possible Answers:**

25

24

cannot be determined

5

180

**Correct answer:**

25

First, let's factor x^{3} – y^{3} using the formula for difference of cubes.

x^{3} – y^{3 }= (x – y)(x^{2} + xy + y^{2})

We are told that x^{2} + xy + y^{2} = 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 x^{2} – 2xy + y^{2}. Notice that if we factor x^{2} – 2xy + y^{2} using the formula for perfect squares, we obtain the following:

x^{2} – 2xy + y^{2 }= (x – y)^{2}.

Since we know that (x – y) = 5, (x – y)^{2} must equal 5^{2}, or 25.

Thus, x^{2} – 2xy + y^{2 }= 25.

The answer is 25.

### Example Question #1 : Equations / Solution Sets

if x – y = 4 and x^{2} – y = 34, what is x?

**Possible Answers:**

6

10

12

15

9

**Correct answer:**

6

This can be solved by substitution and factoring.

x^{2} – y = 34 can be written as y = x^{2} – 34 and substituted into the other equation: x – y = 4 which leads to x – x^{2} + 34 = 4 which can be written as x^{2} – x – 30 = 0.

x^{2} – 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 #2 : Equations / Solution Sets

If *x*^{2} + 2*ax* + 81 = 0. When *a* = 9, what is the value of *x*?

**Possible Answers:**

9

–18

3

–9

0

**Correct answer:**

–9

When *a* = 9, then *x*^{2} + 2*ax* + 81 = 0 becomes

*x*^{2} + 18*x* + 81 = 0.

This equation can be factored as (*x* + 9)^{2} = 0.

Therefore when *a* = 9, *x* = –9.

### Example Question #5 : Factoring Equations

If *f*(*x*) has roots at *x* = –1, 0 and 2, which of the following could be the equation for *f*(*x*)?

**Possible Answers:**

f(x) = x^{3 }– x^{2 }– 2x

f(x) = x^{2 }– x – 2

f(x) = x^{3 }– x^{2 }+ 2x

f(x) = x^{2} + x – 2

f(x) = x^{4} + x^{3} – 2x^{2}

**Correct answer:**

f(x) = x^{3 }– x^{2 }– 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*) = *x*^{2} + *x* – 2, because we cannot factor an *x* out of this polynomial. For the same reason, we can eliminate *f*(*x*) = *x*^{2 }– *x* – 2.

Let's look at the function *f*(*x*) = *x*^{3 }– *x*^{2 }+ 2*x*. When we factor this, we are left with *x*(*x*^{2}^{ }– *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*) = *x*^{4} + *x*^{3} – 2*x*^{2 }.

We can factor out *x*^{2}.

*x*^{2 }(*x*^{2 }+ *x* – 2)

When we factor *x*^{2 }+ *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*) = *x*^{3 }– *x*^{2 }– 2*x*.

When we factor out an *x*, we get (*x*^{2 }– *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*) = *x*^{3 }– *x*^{2 }– 2*x*.

### Example Question #3 : Equations / Inequalities

Factor 36*x*^{2} – 49*y*^{2}.

**Possible Answers:**

6*x*^{2} – 7*y*^{2}

(6*x* + 7*y*)(6*x* – 7*y*)

cannot be factored

(6*x* + 7*y*)(6*x* + 7*y*)

(6*x* – 7*y*)(6*x* – 7*y*)

**Correct answer:**

(6*x* + 7*y*)(6*x* – 7*y*)

This is a difference of squares. The difference of squares formula is *a*^{2} – *b*^{2} = (*a* + *b*)(*a* – *b*). In this problem, *a* = 6*x* and *b* = 7*y*.

So 36*x*^{2} – 49*y*^{2 }= (6*x* + 7*y*)(6*x* – 7*y*).

### Example Question #7 : Factoring Equations

Solve for *x*:

**Possible Answers:**

**Correct answer:**

Find two numbers that add to and multiply to

Factors of

You can use

Then make each factor equal 0.

and

and

### Example Question #1 : How To Factor An Equation

Find the roots of

**Possible Answers:**

**Correct answer:**

Factoring yields giving roots of and .

### Example Question #9 : Factoring Equations

Find the root of the equation above.

**Possible Answers:**

**Correct answer:**

The numerator can be factored into .

Therefore, it can cancel with the denominator. So imples .

### Example Question #1 : Factoring Equations

Solve for .

**Possible Answers:**

**Correct answer:**

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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