### All GRE Math Resources

## Example Questions

### Example Question #1 : Factoring Equations

Solve for x: (–3x + 3) / (x – 1) = x

**Possible Answers:**

No solution possible

5

–3

–3 and 1

**Correct answer:**

–3

Begin by getting all factors to one side of the equal sign.

–3x + 3 = x(x – 1) → -3x + 3 = x^{2 }– x → 0 = x^{2 }+2x – 3.

Now, factor the right side: 0 = (x + 3)(x – 1).

Each of these factors can be set equal to 0 and solved for x. (x + 3) = 0; x = –3.

(x – 1) = 0 → x = 1.

*However*, the answer is not A, because if we return to the original problem, we must note that the denominator of the fraction is (x – 1); therefore, 1 is not a valid answer because this would cause a division by 0. Thus, –3 is the only acceptable answer.

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

x^{2} – 9X + 18 = 0

Find x

**Possible Answers:**

x = 3, –6

x = –3, –6

x = 3, 6

x = –3, 6

x = 3, 9

**Correct answer:**

x = 3, 6

factor the equation:

(x – 3)(x – 6) = 0

set each equal to 0

x = 3, 6

### Example Question #151 : Equations / Inequalities

25*x*^{2} – 36*y*^{2} can be factored into:

**Possible Answers:**

(5*x* + 6*y*)(5*x* + 6*y*)

(5*x* – 6*y*)(5*x* + 6*y*)

5 * 6 * (*x*^{2} – *y*^{2})

(5*x* – 6*y*)(5*x* – 6*y*)

cannot be factored

**Correct answer:**

(5*x* – 6*y*)(5*x* + 6*y*)

This is the difference of squares. You must know this formula for the GRE!

*a*^{2} – *b*^{2} = (*a* – *b*)(*a* + *b*)

Here *a* = 5*x* and *b* = 6*y*, so the difference of squares formula gives us (5*x* – 6*y*)(5*x* + 6*y*).

### Example Question #1 : Factoring Equations

Factor 3*u*^{4} – 24*uv*^{3}.

**Possible Answers:**

3*u*(*u* – 2*v*)(*u*^{2} + 2*uv* + 4*v*^{2})

3*u*(*u*^{3} – 8*v*^{3})

3*u*(*u* – 2*v*)(*u* + 2*v*)

3*u*(*u* – 2*v*)(*u*^{2} – 2*uv* – 4*v*^{2})

3*u*[*u*^{3} – (2*v*)^{3}]

**Correct answer:**

3*u*(*u* – 2*v*)(*u*^{2} + 2*uv* + 4*v*^{2})

First pull out 3u from both terms.

3*u*^{4} – 24*uv*^{3 }= 3*u*(*u*^{3} – 8*v*^{3}) = 3*u*[*u*^{3} – (2*v*)^{3}]

This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember but pat yourself on the back for getting to such hard questions! The difference of cubes formula to remember is *a*^{3} – *b*^{3} = (*a* – *b*)(*a*^{2} + *ab* + *b*^{2}). In our problem, *a* = *u* and *b* = 2*v*, so

3*u*^{4} – 24*uv*^{3 }= 3*u*(*u*^{3} – 8*v*^{3}) = 3*u*[*u*^{3} – (2*v*)^{3}]

= 3*u*(*u* – 2*v*)(*u*^{2} + 2*uv* + 4*v*^{2})

### Example Question #2 : Factoring Equations

Simplify .

**Possible Answers:**

**Correct answer:**

To begin, let's factor the first two terms and the second two terms separately.

*z*^{3} – *z*^{2} – 9*z* + 9 = (*z*^{3} – *z*^{2}) + (–9*z* + 9) = *z*^{2}(*z* – 1) – 9(*z* – 1)

(*z* – 1) can be pulled out because it appears in both terms.

*z*^{3} – *z*^{2} – 9*z* + 9 = (*z*^{3} – *z*^{2}) + (–9*z* + 9) = *z*^{2}(*z* – 1) – 9(*z* – 1) = (*z* – 1)(*z*^{2} – 9)

(*z*^{2} – 9) is a difference of squares, so we can use the formula *a*^{2} – *b*^{2} = (*a* – *b*)(*a* + *b*).

*z*^{3} – *z*^{2} – 9*z* + 9 = (*z*^{3} – *z*^{2}) + (–9*z* + 9) = *z*^{2}(*z* – 1) – 9(*z* – 1)

= (*z* – 1)(*z*^{2} – 9)

= (*z* – 1)(*z* – 3)(*z* + 3)

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

Factor .

**Possible Answers:**

None of the other answers are correct.

**Correct answer:**

We know the equation a^{2} – b^{2} = (a + b)(a – b) for the difference of squares. Since y* ^{2 }*is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).

### Example Question #1 : Factoring Equations

Solve for .

_{}

**Possible Answers:**

**Correct answer:**

Factor the equation by finding two numbers that add to -3 and multiply to -28.

Factors of 28: 1,2,4,7,14,28

-7 and 4 work.

(x-7)(x+4) = 0

Set each equal to zero:

x=7,-4

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