GRE Math : Factoring Equations
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Example Questions
Example Question #152 : Equations / Inequalities
Solve for x: (–3x + 3) / (x – 1) = x
5
–3
–3 and 1
No solution possible
–3
Begin by getting all factors to one side of the equal sign.
–3x + 3 = x(x – 1) → -3x + 3 = x2 – x → 0 = x2 +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 #2 : How To Factor An Equation
x2 – 9X + 18 = 0
Find x
x = –3, –6
x = 3, 6
x = –3, 6
x = 3, 9
x = 3, –6
x = 3, 6
factor the equation:
(x – 3)(x – 6) = 0
set each equal to 0
x = 3, 6
Example Question #153 : Equations / Inequalities
25x2 – 36y2 can be factored into:
(5x – 6y)(5x – 6y)
(5x + 6y)(5x + 6y)
cannot be factored
5 * 6 * (x2 – y2)
(5x – 6y)(5x + 6y)
(5x – 6y)(5x + 6y)
This is the difference of squares. You must know this formula for the GRE!
a2 – b2 = (a – b)(a + b)
Here a = 5x and b = 6y, so the difference of squares formula gives us (5x – 6y)(5x + 6y).
Example Question #1 : Factoring Equations
Factor 3u4 – 24uv3.
3u(u – 2v)(u2 + 2uv + 4v2)
3u(u3 – 8v3)
3u[u3 – (2v)3]
3u(u – 2v)(u + 2v)
3u(u – 2v)(u2 – 2uv – 4v2)
3u(u – 2v)(u2 + 2uv + 4v2)
First pull out 3u from both terms.
3u4 – 24uv3 = 3u(u3 – 8v3) = 3u[u3 – (2v)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 a3 – b3 = (a – b)(a2 + ab + b2). In our problem, a = u and b = 2v, so
3u4 – 24uv3 = 3u(u3 – 8v3) = 3u[u3 – (2v)3]
= 3u(u – 2v)(u2 + 2uv + 4v2)
Example Question #2 : Factoring Equations
Simplify 
To begin, let's factor the first two terms and the second two terms separately.
z3 – z2 – 9z + 9 = (z3 – z2) + (–9z + 9) = z2(z – 1) – 9(z – 1)
(z – 1) can be pulled out because it appears in both terms.
z3 – z2 – 9z + 9 = (z3 – z2) + (–9z + 9) = z2(z – 1) – 9(z – 1) = (z – 1)(z2 – 9)
(z2 – 9) is a difference of squares, so we can use the formula a2 – b2 = (a – b)(a + b).
z3 – z2 – 9z + 9 = (z3 – z2) + (–9z + 9) = z2(z – 1) – 9(z – 1)
= (z – 1)(z2 – 9)
= (z – 1)(z – 3)(z + 3)
Example Question #4 : Factoring Equations
Factor 
None of the other answers are correct.
We know the equation a2 – b2 = (a + b)(a – b) for the difference of squares. Since y2 is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).
Example Question #5 : Factoring Equations
Solve for 
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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