GRE Math : How to factor an equation
Study concepts, example questions & explanations for GRE Math
All GRE Math Resources
Example Questions
Example Question #1 :Factoring Equations
Solve for x: (–3x + 3) / (x – 1) = x
–3 and 1
5
No solution possible
–3
–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 #1 :Factoring Equations
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 #2 :Factoring Equations
25x2– 36y2can be factored into:
cannot be factored
5 * 6 * (x2–y2)
(5x+ 6y)(5x+ 6y)
(5x– 6y)(5x+ 6y)
(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)
Herea= 5xandb= 6y, so the difference of squares formula gives us (5x– 6y)(5x+ 6y).
Example Question #1 :How To Factor An Equation
Factor 3u4– 24uv3.
3u[u3– (2v)3]
3u(u– 2v)(u+ 2v)
3u(u3– 8v3)
3u(u– 2v)(u2+ 2uv+ 4v2)
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 isa3–b3= (a–b)(a2+ab+b2). In our problem,a=uandb= 2v, so
3u4– 24uv3= 3u(u3– 8v3) = 3u[u3– (2v)3]
= 3u(u– 2v)(u2+ 2uv+ 4v2)
Example Question #1 :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 formulaa2–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 #1 :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 y2is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).
Example Question #3 :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
All GRE Math Resources
