How to factor an equation - GRE Quantitative Reasoning

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.

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.

We know the equation a2 – b2 = (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).

factor the equation:

(x – 3)(x – 6) = 0

set each equal to 0

x = 3, 6

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

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_).

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)

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)

We know the equation a2 – b2 = (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).

factor the equation:

(x – 3)(x – 6) = 0

set each equal to 0

x = 3, 6

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

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_).

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)

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)

We know the equation a2 – b2 = (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).

factor the equation:

(x – 3)(x – 6) = 0

set each equal to 0

x = 3, 6

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

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_).

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)

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)