Write an algebraic equation that describes the situation. If x is the number, then x² is the square of the number. Two less than three times the number is 3x – 2. Since the square of the number is equal to two less than three times the number, x² = 3x – 2. Combine like terms on one side of the equation and set the expression equal to 0; x² = 3x – 2, x² – 3x + 2 = 0. Factor the quadratic and set each factor equal to 0 to find the solutions for x. To find the factors of a quadratic, begin by finding two numbers whose product is equal to the constant of the quadratic. Of those numbers, find the pair that adds to the coefficient of the x term of the quadratic; –1 and –2 multiply to 2 and add to –3. Therefore, the factors of x² – 3x + 2 are (x – 1) and (x – 2). Set each factor equal to 0 and solve for x: x – 1 = 0, x = 1, and x – 2 = 0, x = 2.

Use FOIL to multiply the binomials: (x – 7)(x – 5) = x² – 5x – 7x + 35 = x² – 12x + 35. Combine like terms on one side of the equation and set the expression equal to 0; x² – 12x + 35 = –1, x² – 12x + 36 = 0. Factor the quadratic and set each factor equal to 0 to find the solutions for x.To find the factors of a quadratic,begin by finding two numbers whose product is equal to the constant of the quadratic. Of those numbers, find the pair that adds to the coefficient of the x term of the quadratic; –6 and –6 multiply to 36 and add to –12. Therefore, the factors of x² – 12x + 36 are (x – 6) and (x – 6). Since both factors are the same, set either factor equal to 0 and solve for x: x – 6 = 0, x = 6.

Use FOIL to multiply the binomials: (x – 2)(x + 6) = x² + 6x – 2x – 12 = x² + 4x – 12. Combine like terms on one side of the equation and set the expression equal to 0; x² + 4x – 12 = –16, x² + 4x + 4 = 0. Factor the quadratic and set each factor equal to 0 to find the solutions for x. To find the factors of a quadratic, begin by finding two numbers whose product is equal to the constant of the quadratic. Of those numbers, find the pair that adds to the coefficient of the x term of the quadratic; 2 and 2 multiply to 4 and add to 4. Therefore, the factors of x² + 4x + 4 are (x + 2) and (x + 2). Since both factors are the same, set either factor equal to 0 and solve for x: x + 2 = 0, x = –2.

Cross multiply: (x + 5)(x + 4) = (4)(6x – 6), x² + 9x + 20 = 24x – 24. Combine like terms on one side of the equation and set the expression equal to 0; x² + 9x + 20 = 24x – 24, x² – 15x + 44 = 0. Factor the quadratic and set each factor equal to 0 to find the solutions for x. To find the factors of a quadratic, begin by finding two numbers whose product is equal to the constant of the quadratic. Of those numbers, find the pair that adds to the coefficient of the x term of the quadratic; –4 and –11 multiply to 44 and add to –15. Therefore, the factors of x² – 15x + 44 are (x – 4) and (x – 11). Set each factor equal to 0 and solve for x: x – 4 = 0, x = 4, and x – 11 = 0, x = 11.

Combine like terms on one side of the equation and set the expression equal to 0; x² – 3x – 30 = 10, x² – 3x – 40 = 0. Factor the quadratic and set each factor equal to 0 to find the solutions for x. To find the factors of a quadratic, begin by finding two numbers whose product is equal to the constant of the quadratic. Of those numbers, find the pair that adds to the coefficient of the x term of the quadratic; –8 and 5 multiply to –40 and add to –3. Therefore, the factors of x² – 3x – 40 are (x – 8) and (x + 5). Set each factor equal to 0 and solve for x: x – 8 = 0, x = 8, and x + 5 = 0, x = –5; x² – 3x – 30 = 10 when x equals 8 or –5.

Combine like terms on one side of the equation and set the expression equal to 0; x² – x = 12, x² – x – 12 = 0. Factor the quadratic and set each factor equal to 0 to find the solutions for x. To find the factors of a quadratic, begin by finding two numbers whose product is equal to the constant of the quadratic. Of those numbers, find the pair that adds to the coefficient of the x term of the quadratic; –4 and 3 multiply to –12 and add to –1. Therefore, the factors of x² – x – 12 are (x – 4) and (x + 3). Set each factor equal to 0 and solve for x: x – 4 = 0, x = 4, and x + 3 = 0, x = –3; x² – x = 12 when x equals 4 or –3. Trial and error (plugging each answer choice into the equation x² – x = 12) could be used, but only on a multiple-choice SAT question. It is important to be able to solve questions like this that could occur in the grid-in section of the SAT.

Factor the numerator and denominator. The numerator factors into (x + 11)(x + 3) and the denominator factors into (x + 11)(x – 3). Cancel the (x + 11) terms from the numerator and denominator, leaving (x + 3) in the numerator and (x – 3) in the denominator.

Factor the numerator and denominator. The numerator factors into (x + 5)(x – 9) and the denominator factors into (x + 5)(x + 6). Cancel the (x + 5) terms from the numerator and denominator, leaving (x – 9) in the numerator and (x + 6) in the denominator.

Factor the numerator and denominator. The numerator factors into (x – 8)(x + 2) and the denominator factors into (x – 8)(x + 7). Cancel the (x – 8) terms from the numerator and denominator, leaving (x + 2) in the numerator and (x + 7) in the denominator.

Factor the denominator. To find the factors of a quadratic, begin by finding two numbers whose product is equal to the constant of the quadratic. Of those numbers, find the pair that adds to the coefficient of the x term of the quadratic. This quadratic has no x term—the sum of the products of the outside and inside terms of the factors is 0; –9 and 9 multiply to –81 and add to 0. Therefore, the factors of x² – 81 are (x – 9) and (x + 9). Cancel the (x + 9) terms from the numerator and denominator, leaving 1 in the numerator and (x – 9) in the denominator.