If the first integer is x, then the second integer is (x + 1) and the third integer is (x + 2). The product of these integers is x(x + 1)(x + 2); x(x + 1) = x² + x; (x² + x)(x + 2) = x² + 2x² + x2²+ 2x = x³ + 3x² + 2x. This polynomial is equal to the cube of the first integer, x³ , plus 56. Therefore, x³ + 3x² + 2x = x³ + 56. Place all terms on one side of the equation, combining like terms; x³ + 3x² + 2x = x³ + 56, 3x² + 2x – 56 = 0. Factor the polynomial: 3x² + 2x – 56 = (3x + 14)(x – 4). Set each term equal to 0 to find the values of x that make the equation true; 3x + 14 = 0, 3x = –14, x = \(-\frac{14}{3}\), but you are looking for a positive integer; x – 4 = 0, x = 4. If 4 is the first of the consecutive integers, then 5 is the second integer and 6 is the third, and largest, integer. Again, trial and error could be used to find the solution.

If the number is x, then four times the cube of the number is 4x³. That value is equal to 48 times the number (48x) minus four times the square of the number (4x²). Therefore, 4x³ = 48x – 4x². Move all terms onto one side of the equation, and factor the polynomial; 4x³ = 48x – 4x², 4x³ + 4x² – 48x = 0; 4x³ + 4x² – 48x = 4x(x² + x – 12) = 4x(x + 4)(x – 3). Set each factor equal to 0 to find the values of x that make the equation true. 4x = 0, x = 0; x + 4 = 0, x = –4; x – 3 = 0, x = 3. The given situation is true for the numbers 0, –4, and 3, but only 3 is greater than 0. Trial and error could also be used for a multiple-choice problem such as this.

If the number is x, then the cube of the number is x³. Twice the square of the number is 2x². The difference in those values is equal to 80 times the number (80x). Therefore, x³ – 2x² = 80x. Move all terms onto one side of the equation, and factor the polynomial; x³ – 2x² = 80x, x³ – 2x² – 80x = 0; x³ – 2x2²– 80x = x(x² – 2x – 80) = x(x + 8)(x – 10). Set each factor equal to 0 to find the values of x that make the equation true. x = 0; x + 8 = 0, x = –8; x – 10 = 0, x = 10. The given situation is true for the numbers 0, –8, and 10, but only 10 is greater than 0. Trial and error (plugging each answer choice into the equation x³ – 2x² = 80x) 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.

A fraction is undefined when its denominator is equal to 0. Factor the polynomial in the denominator and set each factor equal to 0 to find the values that make the fraction undefined; 4x³ + 44x² + 120x = 4x(x² + 11x + 30) = 4x(x + 5)(x + 6); 4x = 0, x = 0; x + 5 = 0, x = –5; x + 6 = 0, x = –6. The fraction is undefined when x is equal to –6, –5, or 0.

A fraction is undefined when its denominator is equal to 0. Factor the polynomial in the denominator and set each factor equal to 0 to find the values that make the fraction undefined; x³ + 3x² – 4x = x(x + 4)(x – 1); x = 0; x + 4 = 0, x = –4; x – 1 = 0, x = 1. The fraction is undefined when x is equal to –4, 0, or 1.