Bluebook SAT Test Digital SAT Hard Math 5

Question 1

The polynomial p(x) = x³ + ax² + bx − 8 has x = 2 as a root and satisfies p(−1) = −18. What is the value of a + b?

Accepted Answer

3

Answer

Since x = 2 is a root, p(2) = 0: 8 + 4a + 2b − 8 = 0, which simplifies to 2a + b = 0, so b = −2a. From p(−1) = −18: −1 + a − b − 8 = −18, giving a − b = −9. Substituting b = −2a yields 3a = −9, so a = −3 and b = 6. Therefore a + b = −3 + 6 = 3.

Question 2

Let f(x) = |2x − 3| − |x + 1|. How many integers in the closed interval [−5, 5] satisfy f(x) > 0?

Accepted Answer

7

Answer

The critical points are x = −1 and x = 3/2, creating three regions. For x < −1: f(x) = −x + 4 > 0 for all x < 4, so integers −5, −4, −3, −2 all qualify (4 values). For −1 ≤ x < 3/2: f(x) = −3x + 2 > 0 when x < 2/3, so integers −1 and 0 qualify (2 values). For x ≥ 3/2: f(x) = x − 4 > 0 only when x > 4, giving just x = 5 (1 value). Total: 4 + 2 + 1 = 7.

Question 3

If (2 + 3i) / (1 − 2i) = a + bi, where a and b are real numbers and i = √(−1), what is the value of a + b?

Accepted Answer

3/5

Answer

Multiply numerator and denominator by the conjugate (1 + 2i): numerator = (2 + 3i)(1 + 2i) = 2 + 4i + 3i + 6i² = 2 + 7i − 6 = −4 + 7i. Denominator = (1)² + (2)² = 5. So a + bi = (−4 + 7i)/5, meaning a = −4/5 and b = 7/5. Therefore a + b = −4/5 + 7/5 = 3/5.

Question 4

For what positive value of k does the line y = kx − 4 intersect the parabola y = x² − 2x + 1 at exactly one point?

Accepted Answer

2√5 − 2

Answer

Setting kx − 4 = x² − 2x + 1 gives x² − (2 + k)x + 5 = 0. For exactly one intersection, the discriminant must equal zero: (2 + k)² − 4(5) = 0, so (2 + k)² = 20, and 2 + k = 2√5 (taking the positive root to ensure k > 0). Therefore k = 2√5 − 2.

Question 5

A dataset of five numbers has a mean of 10 and a standard deviation of 2. Every value in the dataset is multiplied by 3, and then 5 is subtracted from each result. What is the standard deviation of the transformed dataset?

Accepted Answer

6

Answer

Standard deviation measures spread, not location. Subtracting a constant from every value shifts the entire distribution but does not change how spread out the values are, so the standard deviation is unaffected by that step. Multiplying every value by 3 scales all deviations from the mean by a factor of 3, giving a new standard deviation of 3 × 2 = 6.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Digital SAT Hard Math 5