Bluebook SAT Test Digital SAT Hard Math Practice 6

Question 1

When the polynomial p(x) is divided by (x − 3), the remainder is 7. When p(x) is divided by (x + 2), the remainder is −3. What is the remainder when p(x) is divided by (x − 3)(x + 2)?

Accepted Answer

2x + 1

Answer

When dividing p(x) by a degree-2 polynomial, the remainder is a linear expression r(x) = ax + b. By the Remainder Theorem, r(3) = 3a + b = 7 and r(−2) = −2a + b = −3. Subtracting the second equation from the first gives 5a = 10, so a = 2. Substituting back yields b = 1. Therefore r(x) = 2x + 1.

Question 2

A circle has center (2, −3) and passes through the point (6, 0). What is the slope of the tangent line to the circle at (6, 0)?

Accepted Answer

−4/3

Answer

The radius from center (2, −3) to point (6, 0) has slope (0 − (−3)) / (6 − 2) = 3/4. Since a tangent line is perpendicular to the radius at the point of tangency, the tangent slope is the negative reciprocal: −4/3.

Question 3

A committee of 3 people is randomly chosen from a group of 5 men and 4 women. Given that the committee contains at least one woman, what is the probability it contains exactly 2 women?

Accepted Answer

15/37

Answer

Total ways to choose 3 from 9: C(9,3) = 84. All-male committees: C(5,3) = 10. So committees with at least 1 woman: 84 − 10 = 74. Committees with exactly 2 women: C(4,2) · C(5,1) = 6 · 5 = 30. The conditional probability is 30/74 = 15/37. The common trap is dividing 30 by 84 (ignoring the conditional), which gives 5/14.

Question 4

Let f(x) = x² − 4x − 5. For how many distinct real values of x does |f(x)| = 3?

Accepted Answer

4

Answer

|f(x)| = 3 is equivalent to f(x) = 3 or f(x) = −3. For f(x) = 3: x² − 4x − 8 = 0 has discriminant 16 + 32 = 48 > 0, yielding 2 distinct real solutions. For f(x) = −3: x² − 4x − 2 = 0 has discriminant 16 + 8 = 24 > 0, yielding 2 more distinct real solutions. Total: 4 solutions.

Question 5

What is the real part of the complex number (2 + 3i)² / (1 − i)?

Accepted Answer

−17/2

Answer

First expand (2 + 3i)² = 4 + 12i + 9i² = 4 + 12i − 9 = −5 + 12i. Then multiply numerator and denominator by the conjugate (1 + i): (−5 + 12i)(1 + i) / ((1 − i)(1 + i)) = (−5 − 5i + 12i + 12i²) / 2 = (−5 + 7i − 12) / 2 = (−17 + 7i) / 2. The real part is −17/2.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Digital SAT Hard Math Practice 6