Let f(x) = x ^ 2015 - 6

(a) Find the remainder when f(x) is divided by x + 1

(b) Find the remainder when 32015 divided by 4

Let's solve the two parts of the problem:

### (a) Find the remainder when \( f(x) = x^{2015} - 6 \) is divided by \( x + 1 \)

To find the remainder when dividing a polynomial by a linear divisor, we can use **the Remainder Theorem**, which states that the remainder when a polynomial \( f(x) \) is divided by \( x - a \) is \( f(a) \).

Here, we're dividing by \( x + 1 \), which is equivalent to \( x - (-1) \). Therefore, we need to evaluate \( f(-1) \).

Given that:
\[
f(x) = x^{2015} - 6
\]

We substitute \( x = -1 \) into \( f(x) \):
\[
f(-1) = (-1)^{2015} - 6
\]

Since \( 2015 \) is an odd number, \( (-1)^{2015} = -1 \), so:
\[
f(-1) = -1 - 6 = -7
\]

Thus, the remainder when \( f(x) \) is divided by \( x + 1 \) is \( \boxed{-7} \).

---

### (b) Find the remainder when \( 3^{2015} \) is divided by 4

To find the remainder of \( 3^{2015} \) divided by 4, we can observe the **pattern** in the powers of 3 modulo 4.

Let's compute the first few powers of 3 modulo 4:
\[
3^1 \mod 4 = 3
\]
\[
3^2 \mod 4 = 9 \mod 4 = 1
\]
\[
3^3 \mod 4 = 3 \times 3^2 \mod 4 = 3 \times 1 = 3
\]
\[
3^4 \mod 4 = 3 \times 3^3 \mod 4 = 3 \times 3 = 9 \mod 4 = 1
\]

We notice a repeating cycle: \( 3^1 \equiv 3 \mod 4 \), \( 3^2 \equiv 1 \mod 4 \), and this alternates.

Since \( 2015 \) is odd, we see that:
\[
3^{2015} \equiv 3 \mod 4
\]

Thus, the remainder when \( 3^{2015} \) is divided by 4 is \( \boxed{3} \).

---

Would you like more details on how these concepts work? Here are some follow-up questions you might find interesting:

1. How can we prove the pattern for powers of 3 modulo 4 using induction?
2. What is the Remainder Theorem, and how is it applied in polynomial division?
3. How would you handle higher powers of other numbers modulo small integers?
4. What would happen if the exponent in part (b) was even instead of odd?
5. What is the significance of modular arithmetic in cryptography?

**Tip:** Modular arithmetic is especially useful in number theory, cryptography, and algorithm optimization. Recognizing repeating cycles in powers modulo an integer can significantly reduce the computational work for large exponents.

Let f(x) = x^2015 - 6

(a) Find the remainder when f(x) is divided by x + 1

(b) Find the remainder when 3^2015 is divided by 4

Solve the problem of finding remainders when dividing f(x) = x^2015 - 6 by x + 1 and 3^2015 by 4. Explore the Remainder Theorem and modular arithmetic cyclic patterns with detailed step-by-step solutions. Ideal for Grades 10-12 students studying algebra and number theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation