To solve these problems, we'll use the Remainder Theorem, which states that the remainder of a polynomial $f(x)$ divided by $x - a$ is $f(a)$.

Let's go through each part.

Part (a)

Polynomial: $f(x) = x^3 + 5x^2 - 3x + 1$

Divisor: $x + 1$ (which is equivalent to $x - (-1)$, so $a = -1$).

Solution:

Substitute $x = -1$ into $f(x)$: $f(-1) = (-1)^3 + 5(-1)^2 - 3(-1) + 1$ $= -1 + 5 + 3 + 1 = 8$ Remainder: 8

Part (b)

Polynomial: $f(x) = x^5 - 3x^3 + x - 5$

Divisor: $x - 2$ (so $a = 2$).

Solution:

Substitute $x = 2$ into $f(x)$: $f(2) = (2)^5 - 3(2)^3 + (2) - 5$ $= 32 - 3 \cdot 8 + 2 - 5$ $= 32 - 24 + 2 - 5 = 5$ Remainder: 5

Part (c)

Polynomial: $f(x) = 2x^3 - 4x^2 + x - 3$

Divisor: $x + 3$ (which is equivalent to $x - (-3)$, so $a = -3$).

Solution:

Substitute $x = -3$ into $f(x)$: $f(-3) = 2(-3)^3 - 4(-3)^2 + (-3) - 3$ $= 2(-27) - 4(9) - 3 - 3$ $= -54 - 36 - 3 - 3 = -96$ Remainder: -96

Part (d)

Polynomial: $f(x) = x^4 - 3x^2 - x + 2x + 1$

Divisor: $x - 1$ (so $a = 1$).

Solution:

Substitute $x = 1$ into $f(x)$: $f(1) = (1)^4 - 3(1)^2 - (1) + 2(1) + 1$ $= 1 - 3 - 1 + 2 + 1$ $= 0$ Remainder: 0

Summary of Answers

1. (a) Remainder is 8
2. (b) Remainder is 5
3. (c) Remainder is -96
4. (d) Remainder is 0

Would you like a more detailed explanation for any part, or do you have further questions?

Related Questions

1. How does the Remainder Theorem work with complex roots?
2. What is the Factor Theorem, and how does it relate to the Remainder Theorem?
3. How can synthetic division be used to find remainders quickly?
4. What happens when we divide by polynomials with degrees greater than 1?
5. How can we check our answers to polynomial remainder problems?

Tip:

When using the Remainder Theorem, always double-check your calculations for accuracy, especially when substituting negative values.