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 $f(a)$. This theorem also helps us determine if the divisor is a factor: if the remainder is zero, the divisor is a factor; if not, it isn’t.

Let's solve each one step-by-step.

Problem 1

Polynomial: $p^4 - 8p^3 + 10p^2 + 2p + 4$
Divisor: $p - 2$

1. Set $p = 2$ and substitute $p = 2$ into the polynomial: $f(2) = 2^4 - 8 \cdot 2^3 + 10 \cdot 2^2 + 2 \cdot 2 + 4$
2. Calculate each term:
   • $2^4 = 16$
   • $-8 \cdot 2^3 = -64$
   • $10 \cdot 2^2 = 40$
   • $2 \cdot 2 = 4$
3. Sum these values: $f(2) = 16 - 64 + 40 + 4 + 4 = 0$
4. Conclusion: The remainder is 0, so $p - 2$ is a factor of the polynomial.

Problem 2

Polynomial: $x^5 + 6x^4 - 3x^2 - 22x - 29$
Divisor: $x + 6$

1. Rewrite the divisor as $x - (-6)$, so set $x = -6$.
2. Substitute $x = -6$ into the polynomial: $f(-6) = (-6)^5 + 6 \cdot (-6)^4 - 3 \cdot (-6)^2 - 22 \cdot (-6) - 29$
3. Calculate each term:
   • $(-6)^5 = -7776$
   • $6 \cdot (-6)^4 = 6 \cdot 1296 = 7776$
   • $-3 \cdot (-6)^2 = -3 \cdot 36 = -108$
   • $-22 \cdot (-6) = 132$
4. Sum these values: $f(-6) = -7776 + 7776 - 108 + 132 - 29 = -5$
5. Conclusion: The remainder is -5, so $x + 6$ is not a factor of the polynomial.

Problem 3

Polynomial: $n^5 - 25n^3 - 7n^2 - 37n - 18$
Divisor: $n + 5$

1. Rewrite the divisor as $n - (-5)$, so set $n = -5$.
2. Substitute $n = -5$ into the polynomial: $f(-5) = (-5)^5 - 25 \cdot (-5)^3 - 7 \cdot (-5)^2 - 37 \cdot (-5) - 18$
3. Calculate each term:
   • $(-5)^5 = -3125$
   • $-25 \cdot (-5)^3 = -25 \cdot -125 = 3125$
   • $-7 \cdot (-5)^2 = -7 \cdot 25 = -175$
   • $-37 \cdot (-5) = 185$
4. Sum these values: $f(-5) = -3125 + 3125 - 175 + 185 - 18 = -8$
5. Conclusion: The remainder is -8, so $n + 5$ is not a factor of the polynomial.

Problem 4

Polynomial: $n^4 + 10n^3 + 21n^2 + 6n - 8$
Divisor: $n + 2$

1. Rewrite the divisor as $n - (-2)$, so set $n = -2$.
2. Substitute $n = -2$ into the polynomial: $f(-2) = (-2)^4 + 10 \cdot (-2)^3 + 21 \cdot (-2)^2 + 6 \cdot (-2) - 8$
3. Calculate each term:
   • $(-2)^4 = 16$
   • $10 \cdot (-2)^3 = 10 \cdot -8 = -80$
   • $21 \cdot (-2)^2 = 21 \cdot 4 = 84$
   • $6 \cdot (-2) = -12$
4. Sum these values: $f(-2) = 16 - 80 + 84 - 12 - 8 = 0$
5. Conclusion: The remainder is 0, so $n + 2$ is a factor of the polynomial.

Would you like more details on any of the steps? Here are five questions for further exploration:

1. How is the Remainder Theorem applied in more complex polynomial division?
2. What is synthetic division, and how can it simplify calculations here?
3. Why does a remainder of zero imply that the divisor is a factor of the polynomial?
4. How can we verify these results using alternative methods?
5. What are some practical applications of polynomial division in real-world contexts?

Tip: Always double-check calculations in polynomial division, as minor errors can lead to incorrect conclusions about factors and remainders.