Answer :
Sure! Let's use synthetic division to find the quotient and remainder when [tex]\( x^5 - 3125 \)[/tex] is divided by [tex]\( x - 5 \)[/tex].
To perform synthetic division, follow these steps:
1. Identify the coefficients of the polynomial [tex]\( x^5 - 3125 \)[/tex]. The polynomial can be written as [tex]\( x^5 + 0x^4 + 0x^3 + 0x^2 + 0x - 3125 \)[/tex]. The coefficients are [tex]\([1, 0, 0, 0, 0, -3125]\)[/tex].
2. Set up synthetic division using the coefficients and the root of the divisor [tex]\( x - 5 = 0 \)[/tex] which is [tex]\( x = 5 \)[/tex].
3. Write the coefficients in a row and set up the synthetic division:
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & & & & & \\
\end{array}
\][/tex]
4. Bring down the leading coefficient (the first number) which is 1.
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & & & & & \\
& 1 & & & & & \\
\end{array}
\][/tex]
5. Multiply the root (5) by the number just brought down (1) and write the result under the next coefficient (0).
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & 5 & & & & \\
& 1 & & & & & \\
\end{array}
\][/tex]
6. Add the result to the coefficient above it (0 + 5 = 5) and write the sum below the line.
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & 5 & & & & \\
& 1 & 5 & & & & \\
\end{array}
\][/tex]
7. Repeat the process: Multiply the root (5) by the sum (5) and write the result under the next coefficient (0).
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & 5 & 25 & & & \\
& 1 & 5 & & & & \\
\end{array}
\][/tex]
8. Continue this process until all coefficients have been processed:
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & 5 & 25 & 125 & 625 & 3125 \\
& 1 & 5 & 25 & 125 & 625 & 0 \\
\end{array}
\][/tex]
9. The numbers at the bottom row (besides the last number) represent the coefficients of the quotient. The last number is the remainder.
So, the quotient of [tex]\( x^5 - 3125 \)[/tex] divided by [tex]\( x - 5 \)[/tex] is [tex]\( x^4 + 5x^3 + 25x^2 + 125x + 625 \)[/tex], and the remainder is 0.
Therefore, the quotient is [tex]\( x^4 + 5x^3 + 25x^2 + 125x + 625 \)[/tex] and there is no remainder.
To perform synthetic division, follow these steps:
1. Identify the coefficients of the polynomial [tex]\( x^5 - 3125 \)[/tex]. The polynomial can be written as [tex]\( x^5 + 0x^4 + 0x^3 + 0x^2 + 0x - 3125 \)[/tex]. The coefficients are [tex]\([1, 0, 0, 0, 0, -3125]\)[/tex].
2. Set up synthetic division using the coefficients and the root of the divisor [tex]\( x - 5 = 0 \)[/tex] which is [tex]\( x = 5 \)[/tex].
3. Write the coefficients in a row and set up the synthetic division:
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & & & & & \\
\end{array}
\][/tex]
4. Bring down the leading coefficient (the first number) which is 1.
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & & & & & \\
& 1 & & & & & \\
\end{array}
\][/tex]
5. Multiply the root (5) by the number just brought down (1) and write the result under the next coefficient (0).
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & 5 & & & & \\
& 1 & & & & & \\
\end{array}
\][/tex]
6. Add the result to the coefficient above it (0 + 5 = 5) and write the sum below the line.
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & 5 & & & & \\
& 1 & 5 & & & & \\
\end{array}
\][/tex]
7. Repeat the process: Multiply the root (5) by the sum (5) and write the result under the next coefficient (0).
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & 5 & 25 & & & \\
& 1 & 5 & & & & \\
\end{array}
\][/tex]
8. Continue this process until all coefficients have been processed:
[tex]\[
\begin{array}{r|rrrrrr}
5 & 1 & 0 & 0 & 0 & 0 & -3125 \\
& & 5 & 25 & 125 & 625 & 3125 \\
& 1 & 5 & 25 & 125 & 625 & 0 \\
\end{array}
\][/tex]
9. The numbers at the bottom row (besides the last number) represent the coefficients of the quotient. The last number is the remainder.
So, the quotient of [tex]\( x^5 - 3125 \)[/tex] divided by [tex]\( x - 5 \)[/tex] is [tex]\( x^4 + 5x^3 + 25x^2 + 125x + 625 \)[/tex], and the remainder is 0.
Therefore, the quotient is [tex]\( x^4 + 5x^3 + 25x^2 + 125x + 625 \)[/tex] and there is no remainder.
