Answer :
To calculate \( P(10) \) using synthetic division and the Remainder Theorem for the polynomial \( P(x) = x^4 - 2x^3 - 25x^2 - 26x + 120 \), let's follow these steps systematically:
1. Write down the coefficients of the polynomial:
[tex]\[ [1, -2, -25, -26, 120] \][/tex]
2. Set up synthetic division using 10 as the divisor:
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & & & & \\ \end{array} \][/tex]
3. Bring down the first coefficient (1) as is:
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & & & & \\ & 1 & & & & \\ \end{array} \][/tex]
4. Multiply by 10 and add to the next coefficient:
- Multiply: \( 1 \times 10 = 10 \)
- Add: \( -2 + 10 = 8 \)
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & 10 & & & \\ & 1 & 8 & & & \\ \end{array} \][/tex]
5. Repeat the multiply and add process:
- Multiply: \( 8 \times 10 = 80 \)
- Add: \( -25 + 80 = 55 \)
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & 10 & 80 & & \\ & 1 & 8 & 55 & & \\ \end{array} \][/tex]
6. Continue this process:
- Multiply: \( 55 \times 10 = 550 \)
- Add: \( -26 + 550 = 524 \)
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & 10 & 80 & 550 & \\ & 1 & 8 & 55 & 524 & \\ \end{array} \][/tex]
7. Finally:
- Multiply: \( 524 \times 10 = 5240 \)
- Add: \( 120 + 5240 = 5360 \)
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & 10 & 80 & 550 & 5240 \\ & 1 & 8 & 55 & 524 & 5360 \\ \end{array} \][/tex]
By performing synthetic division, we have obtained the sequence of intermediate results \([1, 8, 55, 524, 5360]\). The last number in this sequence, 5360, represents \( P(10) \). Therefore, the value of the polynomial \( P(x) \) at \( x = 10 \) is:
[tex]\[ P(10) = 5360 \][/tex]
1. Write down the coefficients of the polynomial:
[tex]\[ [1, -2, -25, -26, 120] \][/tex]
2. Set up synthetic division using 10 as the divisor:
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & & & & \\ \end{array} \][/tex]
3. Bring down the first coefficient (1) as is:
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & & & & \\ & 1 & & & & \\ \end{array} \][/tex]
4. Multiply by 10 and add to the next coefficient:
- Multiply: \( 1 \times 10 = 10 \)
- Add: \( -2 + 10 = 8 \)
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & 10 & & & \\ & 1 & 8 & & & \\ \end{array} \][/tex]
5. Repeat the multiply and add process:
- Multiply: \( 8 \times 10 = 80 \)
- Add: \( -25 + 80 = 55 \)
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & 10 & 80 & & \\ & 1 & 8 & 55 & & \\ \end{array} \][/tex]
6. Continue this process:
- Multiply: \( 55 \times 10 = 550 \)
- Add: \( -26 + 550 = 524 \)
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & 10 & 80 & 550 & \\ & 1 & 8 & 55 & 524 & \\ \end{array} \][/tex]
7. Finally:
- Multiply: \( 524 \times 10 = 5240 \)
- Add: \( 120 + 5240 = 5360 \)
[tex]\[ \begin{array}{r|rrrrr} & 1 & -2 & -25 & -26 & 120 \\ 10 & & 10 & 80 & 550 & 5240 \\ & 1 & 8 & 55 & 524 & 5360 \\ \end{array} \][/tex]
By performing synthetic division, we have obtained the sequence of intermediate results \([1, 8, 55, 524, 5360]\). The last number in this sequence, 5360, represents \( P(10) \). Therefore, the value of the polynomial \( P(x) \) at \( x = 10 \) is:
[tex]\[ P(10) = 5360 \][/tex]
