- Identify the polynomial \(P(x) = 5x^{52} - 4x^{15} - 11x^2 + 12\) and the divisor \(x+1\).
- Apply the Polynomial Remainder Theorem by evaluating the polynomial at \(x = -1\).
- Simplify the powers of \(-1\), noting that \((-1)^{52}=1\), \((-1)^{15}=-1\), and \((-1)^2=1\).
- Calculate the final remainder as \(5(1) - 4(-1) - 11(1) + 12 = 10\). The remainder is \(\boxed{10}\).
Explanation
- Understanding the Polynomial Remainder Theorem
To determine the remainder of the division of a polynomial \(P(x) = 5x^{52} - 4x^{15} - 11x^2 + 12\) by a linear divisor of the form \(x - c\), we use the Polynomial Remainder Theorem. This theorem states that the remainder is equal to the value of the polynomial at \(c\), denoted as \(P(c)\). In this problem, the divisor is \(x + 1\), which can be rewritten as \(x - (-1)\). Thus, the value we need to evaluate is \(c = -1\).
- Substituting the Variable into the Polynomial
Following the theorem, the remainder \(R\) is equal to \(P(-1)\). We substitute \(x = -1\) into the expression of the polynomial:
\[P(-1) = 5(-1)^{52} - 4(-1)^{15} - 11(-1)^2 + 12\]
To compute this, we consider the properties of powers of \(-1\):
- For any even power \(n\), \((-1)^n = 1\). Thus, \((-1)^{52} = 1\) and \((-1)^2 = 1\).
- For any odd power \(n\), \((-1)^n = -1\). Thus, \((-1)^{15} = -1\).
- Evaluating the Remainder
Now we substitute these values back into the expression to perform the final calculation:
\[P(-1) = 5(1) - 4(-1) - 11(1) + 12\]
\[P(-1) = 5 + 4 - 11 + 12\]
Adding and subtracting these values step by step:
\[5 + 4 = 9\]
\[9 - 11 = -2\]
\[-2 + 12 = 10\]
Therefore, the remainder of the division is \(10\).
Examples
The Polynomial Remainder Theorem is a powerful tool in algebra that allows us to find the remainder of a division without performing long division. In real life, this can be used in computer science for data error checking or checksum calculations, where we divide a large sequence of numbers represented as a polynomial by a divisor to verify the integrity of the data. By applying the theorem, we save time and computational power compared to full division algorithms.