Jessica determines the remainder of $\frac{5 x^{52}-4 x^{10}-11 x^2+12}{x+1}$ using the remainder theorem.

How does she proceed to the correct answer?

Verified Answer :

  • Identify the polynomial \(P(x) = 5x^{52} - 4x^{10} - 11x^2 + 12\) and the divisor \(x+1\).
  • Apply the Remainder Theorem by setting \(x = -1\).
  • Evaluate the terms \(5(-1)^{52}\), \(-4(-1)^{10}\), and \(-11(-1)^2\) using the property that even powers of \(-1\) are equal to \(1\).
  • Calculate the final result: \(P(-1) = 5 - 4 - 11 + 12 = \boxed{2}\).

Explanation

  1. Understanding the Remainder Theorem

To determine the remainder of the division of the polynomial \(P(x) = 5x^{52} - 4x^{10} - 11x^2 + 12\) by the linear binomial \(x+1\), we apply the Remainder Theorem. The Remainder Theorem states that for any polynomial \(P(x)\), the remainder when dividing by \(x-c\) is simply the value \(P(c)\). In our case, the divisor is \(x+1\), which can be written in the form \(x - (-1)\). Thus, we identify \(c = -1\), and our task is to evaluate the polynomial at \(x = -1\).

  1. Evaluating the Polynomial at \(x = -1\)

Now, we substitute \(x = -1\) into the original polynomial \(P(x)\):
\[P(-1) = 5(-1)^{52} - 4(-1)^{10} - 11(-1)^2 + 12.\]
To evaluate this, we compute the powers of \(-1\). Since any power of \(-1\) raised to an even integer exponent is equal to \(1\), we have:
\((-1)^{52} = 1\)
\((-1)^{10} = 1\)
\((-1)^{2} = 1\)
Substituting these values back into the expression, we get:
\[P(-1) = 5(1) - 4(1) - 11(1) + 12.\]

  1. Final Calculation of the Remainder

Finally, we perform the arithmetic operations:
\[P(-1) = 5 - 4 - 11 + 12.\]
Calculating the sum step by step:
\(5 - 4 = 1\)
\(1 - 11 = -10\)
\(-10 + 12 = 2\)
Thus, the remainder of the division is \(2\), and the correct answer is \(\boxed{2}\).

Examples

The Remainder Theorem is a powerful tool in algebra that simplifies finding remainders in polynomial division, avoiding long, tedious calculations. For instance, if you are designing a digital security algorithm that requires validating large data packets represented as polynomials, you can use the Remainder Theorem to quickly check if a packet is divisible by a specific key without performing the full division, saving significant computational resources.