Understanding the Polynomial Remainder Problem

The question asks us to find the value of \(a^2 + b^2\) based on information about the remainders when two different polynomial expressions involving \(a\) and \(b\) are divided by linear factors.

We are given two polynomials:

• \(P_1(x) = x^2 + ax + b\)
• \(P_2(x) = x^2 + bx + a\)

We are also given two conditions based on polynomial division and remainders:

• When \(P_1(x)\) is divided by \((x - 1)\), the remainder is 15.
• When \(P_2(x)\) is divided by \((x + 1)\), the remainder is -1.

Applying the Remainder Theorem

The Remainder Theorem states that if a polynomial \(P(x)\) is divided by \((x - c)\), then the remainder is \(P(c)\).

Condition 1: Dividing \(P_1(x)\) by \((x - 1)\)

According to the Remainder Theorem, when \(P_1(x) = x^2 + ax + b\) is divided by \((x - 1)\), the remainder is \(P_1(1)\). We are given that this remainder is 15.

So, we evaluate \(P_1(x)\) at \(x = 1\):

\(P_1(1) = (1)^2 + a(1) + b = 1 + a + b\)

Setting this equal to the given remainder:

\(1 + a + b = 15\)

This gives us our first linear equation involving \(a\) and \(b\):

Equation 1: \(a + b = 14\)

Condition 2: Dividing \(P_2(x)\) by \((x + 1)\)

According to the Remainder Theorem, when \(P_2(x) = x^2 + bx + a\) is divided by \((x + 1)\), which can be written as \((x - (-1))\), the remainder is \(P_2(-1)\). We are given that this remainder is -1.

So, we evaluate \(P_2(x)\) at \(x = -1\):

\(P_2(-1) = (-1)^2 + b(-1) + a = 1 - b + a\)

Setting this equal to the given remainder:

\(1 - b + a = -1\)

This gives us our second linear equation involving \(a\) and \(b\):

Equation 2: \(a - b = -2\)

Solving the System of Linear Equations

Now we have a system of two linear equations:

1. \(a + b = 14\)

2. \(a - b = -2\)

We can solve this system using methods like elimination or substitution. Let's use the elimination method by adding the two equations:

\((a + b) + (a - b) = 14 + (-2)\)

\(a + b + a - b = 14 - 2\)

\(2a = 12\)

Dividing both sides by 2:

\(a = \frac{12}{2}\)

\(a = 6\)

Now that we have the value of \(a\), we can substitute it back into either Equation 1 or Equation 2 to find the value of \(b\). Let's use Equation 1:

\(a + b = 14\)

\(6 + b = 14\)

Subtracting 6 from both sides:

\(b = 14 - 6\)

\(b = 8\)

So, the values of \(a\) and \(b\) are \(a = 6\) and \(b = 8\).

Calculating the Value of \(a^2 + b^2\)

The question asks for the value of \(a^2 + b^2\). Now that we have \(a = 6\) and \(b = 8\), we can substitute these values into the expression:

\(a^2 + b^2 = (6)^2 + (8)^2\)

\(a^2 + b^2 = 36 + 64\)

\(a^2 + b^2 = 100\)

The value of \(a^2 + b^2\) is 100.

Revision Table: Key Concepts

Additional Information: Beyond the Polynomial Remainder

Understanding polynomials and theorems like the Remainder Theorem is fundamental in algebra. These concepts help in analyzing polynomial behavior, finding roots, and simplifying expressions.

• Factor Theorem: A special case of the Remainder Theorem. If \(P(c) = 0\), then \((x - c)\) is a factor of \(P(x)\). In our problem, the remainders were non-zero, so \((x-1)\) and \((x+1)\) were not factors of the respective polynomials.
• Polynomial Long Division: While the Remainder Theorem gives the remainder directly without full division, polynomial long division is the process used to divide one polynomial by another, yielding both a quotient and a remainder.
• Applications: Polynomials are used in various fields, including engineering, physics, economics, and computer science, to model curves, surfaces, and relationships between quantities.
• Algebraic Identities: The expression \(a^2 + b^2\) is part of various algebraic identities, such as \((a+b)^2 = a^2 + 2ab + b^2\) and \((a-b)^2 = a^2 - 2ab + b^2\). We could also have found \(a^2 + b^2\) using the identity \(a^2 + b^2 = (a+b)^2 - 2ab\). We know \(a+b = 14\). We found \(a=6\) and \(b=8\), so \(ab = 48\). Then \(a^2 + b^2 = (14)^2 - 2(48) = 196 - 96 = 100\). This provides an alternative verification method.