91Ó°ÊÓ

Polynomial division is an essential concept in algebra that involves dividing a polynomial by another polynomial. When we divide a polynomial by a linear divisor of the form \(x - c\), the Remainder Theorem becomes particularly useful. It's a theorem that helps simplify the division by directly providing the remainder.In the example exercise, we are using the polynomial \(p(x) = (ax + b)^3\). It's divided by two different linear divisors, \(x+1\) and \(x-2\). The Remainder Theorem tells us that the remainder of the division of \(p(x)\) by \(x-c\) is \(p(c)\). Thus:

• When divided by \(x+1\), we find the remainder by calculating \(p(-1)\).
• When divided by \(x-2\), we calculate \(p(2)\) to find the remainder.

By using this theorem, we can transform a potentially complex polynomial division into a much simpler calculation, requiring us only to evaluate the polynomial at a specific point.

A system of equations is a set of multiple equations that are all solved together. In this exercise, we deal with a system of two equations derived from the Remainder Theorem results.To solve the problem, we obtain the following two equations:

• \((-a + b)^3 = -1\), simplifying to \(-a + b = -1\).
• \((2a + b)^3 = 27\), simplifying to \(2a + b = 3\).

Solving a system of linear equations typically involves finding a solution pair \((a, b)\) that satisfies both equations simultaneously. By manipulating these formulas, we find a value for \(a\) and \(b\) to resolve this particular system. Systems of equations are critical in many areas of mathematics, where simultaneous conditions must be satisfied.

Linear equations are equations of the first degree, meaning each term is either a constant or the product of a constant and a single variable. In the context of our exercise, both \(-a + b = -1\) and \(2a + b = 3\) are linear equations.Solving these involves standard techniques:

• First, solve one equation for one variable, such as solving \(-a + b = -1\) for \(b\), leading to \(b = a - 1\).
• Substituting this expression into the second equation \(2a + b = 3\) helps eliminate \(b\).

From there, we simplify to find \(a\), and once \(a\) is known, we substitute back to find \(b\). These steps ensure each variable is precisely determined, and the solution meets all criteria of the original equations. Linear equations are fundamental tools in solving more complex algebraic problems.