Question

find the sum or difference.
\\((4rxt - 8r^2x + x^2) - (6rx^2 + 5rxt - 2x^2)\\)
\\(square x^2 + square rxt + square r^2x + square rx^2\\)

Explanation:

Step1: Distribute the negative sign

First, we distribute the negative sign to each term in the second polynomial:
$$(4rxt - 8r^2x + x^2) - 6rx^2 - 5rxt + 2x^2$$

Step2: Combine like terms for \(x^2\) terms

The \(x^2\) terms are \(x^2\) and \(2x^2\). Combining them:
$$x^2 + 2x^2 = 3x^2$$

Step3: Combine like terms for \(rxt\) terms

The \(rxt\) terms are \(4rxt\) and \(-5rxt\). Combining them:
$$4rxt - 5rxt = -rxt$$

Step4: Combine like terms for \(r^2x\) terms

The only \(r^2x\) term is \(-8r^2x\), so it remains:
$$-8r^2x$$

Step5: Combine like terms for \(rx^2\) terms

The only \(rx^2\) term is \(-6rx^2\), so it remains:
$$-6rx^2$$

Answer:

For the \(x^2\) coefficient: \(3\)
For the \(rxt\) coefficient: \(-1\)
For the \(r^2x\) coefficient: \(-8\)
For the \(rx^2\) coefficient: \(-6\)

So the filled form would be:
\(3x^2 + (-1)rxt + (-8)r^2x + (-6)rx^2\)