Rational Expressions

Product $-6$, sum $1$: $3$ and $-2$.

• Correct: $(x + 3)(x - 2)$:
• $(x - 3)(x + 2)$:
• $(x + 6)(x - 1)$:
• $(x + 3)(x + 2)$:

The denominator is $0$ at $x = 5$.

• Correct: $x = 5$:
• $x = -5$:
• $x = 0$:
• $x = 1$:

$\frac{(x - 3)(x + 3)}{(x + 3)(x - 2)}$$= \frac{x - 3}{x - 2}$ for $x \ne -3$ and $x \ne 2$. The original expression is undefined at both, so both stay excluded.

• $\frac{x - 3}{x - 2}$, with $x \ne 2$: The canceled factor still counts: $x = -3$ makes the original denominator $0$.
• Correct: $\frac{x - 3}{x - 2}$, with $x \ne -3$ and $x \ne 2$: Right: both zeros of the original denominator stay excluded.
• $\frac{-9}{x - 6}$, with $x \ne 6$: Only factors cancel, never terms: $x^2$ cannot be canceled from a sum.
• $\frac{x + 3}{x - 2}$, with $x \ne 2$: $x^2 - 9 = (x - 3)(x + 3)$; the factor that cancels is $x + 3$, leaving $x - 3$.

Common denominator $(x - 1)(x + 1) = x^2 - 1$: $\frac{(x + 1) + 2(x - 1)}{x^2 - 1}$$= \frac{3x - 1}{x^2 - 1}$.

• Answer: $\frac{3x - 1}{x^{2} - 1}$

A fraction is undefined where its denominator is $0$: $x(x - 4) = 0$ at $x = 0$ and $x = 4$. The numerator's zeros do not matter for the domain.

• Correct: $x = 0$ and $x = 4$: Right: $x^2 - 4x = x(x - 4)$ is $0$ at both.
• $x = -2$: At $x = -2$ the numerator is $0$, which is allowed: the value is $0$.
• $x = 4$: $x = 0$ also makes $x(x - 4)$ zero.
• $x = 2$ and $x = -2$: These solve $x^2 - 4 = 0$; the denominator is $x^2 - 4x$.

$\frac{x^2 - 4}{x - 2}$$= \frac{(x - 2)(x + 2)}{x - 2}$$= x$$+ 2$ only when $x \ne 2$. At $x = 2$ the fraction has no value; its graph is the line $y = x + 2$ with that one point missing.

• $x = -2$: Both are $0$ at $x = -2$: $\frac{0}{-4} = 0$ and $-2 + 2 = 0$.
• $x = 4$: At $x = 4$ both give $6$.
• $x = 0$: At $x = 0$ both give $2$.
• Correct: $x = 2$: Right: $\frac{x^2 - 4}{x - 2}$ is undefined at $x = 2$, while $x + 2 = 4$ there.

Cancel only common factors of the whole numerator and denominator: $\frac{x(x + 1)}{x} = x + 1$. Terms of a sum cannot be canceled.

• $\frac{x + 6}{x + 3} = 2$: $x$ and $3$ are terms of a sum, not factors; check $x = 1$: $\frac{7}{4} \ne 2$.
• $\frac{3x + 6}{3} = x + 6$: Divide both terms by $3$: $x + 2$.
• Correct: $\frac{x^2 + x}{x} = x + 1$, for $x \ne 0$: Right: $x$ is a factor of the top, $x(x + 1)$, so it cancels.
• $\frac{x + 1}{x} = 1$, for $x \ne 0$: Check $x = 1$: $\frac{2}{1} = 2$, not $1$.

Write the top around the denominator: $x^2 + 2x + 3$$= (x + 1)^2 + 2$. Then $\frac{(x + 1)^2 + 2}{x + 1}$$= x$$+ 1$$+ \frac{2}{x + 1}$.

• $x + 2 + \frac{3}{x + 1}$: Check $x = 0$: $3$ on the left but $5$ here.
• Correct: $x + 1 + \frac{2}{x + 1}$: Right: $x^2 + 2x + 3$$= (x + 1)^2 + 2$, so divide each part by $x + 1$.
• $x + 3$: Check $x = 0$: $3 = 3$, but at $x = 1$ the left is $3$ and this is $4$.
• $x + 1 + \frac{3}{x + 1}$: $(x + 1)^2 = x^2 + 2x + 1$, so the leftover is $2$, not $3$.