Exercises — Rational expressions — simplification, operations

What we do: a fraction is undefined only when its bottom is $0$. So we set each factor of the denominator to zero. $x-4=0\Rightarrow x=4,x+1=0\Rightarrow x=-1$ Why the top doesn't matter: the numerator being $0$ just makes the whole fraction $0$ — that's perfectly allowed. Only a zero denominator breaks the rule "never divide by zero." Answer: undefined at $x=4$ and $x=-1$.

Top: $x^2-9$ is a difference of squares, $a^2-b^2=(a-b)(a+b)$ with $a=x,b=3$: $x^2-9=(x-3)(x+3)$ Bottom: find two numbers multiplying to $12$ and adding to $7$ → $3$ and $4$: $x^2+7x+12=(x+3)(x+4)$ Common factor: the piece appearing in both is $\boxed{(x+3)}$.

False. The $x$ on top is added to $5$, not multiplied. You may only cancel multiplied pieces (factors). Here $x+5$ and $x+2$ are single lumps that cannot be split. Number test: let $x=3$. Then $\frac{3+5}{3+2}=\frac{8}{5}=1.6$, but $\frac{5}{2}=2.5$. Not equal, so the "cancel the $x$'s" move is wrong.

Factor top (difference of squares): $x^2-4=(x-2)(x+2)$. Factor bottom: two numbers multiplying to $-6$, adding to $-1$ → $-3$ and $2$: $x^2-x-6=(x-3)(x+2)$ Restrictions first (from original bottom): $x\ne 3$ and $x\ne -2$. Cancel the shared factor $(x+2)$: $\frac{(x-2)(x+2)}{(x-3)(x+2)}=\frac{x-2}{x-3}$ Answer: $\frac{x-2}{x-3}$, where $x\ne 3,-2$.

Factor everything first (so we can cancel across the two fractions before multiplying): $\frac{(x-5)(x+5)}{x(x+6)}\cdot \frac{x}{x-5}$ Restrictions (every original denominator): $x\ne 0,x\ne -6,x\ne 5$. Write as one fraction and cancel the $x$ and the $(x-5)$: $\frac{(x-5)(x+5)\cdot x}{x(x+6)(x-5)}=\frac{x+5}{x+6}$ Answer: $\frac{x+5}{x+6}$, where $x\ne 0,5,-6$.

Flip and multiply (dividing by a fraction = multiplying by its reciprocal): $\frac{x^2-1}{x+4}\cdot \frac{x^2-16}{x-1}$ Factor: $x^2-1=(x-1)(x+1)$ and $x^2-16=(x-4)(x+4)$: $\frac{(x-1)(x+1)}{x+4}\cdot \frac{(x-4)(x+4)}{x-1}$ Restrictions: the original bottom $x+4$ gives $x\ne -4$; the divisor $\frac{x-1}{x^2-16}$ has bottom $x^2-16$ so $x\ne 4,-4$; and dividing by that fraction needs its top $x-1\ne 0$, so $x\ne 1$. Altogether $x\ne 1,4,-4$. Cancel $(x-1)$ and $(x+4)$: $(x+1)(x-4)=x^2-3x-4$ Answer: $x^2-3x-4$, where $x\ne 1,4,-4$.

LCD = product of the two distinct denominators $=(x-3)(x+2)$. Rewrite each fraction by multiplying by a clever form of $1$: $\frac{4(x+2)}{(x-3)(x+2)}+\frac{5(x-3)}{(x-3)(x+2)}$ Combine numerators: $4(x+2)+5(x-3)=4x+8+5x-15=9x-7$. Answer: $\frac{9x-7}{(x-3)(x+2)}$, where $x\ne 3,-2$.

Restrictions first: original bottom is $(x-3)(x+5)$, so $x\ne 3$ and $x\ne -5$. Cancel $(x-3)$: $\frac{1}{x+5}$. Key insight: even though $\frac{1}{x+5}$ looks fine at $x=3$, the original expression was never defined there. We keep $x\ne 3$. Picture: at $x=3$ the factor cancelled → a hole (a single missing dot). At $x=-5$ the factor $(x+5)$ survives in the bottom → a vertical wall (asymptote). See the figure below. Answer: $\frac{1}{x+5}$, where $x\ne 3,-5$.

Factor: $x^2-9=(x-3)(x+3)$, so the LCD is $(x-3)(x+3)$. $\frac{x}{(x-3)(x+3)}-\frac{3(x+3)}{(x-3)(x+3)}$ Subtract the whole second numerator — wrap it in brackets so the minus hits every term: $\frac{x-3(x+3)}{(x-3)(x+3)}=\frac{x-3x-9}{(x-3)(x+3)}=\frac{-2x-9}{(x-3)(x+3)}$ Answer: $\frac{-2x-9}{(x-3)(x+3)}$, where $x\ne 3,-3$.

Recognise the flip: $3-x=-(x-3)$. This is the classic "opposite of a factor" move. Factor bottom: $(x-3)(x+3)$. $\frac{-(x-3)}{(x-3)(x+3)}$ Restrictions: $x\ne 3,-3$. Cancel $(x-3)$, keeping the minus sign: $\frac{-1}{x+3}$ Answer: $-\frac{1}{x+3}$, where $x\ne 3,-3$.

Factor each of the four polynomials:

• $x^2+5x+6=(x+2)(x+3)$
• $x^2-4=(x-2)(x+2)$
• $x^2-5x+6=(x-2)(x-3)$
• $x^2+6x+9=(x+3{)}^2$ $\frac{(x+2)(x+3)}{(x-2)(x+2)}\cdot \frac{(x-2)(x-3)}{(x+3{)}^2}$ Restrictions: $x\ne 2,-2,-3$. Cancel $(x+2)$, $(x-2)$, and one $(x+3)$: $\frac{x-3}{x+3}$ Answer: $\frac{x-3}{x+3}$, where $x\ne 2,-2,-3$.

Strategy: multiply top and bottom by the LCD of all small fractions, which is $2x$. This clears every mini-fraction at once (multiplying by $\frac{2x}{2x}=1$ changes nothing). Top: $\left(\frac{1}{x}+\frac{1}{2}\right)\cdot 2x=2+x$. Bottom: $\left(\frac{1}{x}-\frac{1}{2}\right)\cdot 2x=2-x$. $\frac{2+x}{2-x}$ Restrictions: $x\ne 0$ (the inner fractions need it), and the bottom $2-x\ne 0\Rightarrow x\ne 2$. Answer: $\frac{x+2}{2-x}$, where $x\ne 0,2$.

Do the division first (flip and multiply): $\frac{(x-2)(x+2)}{(x-3)(x+3)}\cdot \frac{x+3}{x+2}=\frac{x-2}{x-3}$ after cancelling $(x+2)$ and $(x+3)$. Restrictions so far: $x\ne 3,-3,-2$. Now add $\frac{1}{x-3}$ (same denominator $x-3$): $\frac{x-2}{x-3}+\frac{1}{x-3}=\frac{(x-2)+1}{x-3}=\frac{x-1}{x-3}$ Answer: $\frac{x-1}{x-3}$, where $x\ne 3,-3,-2$.

Factor both bottoms: $x^2-x=x(x-1)$ and $x^2-1=(x-1)(x+1)$. LCD: take each distinct factor to its highest power → $x(x-1)(x+1)$. Restrictions: $x\ne 0,x\ne 1,x\ne -1$. Rewrite each fraction over the LCD. First needs an extra $(x+1)$, second needs an extra $x$: $\frac{2(x+1)}{x(x-1)(x+1)}-\frac{3x}{x(x-1)(x+1)}$ Combine (bracket the subtraction): $\frac{2(x+1)-3x}{x(x-1)(x+1)}=\frac{2x+2-3x}{x(x-1)(x+1)}=\frac{-x+2}{x(x-1)(x+1)}$ Answer: $\frac{2-x}{x(x-1)(x+1)}$, where $x\ne 0,1,-1$.

Factor: top $=(x-3)(x+2)$, bottom $=(x-3)(x+3)$. Restrictions from original: $x\ne 3,x\ne -3$. Cancel $(x-3)$: $f(x)=\frac{x+2}{x+3}$. Classify:

• $x=3$: the factor $(x-3)$ cancelled → the discontinuity is a hole. Its height is the simplified value $\frac{3+2}{3+3}=\frac{5}{6}$, a missing point at $\left(3,\frac{5}{6}\right)$.
• $x=-3$: the factor $(x+3)$ survives in the bottom → vertical wall (asymptote), since the bottom → $0$ while the top → $-1\ne 0$. Answer: $f(x)=\frac{x+2}{x+3}$, domain $x\ne 3,-3$; hole at $x=3$, wall at $x=-3$.

Top: $4-x^2=-(x^2-4)=-(x-2)(x+2)$. (Pull the minus out so factors line up.) Bottom: $x^2+x-6=(x+3)(x-2)$. $\frac{-(x-2)(x+2)}{(x+3)(x-2)}$ Restrictions: $x\ne 2,x\ne -3$. Cancel $(x-2)$, keep the minus: $\frac{-(x+2)}{x+3}=\frac{-x-2}{x+3}$ Equals zero: when the numerator $-x-2=0\Rightarrow x=-2$ (and $x=-2$ is allowed, so it's a genuine zero). Undefined: at $x=2$ (hole, cancelled factor) and $x=-3$ (wall). Answer: $-\frac{x+2}{x+3}$; zero at $x=-2$; undefined at $x=2,-3$.

The one habit that fixes 90% of errors here ::: Write down all denominator-zero restrictions before you cancel, and only cancel multiplied factors. Hole vs wall in one sentence ::: A factor that cancels leaves a hole; a factor that survives in the denominator is a wall (vertical asymptote). Difference of squares pattern ::: $a^2-b^2=(a-b)(a+b)$. How to kill a complex fraction fast ::: Multiply top and bottom by the LCD of all the little fractions.