EXAM TWO fresh from today

Polynomial & Rational Inequalities

You came into today sure these were going to sink you, and you left saying it finally clicked — one of your strongest sessions in a while. This page is for the next time you're doing them alone and that old voice creeps back. It's one method, every single time: find the zeros, split the number line into regions, and test each region for nothing but its sign. You're not solving. You're just hunting for a plus or a minus.

‹ Back to Exam Two

Spot the type

Which Problem Is This?

Type 1 — A polynomial inequality. Something like $6x^2-13x<28$ or $6x^3-6x^2-12x<0$ — a quadratic or a cubic compared to a number. Get everything to one side, factor, and the zeros are your boundary points.

Type 2 — A rational inequality. A fraction compared to $0$, like $\dfrac{x+10}{x-10}\le 0$. Same method — but now the bottom gives you boundary points too, and those are special: they're never allowed in the answer.

Both types run the exact same four moves. Learn it once on the polynomials, add one rule for the fractions, and you're done.

The method "you're not solving — you're just getting signs" — the whole trick

Zeros → Number Line → Test the Regions

1 · Get everything to one side. Move it all over so the other side is $0$. $6x^2-13x<28$ becomes $6x^2-13x-28<0$. Now $0$ is the thing you're comparing to.

2 · Factor to find the zeros. The zeros are where the expression equals $0$ — the boundary points where the sign can flip. For a fraction, get zeros from the top AND the points that make the bottom zero.

3 · Set up the number line. Mark the zeros. They chop the line into regions. Draw it out — it makes it easier.

4 · Test each region for the sign only. Pick one easy number from a region ($-10$, $0$, $10$ are your friends) and check whether the expression comes out $+$ or $-$. Don't finish the arithmetic — you only care about the sign. Keep the regions whose sign matches the inequality.

The sign shortcut — count the negatives. You don't multiply anything out. Look at each factor, decide if it's $+$ or $-$, and count how many are negative. An even number of negatives → the whole thing is $+$. An odd number → it's $-$. That's the "I don't care what the answer is, I only care if it's positive or negative" move.

Then hand back the right bracket. This is the part people drop.

• Strict ($<$ or $>$) → parentheses. "They gave us less-than-but-not-equal, so we give them back parentheses."

• Not strict ($\le$ or $\ge$) → put brackets on the zeros that came from the top (or from a plain polynomial). Those points are allowed in.

• A zero from the bottom of a fraction is ALWAYS a parenthesis — no matter the sign. More on that below.

filled = included = bracket hollow = not included = parenthesis from the bottom = always hollow

Type 1 · Polynomial Inequalities

Quadratics and cubics. Get it to one side, factor, and run the number line. The only thing that changes between a quadratic and a cubic is the number of regions — the method is identical.

Worked Example — a quadratic

$$\text{Solve}\quad 6x^2 - 13x < 28$$

1 · Everything to one side. $6x^2 - 13x - 28 < 0$. Bring the $28$ over first — you can't find zeros until one side is $0$.

2 · Factor. Two numbers that multiply to $6\cdot(-28)=-168$ and add to $-13$: that's $-21$ and $+8$. So $6x^2-13x-28=(3x+4)(2x-7)$.

3 · Zeros. $3x+4=0\Rightarrow x=-\tfrac{4}{3}$, and $2x-7=0\Rightarrow x=\tfrac{7}{2}$. Two boundary points, three regions.

4 · Test each region for sign.

$-\infty$ + $-\tfrac{4}{3}$ − $\tfrac{7}{2}$ + $\infty$

Left ($x=-10$): $(3(-10)+4)(2(-10)-7)=(-)(-)$ → two negatives → $\mathbf{+}$.

Middle ($x=0$): $(3(0)+4)(2(0)-7)=(+)(-)$ → one negative → $\mathbf{-}$. ✓ matches $<0$.

Right ($x=10$): $(+)(+)$ → zero negatives → $\mathbf{+}$.

5 · Answer. Only the middle region is negative, and $<$ is strict, so parentheses on both ends:

$$\left(-\tfrac{4}{3},\ \tfrac{7}{2}\right)$$

Worked Example — a cubic

$$\text{Solve}\quad 6x^3 - 6x^2 - 12x < 0$$

1 · Already on one side. Pull out the common $6x$: $\;6x(x^2 - x - 2)<0$.

2 · Factor the rest. $x^2-x-2=(x-2)(x+1)$, so $\;6x(x-2)(x+1)<0$.

3 · Zeros. $x=0,\ x=2,\ x=-1$. Three boundary points, four regions.

Here's the "$0$" trap you fixed today. $0$ is a boundary point on this one — but only because $6x$ is a real factor, so $x=0$ genuinely makes the expression zero. That's different from plugging in $x=0$ to test a region. Mark $0$ on the line when it's a factor; use it as a test number when it isn't. Two different jobs.

$-\infty$ − $-1$ + $0$ − $2$ + $\infty$

Test $x=-2$: $6(-)(-)(-)$ → three negatives → $\mathbf{-}$. ✓

Test $x=-\tfrac12$: $6(-)(-)(+)$ → two negatives → $\mathbf{+}$.

Test $x=1$: $6(+)(-)(+)$ → one negative → $\mathbf{-}$. ✓

Test $x=10$: $6(+)(+)(+)$ → $\mathbf{+}$.

4 · Answer. Two negative regions, strict $<$, so parentheses. Two separate stretches → join them with $\cup$:

$$(-\infty,\ -1)\ \cup\ (0,\ 2)$$

Now you

Practice — Polynomial Inequalities

Work each on paper: one side → factor → number line → test for sign → hand back the right bracket. Reveal to check. Watch problems 1 and 2 — same expression, different sign.

1 · $x^2 - 2x - 8 < 0$

Check it ›

$(x-4)(x+2)<0$. Zeros $-2,4$. Middle region is negative, strict $<$ → parentheses. $\;(-2,\ 4)$.

2 · Same expression, flipped sign. $x^2 - 2x - 8 \ge 0$

Check it ›

Same zeros $-2,4$ — but now you keep the positive outer regions, and $\ge$ lets the zeros in, so brackets. $\;(-\infty,\ -2]\ \cup\ [4,\ \infty)$. Notice: only the brackets changed.

3 · Non-monic, fractional zero. $2x^2 - 5x - 3 \ge 0$

Check it ›

$(2x+1)(x-3)\ge 0$. Zeros $-\tfrac12,\ 3$. Keep the positive outer regions; $\ge$ → brackets. $\;\left(-\infty,\ -\tfrac{1}{2}\right]\ \cup\ [3,\ \infty)$.

4 · Bring the number over first. $x^2 + 3x < 10$

Check it ›

$x^2+3x-10<0 \Rightarrow (x+5)(x-2)<0$. Zeros $-5,2$. Middle region negative, strict → parentheses. $\;(-5,\ 2)$.

5 · Cubic — pull out the common factor. $2x^3 - 8x > 0$

Check it ›

$2x(x^2-4)=2x(x-2)(x+2)>0$. Zeros $-2,0,2$. Test the four regions: keep the positive ones, strict $>$ → parentheses. $\;(-2,\ 0)\ \cup\ (2,\ \infty)$.

The Only New Thing in a Fraction

Watch the same top with and without a bottom. Both use the identical sign method — the denominator just adds one boundary point that can never be in the answer, and it changes the whole shape.

$(x+6)(x-2)\le 0 \;\Rightarrow\; [-6,\ 2]$ — a clean closed interval.

$\dfrac{(x+6)(x-2)}{x+1}\le 0 \;\Rightarrow\; (-\infty,\ -6]\ \cup\ (-1,\ 2]$ — the $x=-1$ from the bottom punches a hole and splits it in two. Bracket on the top zeros ($-6,2$), parenthesis on the bottom zero ($-1$).

Type 2 · Rational Inequalities

"The first time I can see a fraction and not care anymore." Same four moves. The top gives zeros like always. The bottom gives boundary points too — but those are places the expression is undefined, so they get a parenthesis every time, even when the sign is $\le$ or $\ge$.

The one rule to burn in: a zero from the bottom is always excluded — always a parenthesis, no matter what the inequality sign is. You can't divide by zero, so that $x$-value can never be an answer. A zero from the top plays by the normal rule: bracket if $\le/\ge$, parenthesis if $</>$.

Worked Example — one bracket, one parenthesis

$$\text{Solve}\quad \frac{x+10}{x-10}\le 0$$

1 · Already a fraction vs. $0$. Good to go.

2 · Boundary points. Top zero: $x+10=0\Rightarrow x=-10$ (a real zero — allowed in). Bottom zero: $x-10=0\Rightarrow x=10$ (undefined — never allowed).

3 · Number line + test for sign.

$-\infty$ + $-10$ − $10$undef + $\infty$

$x=-20$: $\dfrac{(-)}{(-)}$ → two negatives → $\mathbf{+}$.

$x=0$: $\dfrac{(+)}{(-)}$ → one negative → $\mathbf{-}$. ✓ matches $\le 0$.

$x=20$: $\dfrac{(+)}{(+)}$ → $\mathbf{+}$.

4 · Answer. Middle region. The top zero $-10$ is allowed and $\le$ → bracket. The bottom zero $10$ is always a parenthesis:

$$[-10,\ 10)$$

Worked Example — a product on top

$$\text{Solve}\quad \frac{(x+8)(x-1)}{x-4}\ge 0$$

1 · Boundary points. Top zeros: $x=-8$ and $x=1$ (allowed in). Bottom zero: $x=4$ (undefined — out). Three boundaries, four regions.

2 · Test each region for sign — count the negatives across all three factors.

$-\infty$ − $-8$ + $1$ − $4$undef + $\infty$

$x=-10$: $\dfrac{(-)(-)}{(-)}$ → three negatives → $\mathbf{-}$.

$x=0$: $\dfrac{(+)(-)}{(-)}$ → two negatives → $\mathbf{+}$. ✓

$x=2$: $\dfrac{(+)(+)}{(-)}$ → one negative → $\mathbf{-}$.

$x=10$: $\dfrac{(+)(+)}{(+)}$ → $\mathbf{+}$. ✓

3 · Answer. Keep the two positive regions. Top zeros $-8,1$ with $\ge$ → brackets; bottom zero $4$ → parenthesis:

$$[-8,\ 1]\ \cup\ (4,\ \infty)$$

Now you

Practice — Rational Inequalities

Mark the top zeros and the bottom zeros differently before you start. Bottom zeros are always parentheses. Reveal to check.

1 · $\dfrac{x-2}{x+3}\le 0$

Check it ›

Top zero $2$ (allowed, $\le$ → bracket), bottom zero $-3$ (undefined → parenthesis). Middle region is negative. $\;(-3,\ 2]$.

2 · $\dfrac{x+5}{x-2}>0$

Check it ›

Top zero $-5$, bottom zero $2$. Outer regions are positive; strict $>$ → parentheses (and the bottom zero would be one anyway). $\;(-\infty,\ -5)\ \cup\ (2,\ \infty)$.

3 · Product on top. $\dfrac{(x+3)(x-2)}{x+1}<0$

Check it ›

Top zeros $-3,2$; bottom zero $-1$. Boundaries $-3,-1,2$. Negative regions are the far left and the one just right of $-1$; strict $<$ → all parentheses. $\;(-\infty,\ -3)\ \cup\ (-1,\ 2)$.

4 · Product on top, not strict. $\dfrac{(x-4)(x+1)}{x-2}\ge 0$

Check it ›

Top zeros $4,-1$ (brackets, $\ge$); bottom zero $2$ (parenthesis). Boundaries $-1,2,4$. Positive regions: $(-1,2)$ and right of $4$. $\;[-1,\ 2)\ \cup\ [4,\ \infty)$.

Mixed review

No Labels This Time

Polynomial or rational? Strict or not? Name it, name your first move, then solve. Reveal to check.

1 · $x^2 - 9 \le 0$

Check it ›

Polynomial. $(x-3)(x+3)\le 0$. Zeros $\pm3$, middle region negative, $\le$ → brackets. $\;[-3,\ 3]$.

2 · $\dfrac{x+7}{x-1}\ge 0$

Check it ›

Rational. Top zero $-7$ (bracket), bottom zero $1$ (parenthesis). Outer regions positive. $\;(-\infty,\ -7]\ \cup\ (1,\ \infty)$.

3 · $x^3 - 4x > 0$

Check it ›

Polynomial. $x(x-2)(x+2)>0$. Zeros $-2,0,2$. Positive regions are $(-2,0)$ and right of $2$; strict → parentheses. $\;(-2,\ 0)\ \cup\ (2,\ \infty)$.

4 · $\dfrac{(x-5)(x+2)}{x-1}\le 0$

Check it ›

Rational. Top zeros $5,-2$ (brackets); bottom zero $1$ (parenthesis). Boundaries $-2,1,5$. Negative regions: far left and $(1,5)$. $\;(-\infty,\ -2]\ \cup\ (1,\ 5]$.

This one has it all

Capstone — Every Rule at Once

A fraction, a product on top, four regions, and a not-strict sign — so it forces every decision: which regions to keep, brackets on the top zeros, a parenthesis on the bottom zero.

$$\frac{(x-1)(x+4)}{x-3}\le 0$$

1 · Top zeros $x=1,\ x=-4$ (allowed, $\le$ → brackets). Bottom zero $x=3$ (undefined → parenthesis).

2 · Boundaries in order: $-4,\ 1,\ 3$. Four regions — test each for sign.

$-\infty$ − $-4$ + $1$ − $3$undef + $\infty$

Check it ›

$x=-10:\dfrac{(-)(-)}{(-)}=\mathbf{-}$ ✓ · $x=0:\dfrac{(-)(+)}{(-)}=\mathbf{+}$ · $x=2:\dfrac{(+)(+)}{(-)}=\mathbf{-}$ ✓ · $x=10:\dfrac{(+)(+)}{(+)}=\mathbf{+}$

Keep the two negative regions. Top zeros $-4,1$ → brackets; bottom zero $3$ → parenthesis:

$(-\infty,\ -4]\ \cup\ [1,\ 3)$

Now you

Practice — Capstone Level

No scaffolding. Run all four moves and mind every bracket. Reveal to check.

1 · $\dfrac{(x+2)(x-5)}{x+1}\ge 0$

Check it ›

Top zeros $-2,5$ (brackets); bottom zero $-1$ (parenthesis). Boundaries $-2,-1,5$. Positive regions: $(-2,-1)$ and right of $5$. $\;[-2,\ -1)\ \cup\ [5,\ \infty)$.

2 · Cubic, not strict. $3x^3 - 12x \ge 0$

Check it ›

$3x(x-2)(x+2)\ge 0$. Zeros $-2,0,2$ — all from a polynomial, so $\ge$ → brackets. Positive regions: $[-2,0]$ and right of $2$. $\;[-2,\ 0]\ \cup\ [2,\ \infty)$.

3 · Bring it to one side first. $6x^2 < 7x + 3$

Check it ›

$6x^2-7x-3<0 \Rightarrow (3x+1)(2x-3)<0$. Zeros $-\tfrac13,\ \tfrac32$. Middle region is negative, strict $<$ → parentheses. $\;\left(-\tfrac{1}{3},\ \tfrac{3}{2}\right)$.

4 · Product on top, strict. $\dfrac{(x-3)(x+5)}{x-1}<0$

Check it ›

Top zeros $3,-5$; bottom zero $1$. Boundaries $-5,1,3$. Negative regions: far left and $(1,3)$; strict $<$ → all parentheses. $\;(-\infty,\ -5)\ \cup\ (1,\ 3)$.

5 · Product on top, not strict. $\dfrac{(x+2)(x-4)}{x+6}\le 0$

Check it ›

Top zeros $-2,4$ (brackets, $\le$); bottom zero $-6$ (parenthesis). Boundaries $-6,-2,4$. Negative regions: far left and $[-2,4]$. $\;(-\infty,\ -6)\ \cup\ [-2,\ 4]$.

6 · Cubic — pull out the common factor. $2x^3 - 2x^2 - 12x > 0$

Check it ›

$2x(x^2-x-6)=2x(x-3)(x+2)>0$. Zeros $-2,0,3$. Positive regions: $(-2,0)$ and right of $3$; strict $>$ → parentheses. $\;(-2,\ 0)\ \cup\ (3,\ \infty)$.

7 · Two undefined points on the bottom. $\dfrac{x-1}{(x+2)(x-4)}\ge 0$

Check it ›

Top zero $1$ (bracket, $\ge$). Both $-2$ and $4$ are bottom zeros — both parentheses, both left out. Boundaries $-2,1,4$. Positive regions: $(-2,1]$ and right of $4$. $\;(-2,\ 1]\ \cup\ (4,\ \infty)$.

Before You Leave — Your Checklist Card

The card you said you'd build. Copy these lines onto your cheat sheet, in your order:

① What is it asking? $>,\ <,\ \ge,\ \le$ — this decides parenthesis vs. bracket before you even start.

② Everything to one side, factor, find the zeros. Then number line.

③ Plug in, get the sign, does it match? Count the negatives — even is $+$, odd is $-$. Don't finish the arithmetic.

Then the three checks that catch your own slips before you hand it in:

✓ Negatives — did you keep every negative sign through the factoring?

✓ Is $0$ really a zero? Only put it on the line if it's an actual factor — otherwise it's just a test number.

✓ Bottom zeros — always a parenthesis, always left out, even with $\le$ or $\ge$.

‹ Back to Exam Two