Polynomial & Rational Expressions

The one operation polynomials were still missing

Divide $17$ by $5$ and you get $3$ with $2$ left over. Before moving on, notice what that sentence really claims: $17 = 5 \cdot 3 + 2$. Division-with-remainder is a fact about multiplication — the divisor times the quotient, plus a leftover too small to divide again. You can add, subtract, and multiply polynomials already (that’s what makes an expression like $6x^2 + 5x - 4$ yours to rearrange — M1’s whole game). Division is the gear that was missing, and it works exactly like the $17$ and the $5$:

Written back as multiplication, just like the numbers:

That second shape is worth staring at, because the SAT loves it: “which of the following is equivalent to $\frac{6x^2+5x-4}{2x+3}$?” is this division in costume. The fraction of the leftover over the divisor is the tell.

The shortcut hiding in the identity

Now divide by something simpler — a monic $x - a$ — and look at the identity with a detective’s eye:

The remainder $r$ is a constant (it must have smaller degree than $x - a$). And this equation holds for every $x$ — including $x = a$. Set it there and watch the first piece die: $(a - a)\,q(a) = 0$, leaving $p(a) = r$. Read that again: the remainder is just the value of the polynomial at $a$. No division required. Take $p(x) = x^3 - 2x^2 - 5x + 6$ divided by $x - 2$:

$p(2) = (2)^3 - 2(2)^2 - 5(2) + 6 = 8 - 8 - 10 + 6 = -4$

One plug-in, and you know a whole long division’s leftover: remainder $-4$. That is the remainder theorem, and on the SAT it is a speed weapon — the question says “divided by”, the solve is a substitution.

Remainder Dial makes the theorem physical: the remainder is a height.

The dial opens exactly at the division above — the cubic, divided at $a = 2$, bar at $-4$. Predict before you drag: the curve seems to cross the axis near $x = 3$. If it really does, what must the remainder of dividing by $x - 3$ be? Drag the dot there and watch the bar. Then hunt the other two crossings ($x = 1$ and $x = -2$), and finally load never lands: that curve refuses to touch the axis, so no integer $a$ kills the remainder — $p$ has no factor of the form $x - a$ at all.

A zero remainder is a certificate

What you just felt has a name. If $p(a) = 0$, the remainder of dividing by $x - a$ is zero — the division comes out exact, so $x - a$ is a factor of $p(x)$, and the graph crosses the axis at $a$. Three statements, one fact. The SAT’s favorite dress for it is a table:

“Which of the following must be a factor of $p$?” The table whispers it: $p(1) = 0$, so $x - 1$ divides in exactly. Watch the sign — a zero at $x = -4$ certifies the factor $x - (-4) = x + 4$. Plus inside, for a zero on the negative side.

Fractions made of polynomials

Numbers reduce by shared factors: $\dfrac{6}{8}$ is $\dfrac{2 \cdot 3}{2 \cdot 4}$, the $2$s cancel, $\dfrac{3}{4}$ remains. A rational expression is a fraction whose top and bottom are polynomials, and it plays by the same rule — with one extra law the numbers never needed. Here is the SAT’s usual costume:

The points that stay banned — walls and pinholes

Both banned values are undefined, but they fail in two very different ways, and the SAT asks about the difference. At $x = 2$ the reduced bottom $x - 2$ still dies: right next to $2$, the top sits near $-1$ while the bottom is tiny — and a nonzero number divided by a tiny number is enormous. The graph shoots up a vertical asymptote: a wall it climbs forever and never touches. At $x = -2$ the guilty factor cancelled — nearby, the expression behaves exactly like $\frac{x-3}{x-2}$, calmly approaching $\frac{-5}{-4} = \frac{5}{4}$. The graph is a smooth curve with a single point missing: a hole. Cancelling removed the wall. It did not remove the ban.

It opens on the exact fraction above. Predict before you touch: which banned $x$ has the dashed wall, and which the open circle? Now tap the top row’s $(x+2)$ off — the bottom’s $(x+2)$ just lost its cancelling partner, so what must replace the pinhole at $x = -2$? Check, then tap it back on and drag the probe from $x = 1$ toward the wall at $2$ and read the values: the strip keeps halving your distance, and the values keep roughly doubling — no ceiling. Park the probe near the pinhole instead and the same strip calmly settles toward $\frac{5}{4}$. Blow-up versus settle: your hands now know the difference the vocabulary is naming.

The one thing to remember

Division leaves a fingerprint: $p(x) = (x-a)\,q(x) + r$, and setting $x = a$ reads the fingerprint without dividing — $r = p(a)$, with $r = 0$ certifying a factor. Fractions of polynomials reduce like fractions of numbers — by whole factors only — and the denominator’s zeros are banned before the cancelling starts: a surviving factor below is a wall, a cancelled one leaves a pinhole.

Division and its two theorems

Long division: divide the leading terms → multiply back → subtract → repeat. The SAT’s “equivalent expression” form: $\dfrac{p}{d} = q + \dfrac{r}{d}$.

Sign check: a zero at $x = -4$ gives the factor $x + 4$. Zero negative, sign inside positive.

Rational expressions

Simplify: factor top and bottom → note every $x$ that zeroes the bottom → cancel whole shared factors → keep the bans.

Undefined: exactly where the denominator is $0$ — including factors that cancel.

At a hole, the missing value is the reduced expression evaluated there.