Develop the structural consequences of the factor theorem: distinct zeroes give distinct linear factors and a degree-n polynomial has at most n zeroes; a polynomial agreeing with another at n + 1 points is identical, so a degree-n graph is fixed by n + 1 points; the number of intersections of two curves is bounded by degree via F(x) = P(x) - Q(x); and re-express a polynomial in powers of (x - a)

What this dot point is asking

The factor theorem on the previous page is a single fact: $x - a$ is a factor of $P(x)$ exactly when $P(a) = 0$. This page works out what that one fact forces to be true about every polynomial. Each zero is a factor, so several zeroes mean several factors, and factors take up degree fast, which is why a polynomial cannot have more zeroes than its degree. From there everything else follows: if two polynomials of degree $n$ agree at $n + 1$ points they must be the same polynomial, so a curve is completely fixed by enough points; and two graphs cannot cross more times than their degree allows, a fact you unlock by studying the single difference curve $F(x) = P(x) - Q(x)$.

These are the structural consequences NESA groups under ME-F2, and they are quietly powerful: they turn "find the polynomial" into a short calculation, let you prove two expressions are identical without grinding through algebra, and explain at a glance why a line meets a cubic at most three times. This is the Year 11 first-contact treatment. The applied versions, tangents and chords handled with these ideas, are developed at geometry using polynomials and in the Year 12 page on roots and coefficients; here the goal is to make the structure itself second nature.

The answer

Distinct zeroes give distinct factors

Suppose you have found several different zeroes of a polynomial $P(x)$, say $\alpha_1, \alpha_2, \dots, \alpha_s$, all distinct. The factor theorem makes each one a factor: $x - \alpha_1$ divides $P(x)$, so $P(x) = (x - \alpha_1)P_1(x)$. Now $\alpha_2$ is also a zero, so $P(\alpha_2) = 0$; but $(\alpha_2 - \alpha_1) \ne 0$ because the zeroes are distinct, so the other factor $P_1(\alpha_2)$ must be zero. That makes $x - \alpha_2$ a factor of $P_1(x)$, and continuing this way each distinct zero peels off its own linear factor.

This is exactly why the integer-zero hunt from the previous page pays off so well. Every separate zero you find is a separate factor, so finding several zeroes builds up a quadratic or cubic factor for free and shrinks, or removes, the long division you would otherwise face.

A degree-n polynomial has at most n zeroes

Push the previous idea to its limit. If a polynomial of degree $n$ had $n + 1$ distinct zeroes, then by the rule above it would be divisible by a product of $n + 1$ distinct linear factors, a polynomial of degree $n + 1$. But a degree-$n$ polynomial cannot have a factor of higher degree than itself. That is a contradiction, so the supply of zeroes runs out.

Two consequences are worth pulling out straight away. First, if you can find $n$ distinct zeroes of a degree-$n$ polynomial, you have found all of them and the factoring is complete: no division is required. Second, the only polynomial with more zeroes than its degree, indeed with infinitely many zeroes, is the zero polynomial $Z(x) = 0$, which is zero for every $x$ and is the one polynomial to which the degree bound does not apply.

Finding all the zeroes when they are distinct

The first bound turns into a fast factoring method whenever a polynomial has only simple (distinct) zeroes. Instead of finding one factor, dividing, and repeating, you simply keep testing divisors of the constant term until you have collected as many distinct zeroes as the degree.

1. List the candidates. Write the divisors of the constant term (both signs); these are the only possible integer zeroes.
2. Collect zeroes. Test them with the factor theorem, noting each $a$ for which $P(a) = 0$.
3. Stop at the degree. Once you have $n$ distinct zeroes of a degree-$n$ polynomial, you have them all. The polynomial is $a$ times the product of the linear factors, where $a$ is the leading coefficient.

No long division at all is needed in this best case, which is the whole point of the maximum-zeroes bound. (When a zero is repeated you will fall one factor short and still need a single division to recover the repeated factor, the situation handled on the multiplicity page.)

Agreement at n + 1 points forces identity

The maximum-zeroes bound has a striking restatement. Suppose two polynomials $P(x)$ and $Q(x)$, each of degree at most $n$, take the same value at $n + 1$ different values of $x$. Look at their difference $F(x) = P(x) - Q(x)$. Wherever $P$ and $Q$ agree, $F$ is zero, so $F$ has $n + 1$ zeroes. But $F$ has degree at most $n$, and a polynomial of degree at most $n$ cannot have $n + 1$ zeroes unless it is the zero polynomial. Therefore $F(x) = 0$ for all $x$, which means $P(x) = Q(x)$ for all $x$: the two polynomials are identically equal, and their corresponding coefficients match.

This is the rigorous reason "equating coefficients" is allowed: two polynomial expressions that are equal for enough values must have identical coefficients. It also means you can prove an identity by checking it at enough points rather than expanding both sides, a genuine shortcut.

A graph is fixed by n + 1 points

Read the identity condition geometrically. A polynomial graph of degree $n$ is the graph of a degree-$n$ polynomial, and that polynomial is pinned down by $n + 1$ of its values. So the curve cannot be nudged: once $n + 1$ points are specified, exactly one polynomial of that degree passes through them.

So a parabola is fixed by $3$ points, a cubic by $4$, and a line ($n = 1$) meets a cubic ($n = 3$) at most $3$ times. The cubic below is the one from the first worked example, $y = -3x^3 + 6x^2 + 33x - 36$; its three zeroes and the single point $(2, 30)$ are four points, and they admit no other cubic.

Counting intersections with the difference F(x) = P(x) - Q(x)

The single most useful tool on this page is the difference function $F(x) = P(x) - Q(x)$. Two curves $y = P(x)$ and $y = Q(x)$ meet exactly where $P(x) = Q(x)$, that is, exactly where $F(x) = 0$. So the intersections of the two curves are precisely the zeroes of the one polynomial $F$, and the whole intersection question collapses to factoring $F$.

This immediately bounds the number of intersections: $F$ has degree at most the larger of the two degrees, so it has at most that many zeroes, so the curves meet at most that many times. It also locates them, because factoring $F$ gives every intersection $x$-value at once. Best of all, the nature of each meeting is read off the multiplicity of the corresponding factor of $F$, exactly as a single curve behaves at its own zeroes:

The reason mirrors the single-curve case. A simple factor $(x - \alpha)$ lets $F$ change sign at $\alpha$, so $P - Q$ swaps from positive to negative, meaning the curves swap which one is on top, a crossing. A squared factor $(x - \alpha)^2$ is never negative, so $F$ touches zero at $\alpha$ and returns on the same side; $P - Q$ keeps its sign, so neither curve overtakes the other and they only touch, a tangency.

Take an original pair built so the difference is a clean cubic: let

Both are quartics with the same leading term $x^4$, so the $x^4$ cancels in the difference and

The double factor $(x - 1)^2$ predicts a tangency at $x = 1$, and the simple factor $(x + 2)$ predicts a crossing at $x = -2$. Plotting both curves confirms it exactly.

The clearest way to see why one root touches and the other crosses is to plot the difference curve $y = F(x)$ by itself. Its zeroes are the intersection $x$-values, and how it behaves at each zero, touch or cross, against the $x$-axis is exactly how the two original curves behave at that meeting.

Re-expressing a polynomial in powers of (x - a)

The identity condition also justifies a useful change of viewpoint: any polynomial can be rewritten in powers of $(x - a)$ rather than powers of $x$. For a cubic and $a$ a chosen number, you write

and find the constants $A$, $B$, $C$, $D$. This is allowed because both sides are cubics; if they agree at four points they are identical, so the constants are uniquely determined. It is the algebra behind shifting a curve's "centre" to $x = a$, and it makes the value and local behaviour at $x = a$ transparent: $D = P(a)$ is the height there, and the later terms describe how the curve departs from that point.

The reliable way to find the constants combines two of this page's ideas. Equate the leading coefficients to get $A$ (only the $A(x - a)^3$ term produces $x^3$, so $A$ equals the leading coefficient of $P$). Substitute $x = a$ to get $D$, because every $(x - a)$ term vanishes there, leaving $D = P(a)$. Then substitute two more convenient values to get two equations in $B$ and $C$, and solve. A worked example below carries this out in full.

How exam questions ask about these consequences

The wordings are predictable once mapped to the right consequence.

• "Write down / find the polynomial with zeroes ... (and passing through ...)." Factor theorem in reverse: $P(x) = a\prod(x - \alpha_i)$, then use the extra point to find $a$ and expand.
• "Find all the zeroes" or "factor completely" with several easy zeroes. Test divisors of the constant term; once you have as many distinct zeroes as the degree, stop, and quote "at most $n$ zeroes".
• "Show that two expressions are equal for all $x$" or "find the constants so that ... for all $x$." Identity condition: equate coefficients, or substitute enough values, justified by agreement at $n + 1$ points.
• "A cubic passes through these four points; find it." A degree-$n$ curve is fixed by $n + 1$ points; set up the polynomial and solve.
• "By considering $P(x) - Q(x)$, describe the intersections / show the curves are tangent / find where $P$ is above $Q$." Factor the difference; multiplicity gives touch versus cross; the sign of the difference gives which curve is on top.
• "Express $P(x)$ in powers of $(x - a)$" or "in the form $A(x - a)^3 + \dots$." Equate the leading coefficient for $A$, substitute $x = a$ for $D$, then two more values for $B$ and $C$.
• "Explain why a line meets this cubic at most three times." A line is degree $1$ and the cubic degree $3$; their difference has degree $3$, so at most $3$ zeroes, hence at most $3$ intersections.