1The idea in one paragraph

A polynomial is a sum of whole-number powers of x with constant coefficients, such as 2x³ − 3x² − 11x + 6. Every polynomial can in principle be written as a product of linear factors, one for each root, and the two forms carry the same information. The factor theorem says that (x − a) is a factor exactly when P(a) = 0, so finding a root and finding a factor are the same job. The remainder theorem says that dividing P(x) by (x − a) leaves the remainder P(a), so a remainder is just a value of the function. And when the product of factors is multiplied out, the coefficients turn out to be built from the roots, which is why the sum of the roots and the product of the roots can be read straight off the equation without solving it.

2Polynomials and their graphs

A polynomial of degree n is

P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, with aₙ ≠ 0

The powers are whole numbers 0, 1, 2, …, so 1/x and √x are not allowed. aₙ is the leading coefficient and a₀ is the constant term, which is also the y-intercept, since P(0) = a₀. The formula booklet writes the same thing as a sum, ∑ aᵣxʳ from r = 0 to n.

Where the ends go. For large x the highest power swamps everything else, so the ends of the graph behave like aₙxⁿ. Figure 1 shows the four cases. An odd degree sends the two ends in opposite directions; an even degree sends them the same way. A positive leading coefficient sends the right-hand end up.

How many roots and turns. A polynomial of degree n has at most n real roots and at most n − 1 turning points. An odd-degree polynomial must cross the x-axis at least once, because its ends are on opposite sides of it. An even-degree polynomial need not cross at all: x⁴ + 1 never does.

Zeros, roots, factors. The zeros of P, the roots of P(x) = 0 and the x-intercepts of y = P(x) are the same numbers. Each real root a gives a linear factor (x − a). If a factor appears twice, (x − a)², the root a is a double root (a repeated root, of multiplicity 2); three times, a triple root.

The multiplicity decides what the graph does at the root, as Figure 2 shows:

• Single root: the graph crosses the axis. The factor (x − a) changes sign at a.
• Double root: the graph touches the axis and turns back. (x − a)² is never negative, so there is no change of sign; the root is a turning point on the axis.
• Triple root: the graph crosses and flattens as it does so. (x − a)³ changes sign, but slowly, so the curve has a horizontal tangent at the crossing.

Sketching from factors. Take y = −(x + 2)(x − 1)²(x − 3). Four readings give the whole sketch in Figure 3.

1. Degree: 1 + 2 + 1 = 4, even. Leading coefficient: −1, negative. So both ends go down.
2. Roots: −2 (single, crosses), 1 (double, touches), 3 (single, crosses).
3. y-intercept: x = 0 gives −(2)(1)(−3) = 6.
4. Join them up. Coming in from the bottom left, the curve crosses at −2, rises through (0, 6), comes down to touch at (1, 0), rises again, and crosses down through 3 to the bottom right.

A sketch is marked on shape, intercepts and behaviour at the roots. The exact height of the turning points is not needed unless it is asked for; on Paper 2 the GDC gives it, here about (−1.15, 16.3) and (2.40, 5.17).

3Dividing a polynomial by a linear factor

If you know one factor, dividing it out leaves a polynomial one degree lower, which is often a quadratic you can finish. Take P(x) = 2x³ − 3x² − 11x + 6 and the factor (x − 3). There are three standard methods. They give the same answer, and any of them earns the marks.

Comparing coefficients. The quotient must be a quadratic, so write P(x) = (x − 3)(Ax² + Bx + C) and match terms.

Synthetic division. Write the root 3 and the coefficients. Bring down the first coefficient; multiply by 3 and add to the next; repeat.

The bottom row gives the quotient 2x² + 3x − 2 and, in the last place, the remainder 0.

Long division works exactly like numerical long division, dividing the leading term at each stage; it is slower on the page but the same idea.

So P(x) = (x − 3)(2x − 1)(x + 2), and the roots are 3, ½ and −2.

4The remainder theorem and the factor theorem

Divide any polynomial P(x) by (x − a). The remainder has a lower degree than (x − a), so it is a constant R, and

P(x) = (x − a)·Q(x) + R

Put x = a. The first term vanishes, because it contains (x − a), and what is left is P(a) = R. That is the remainder theorem:

The remainder when P(x) is divided by (x − a) is P(a).

So you never need to divide to find a remainder. Evaluate. For P(x) = x⁴ − 3x² + 2x − 5 divided by (x + 2), the remainder is P(−2) = 16 − 12 − 4 − 5 = −5.

For a divisor (ax − b), the root is x = b/a, so the remainder is P(b/a). Dividing the same P(x) by (2x − 1):

The factor theorem is the case R = 0:

(x − a) is a factor of P(x) if and only if P(a) = 0.

Figure 4 makes both theorems visible. The remainder on dividing by (x − a) is the height of the graph at x = a; a factor is a root, where that height is zero.

Finding a root to start with. On Paper 1 the first root must be found by trial. For a polynomial with integer coefficients and leading coefficient 1, any integer root divides the constant term, so for x³ − 7x − 6 try ±1, ±2, ±3, ±6. With a leading coefficient other than 1, fractions p/q are possible, where p divides a₀ and q divides aₙ; that is how ½ turned up above. Try the small ones first.

5Finding unknown coefficients

The most common Paper 1 question gives you a polynomial with two unknown coefficients and two facts about it. Each fact becomes one equation, and you solve the pair.

Worked example 1. P(x) = x³ + ax² + bx − 6. (x − 2) is a factor of P(x), and the remainder when P(x) is divided by (x + 1) is −12. Find a and b, and hence factorise P(x).

The quadratic x² + 3 has no real roots. So P(x) = 0 has one real root, x = 2, and two complex roots, x = ±i√3.

Two slips account for most lost marks here: substituting x = 1 for the divisor (x + 1), when the root is −1, and forgetting to set the remainder equation equal to −12 rather than 0.

6Complex roots of real polynomials

From 1.14, a polynomial of degree n has exactly n roots when you count complex roots and count each repeated root as often as it is repeated. When the coefficients are all real, the non-real roots come in conjugate pairs: if 2 − i is a root, so is 2 + i. That is because each pair multiplies out to a real quadratic:

Figure 5 shows what this means on a graph. The cubic x³ − 3x + k has three real roots for k = 0, a repeated root at k = 2, and for k = 4 only one crossing; the other two roots have become a complex conjugate pair. A cubic with real coefficients always has at least one real root, because complex roots arrive two at a time.

7The sum and the product of the roots

Multiply out the factor form of a cubic with roots α, β and γ:

Compare with ax³ + bx² + cx + d. The x² coefficients give b = −a(α + β + γ), and the constants give d = −aαβγ. So

The same expansion works for any degree, and the formula booklet gives the general result. For aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀ = 0:

sum of the roots = −aₙ₋₁ ÷ aₙ product of the roots = (−1)ⁿ a₀ ÷ aₙ

In words: the sum comes from the top two coefficients, and the product from the constant and the leading coefficient, with a sign that is + for even degree and − for odd degree. For a quadratic ax² + bx + c = 0 these are the familiar α + β = −b/a and αβ = c/a.

Three rules make the formulas safe to use.

• Count every root, complex ones and repeated ones included. x³ − 2x² + 3x − 6 = 0 has roots 2, i√3 and −i√3. Their sum is 2 = −(−2)/1 and their product is 2 × 3 = 6 = (−1)³(−6)/1. Both check.
• A missing term has coefficient 0. For x⁴ + 3x² − 2x + 7 = 0 the x³ term is missing, so the sum of the four roots is 0, and the product is (+1)(7)/1 = 7.
• Mind the sign for the degree. For 2x⁴ − 5x³ + x − 6 = 0, the sum is −(−5)/2 = 5/2 and the product is (−1)⁴(−6)/2 = −3.

Worked example 2 · a condition on the roots. The equation x³ − 3x² + kx + 12 = 0 has two roots that add up to zero. Find k and all three roots.

Call the roots α, −α and β.

Check: (x + 2)(x − 2)(x − 3) = (x² − 4)(x − 3) = x³ − 3x² − 4x + 12. That is the equation, with k = −4.

Worked example 3 · a complex root given. x³ + px² + qx + 10 = 0, with p and q real, has a root 1 − 2i. Find the other roots and the values of p and q.

The coefficients are real, so 1 + 2i is also a root. Call the third root γ.

The roots are 1 − 2i, 1 + 2i and −2, with p = 0, q = 1. The pair (x² − 2x + 5) came from the pair of roots, the same way as in section 6.

8With technology

Paper 2 may give a polynomial whose roots are not rational, or ask for roots of a quartic. Use the polynomial solver or the zero tool on the graph, one root at a time, and give each to 3 significant figures. The guide expects HL candidates to use technology whenever no analytic method is appropriate. The sum and product still help: if the GDC gives the two real roots of a quartic, the sum and product of all four roots let you build the quadratic whose roots are the other two, as Q5 below does.