Chapter 4 — Polynomials

An algebraic expression in one variable x is called a polynomial if the powers of x that appear are all non-negative integers (0, 1, 2, 3, ...) and the coefficients are real numbers. So 4x + 2, 2y² − 3y + 4, 5x³ − 4x² + x − 2 and 7 are all polynomials. But 1/x = x⁻¹, √x = x^1/2, x + 1/x and 3/(x − 2) are not polynomials because they involve negative or fractional powers, or a variable in the denominator.

Each part of a polynomial separated by + or − signs is a term. In 5x³ − 4x² + x − 2 the terms are 5x³, −4x², x and −2, and the coefficients are 5, −4, 1 and −2. The term without x, here −2, is the constant term. A polynomial in x is usually written p(x), q(x), and so on, so that its value at a number can be written p(2) or q(−1).

The degree of a polynomial is the highest power of the variable that appears with a non-zero coefficient. 4x + 2 has degree 1; 2y² − 3y + 4 has degree 2; 5x³ − 4x² + x − 2 has degree 3; the constant 7 has degree 0 because 7 = 7x⁰. The zero polynomial 0 has no degree defined. Take care when a polynomial is not written in descending powers: 3 − x⁴ + 2x has degree 4.

Polynomials are named by degree. A linear polynomial has degree 1 and general form ax + b with a ≠ 0. A quadratic polynomial has degree 2 and general form ax² + bx + c with a ≠ 0 (the word comes from the Latin quadratus, square). A cubic polynomial has degree 3 and general form ax³ + bx² + cx + d with a ≠ 0. The condition a ≠ 0 is essential: if a = 0 in ax² + bx + c the expression is only linear.

Polynomials are also named by the number of terms: a monomial has one term (5x²), a binomial two (x + 3), a trinomial three (x² + 2x + 1).

A polynomial may have any variable letter: 3t² − t + 1 is a quadratic in t, and 2u³ + u is a cubic in u. Polynomials in two variables, such as x² + xy + y², exist but this chapter deals only with polynomials in one variable.

Two more facts are used constantly. Two polynomials are equal if and only if the coefficients of like powers are equal; this is how unknown coefficients are found by comparing. And the degree of a product is the sum of the degrees: (x + 1)(x² + 2) has degree 3.

If p(x) is a polynomial and k is a real number, the number obtained by replacing x by k in p(x) is called the value of p(x) at k and written p(k). For p(x) = x² − 3x − 4: p(2) = 4 − 6 − 4 = −6; p(0) = −4; p(−1) = 1 + 3 − 4 = 0; p(4) = 16 − 12 − 4 = 0.

A real number k is called a zero of the polynomial p(x) if p(k) = 0. From the calculations above, −1 and 4 are zeroes of x² − 3x − 4, while 2 and 0 are not. Checking whether a given number is a zero is therefore a matter of substitution.

Zero of a linear polynomial. For p(x) = ax + b with a ≠ 0, setting ax + b = 0 gives x = −b/a. So a linear polynomial has exactly one zero, equal to minus the constant term divided by the coefficient of x. The zero of 2x + 3 is −3/2; of 5x − 10 is 2; of x + 7 is −7.

Zeroes and factors. If k is a zero of p(x), then (x − k) is a factor of p(x), and conversely. For x² − 3x − 4 = (x − 4)(x + 1), the factors x − 4 and x + 1 correspond to the zeroes 4 and −1. This link is the factor theorem from Class 9 and is the reason factorisation finds zeroes.

Number of zeroes. A polynomial of degree n has at most n real zeroes. So a linear polynomial has exactly one, a quadratic at most two, a cubic at most three. A quadratic may have fewer than two: x² − 4x + 4 = (x − 2)² has only the zero 2 (counted twice), and x² + 1 has no real zero because x² + 1 is never zero for a real x. The geometric reason is given in the next section.

Zeroes of a constant polynomial. A non-zero constant such as 7 has no zero, since 7 is never 0. The zero polynomial, on the other hand, has every real number as a zero, which is why its degree is left undefined.

Working examples. Is 1 a zero of x³ − 1? p(1) = 1 − 1 = 0, yes. Is −2 a zero of x² + 4? p(−2) = 4 + 4 = 8, no. Is 1/2 a zero of 2x² − 3x + 1? p(1/2) = 2(1/4) − 3/2 + 1 = 1/2 − 3/2 + 1 = 0, yes. Is 3 a zero of x² − 9? p(3) = 0, yes, and so is −3.

Care with signs. Substituting a negative number needs brackets: for p(x) = x² − 2x, p(−3) = (−3)² − 2(−3) = 9 + 6 = 15, not 9 − 6. Most substitution errors in the examination come from dropping these brackets.

The zeroes of a polynomial have a picture. If we draw the graph of y = p(x), the zeroes are the x-coordinates of the points where the graph meets the x-axis, because on the x-axis y = 0, that is, p(x) = 0.

Graph of a linear polynomial. For p(x) = ax + b, the graph of y = ax + b is a straight line. To draw it, choose two or three values of x, find y, plot the points and join them with a ruler. For y = 2x + 3: at x = −2, y = −1; at x = 0, y = 3; at x = 2, y = 7. The line through (−2, −1), (0, 3), (2, 7) crosses the x-axis at x = −3/2, which is exactly the zero of 2x + 3 found by algebra. The line crosses the y-axis at (0, 3), the constant term.

A straight line that is not horizontal crosses the x-axis at exactly one point, which is why a linear polynomial has exactly one zero. The sign of a decides the direction: if a > 0 the line rises to the right; if a < 0 it falls to the right. For y = −x + 2 the line falls and crosses the x-axis at x = 2.

A worked graph. Draw y = x − 3 and find the zero. Table: x = 0, y = −3; x = 3, y = 0; x = 5, y = 2. Plot (0, −3), (3, 0), (5, 2) and join. The line meets the x-axis at (3, 0), so the zero is 3. Check: p(3) = 3 − 3 = 0.

Reading a graph. If a graph of a linear polynomial is given and passes through (−4, 0), the zero is −4. If it passes through (0, 5) and (5, 0), then the polynomial is y = −x + 5, since the slope is (0 − 5)/(5 − 0) = −1 and the y-intercept is 5; its zero is 5.

Special case. If a = 0, the expression is the constant b and the graph y = b is a horizontal line. If b ≠ 0 the line never meets the x-axis, so there is no zero, matching the algebra. If b = 0 the line is the x-axis itself, and every x is a zero, which is the zero polynomial.

Examination questions ask you to draw the graph of a given linear polynomial on graph paper, mark the zero, and confirm it algebraically. Use a scale that shows the x-intercept clearly, label the axes, write the table of values beside the graph, and state the zero in a sentence.

The graph of a quadratic polynomial y = ax² + bx + c is a U-shaped curve called a parabola. If a > 0 the parabola opens upwards (like a cup); if a < 0 it opens downwards (like a cap). The lowest point of an upward parabola, or the highest point of a downward one, is called the vertex. The parabola is symmetric about the vertical line through its vertex.

Drawing y = x² − 3x − 4. Make a table: x = −2, y = 6; x = −1, y = 0; x = 0, y = −4; x = 1, y = −6; x = 2, y = −6; x = 3, y = −4; x = 4, y = 0; x = 5, y = 6. Plot and join with a smooth curve. The curve crosses the x-axis at x = −1 and x = 4, which are the zeroes. The vertex is at x = 3/2, midway between the zeroes, where y = −25/4.

Three cases. A parabola can meet the x-axis in three ways, and this gives the three cases for the zeroes of a quadratic. Case 1: the parabola cuts the x-axis at two distinct points A and A'. Then the quadratic has two distinct real zeroes, the x-coordinates of A and A'. Example: x² − 3x − 4. Case 2: the parabola touches the x-axis at exactly one point (the vertex lies on the axis). Then the two zeroes coincide and the quadratic has one real zero, counted twice. Example: x² − 4x + 4 = (x − 2)², which touches at x = 2. Case 3: the parabola lies entirely above or entirely below the x-axis and never meets it. Then the quadratic has no real zeroes. Example: x² + 1, whose lowest point is (0, 1), or −x² − 2, whose highest point is (0, −2).

Since a parabola can meet a line in at most two points, a quadratic polynomial has at most two zeroes. This is the geometric explanation of the algebraic fact.

Drawing y = −x² + 2x + 8. Table: x = −2, y = 0; x = −1, y = 5; x = 0, y = 8; x = 1, y = 9; x = 2, y = 8; x = 3, y = 5; x = 4, y = 0. The parabola opens downwards, has vertex (1, 9), and crosses the x-axis at x = −2 and x = 4, the zeroes. Check: −x² + 2x + 8 = −(x² − 2x − 8) = −(x − 4)(x + 2).

Reading zeroes from a graph. If a graph is given and cuts the x-axis at (−3, 0) and (1, 0), the zeroes are −3 and 1. If it touches at (2, 0), the only zero is 2. If it stays above the axis, there are no zeroes. Board questions frequently show a picture and ask for the number of zeroes; count the meeting points with the x-axis, not with the y-axis.

When drawing on graph paper, use at least seven points including the vertex, choose the scale so that both zeroes fall on the sheet, draw a smooth curve without straight segments, and state the zeroes in a sentence.

The graph of a cubic polynomial y = ax³ + bx² + cx + d is a curve that rises at one end and falls at the other, usually with one hump and one dip in between. If a > 0 it comes from the bottom left and goes to the top right; if a < 0 the reverse. Because of this shape, the graph must cross the x-axis at least once, so every cubic has at least one real zero; and it can cross at most three times, so a cubic has at most three real zeroes.

Drawing y = x³ − 4x. Table: x = −2, y = 0; x = −1, y = 3; x = 0, y = 0; x = 1, y = −3; x = 2, y = 0. Also x = −3, y = −15 and x = 3, y = 15 to show the ends. The curve crosses the x-axis at −2, 0 and 2: three zeroes. Algebraically, x³ − 4x = x(x − 2)(x + 2), confirming them.

Drawing y = x³. Table: x = −2, y = −8; x = −1, y = −1; x = 0, y = 0; x = 1, y = 1; x = 2, y = 8. The curve passes through the origin and flattens there; it meets the x-axis only at x = 0. So x³ has one zero (counted three times).

Drawing y = x³ − x². Table: x = −1, y = −2; x = 0, y = 0; x = 1/2, y = −1/8; x = 1, y = 0; x = 2, y = 4. It meets the x-axis at 0 (touching) and at 1 (crossing). Algebraically x³ − x² = x²(x − 1), so the zeroes are 0 and 1, with 0 repeated.

General principle. For a polynomial of degree n, the graph meets the x-axis at most n times, so there are at most n zeroes. The number of zeroes equals the number of distinct meeting points. This holds for every degree, and the examination tests it by showing a graph and asking how many zeroes the polynomial has: count the distinct points where the curve meets or touches the x-axis.

Reading graphs. A curve crossing the x-axis at −1, 1 and 3 belongs to a cubic with zeroes −1, 1, 3, for example (x + 1)(x − 1)(x − 3). A curve that crosses once and never returns to the axis belongs to a cubic with one real zero, such as x³ + x. A wavy curve crossing the axis four times cannot be a cubic; it would need degree at least 4.

A practical note on drawing cubics: the values grow quickly, so choose a compressed scale on the y-axis (for example 1 cm = 5 units) and a normal scale on the x-axis, and take points close together near the zeroes to show the crossings clearly.

To find the zeroes of a quadratic polynomial ax² + bx + c, factorise it into two linear factors and set each to zero. The method of splitting the middle term is used: find two numbers whose product is a × c and whose sum is b, rewrite bx as the sum of two terms using these numbers, and factor by grouping.

Example 1. x² + 7x + 10. Here a × c = 10 and b = 7; the numbers are 2 and 5. So x² + 2x + 5x + 10 = x(x + 2) + 5(x + 2) = (x + 2)(x + 5). Zeroes: x + 2 = 0 gives −2, x + 5 = 0 gives −5.

Example 2. x² − 2x − 8. a × c = −8, b = −2; the numbers are −4 and 2. x² − 4x + 2x − 8 = x(x − 4) + 2(x − 4) = (x − 4)(x + 2). Zeroes: 4 and −2.

Example 3. 3x² − x − 4. a × c = −12, b = −1; the numbers are −4 and 3. 3x² − 4x + 3x − 4 = x(3x − 4) + 1(3x − 4) = (3x − 4)(x + 1). Zeroes: 4/3 and −1.

Example 4. 4s² − 4s + 1. a × c = 4, b = −4; the numbers are −2 and −2. 4s² − 2s − 2s + 1 = 2s(2s − 1) − 1(2s − 1) = (2s − 1)². Only one zero, 1/2, repeated.

Example 5. x² − 3. This is a difference of squares: x² − (√3)² = (x − √3)(x + √3). Zeroes: √3 and −√3. The zeroes need not be rational.

Example 6. 6x² − 7x − 3. a × c = −18, b = −7; the numbers are −9 and 2. 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (2x − 3)(3x + 1). Zeroes: 3/2 and −1/3.

Example 7. x² + 4. There are no two real numbers with product 4 and sum 0, and x² + 4 ≥ 4 for every real x, so there is no real zero.

Verification. After finding zeroes, substitute back. For 3x² − x − 4 at x = 4/3: 3(16/9) − 4/3 − 4 = 16/3 − 4/3 − 12/3 = 0. At x = −1: 3 + 1 − 4 = 0. Both check.

A note on the leading coefficient. When a ≠ 1, the zeroes are the values that make each linear factor zero, so the factor 3x − 4 gives the zero 4/3, not 4. Also, factorising by taking a common factor first can simplify: 2x² − 8 = 2(x² − 4) = 2(x − 2)(x + 2), zeroes ±2; the constant 2 does not affect the zeroes.

The examination asks for zeroes of a given quadratic followed by verification of the relations with the coefficients, which is the subject of the next section. Practise until splitting the middle term takes seconds; it is also the main tool of the chapter on quadratic equations.

Let α and β (alpha and beta) be the zeroes of the quadratic polynomial p(x) = ax² + bx + c, a ≠ 0. Then (x − α) and (x − β) are factors, so ax² + bx + c = k(x − α)(x − β) for some constant k. Expanding the right side, k[x² − (α + β)x + αβ] = kx² − k(α + β)x + kαβ. Comparing coefficients with ax² + bx + c: k = a, −k(α + β) = b, kαβ = c. Hence:

Sum of zeroes: α + β = −b/a. Product of zeroes: αβ = c/a.

In words, the sum of the zeroes is minus the coefficient of x divided by the coefficient of x², and the product is the constant term divided by the coefficient of x².

Verification 1. x² + 7x + 10 has zeroes −2 and −5. Sum = −7 = −(7)/1 = −b/a. Product = 10 = 10/1 = c/a. Both relations hold.

Verification 2. 3x² − x − 4 has zeroes 4/3 and −1. Sum = 4/3 − 1 = 1/3, and −b/a = −(−1)/3 = 1/3. Product = −4/3, and c/a = −4/3. Both hold.

Verification 3. x² − 3 has zeroes √3 and −√3. Sum = 0 = −0/1. Product = −3 = −3/1. Both hold.

Verification 4. 4u² + 8u = 4u(u + 2) has zeroes 0 and −2. Sum = −2 = −8/4. Product = 0 = 0/4. Both hold.

Using the relations to find unknowns. If one zero of x² − 5x + k is 2, find k and the other zero. Let the zeroes be 2 and β. Sum: 2 + β = 5, so β = 3. Product: 2 × 3 = k, so k = 6. If the zeroes of x² + px + 12 are in the ratio 1 : 3, find p. Let the zeroes be α and 3α. Product: 3α² = 12, so α² = 4, α = ±2. Sum: 4α = −p, so p = ∓8; that is, p = −8 or p = 8.

Symmetric expressions. Many expressions in α and β can be found from the sum and product without knowing the zeroes themselves. α² + β² = (α + β)² − 2αβ. 1/α + 1/β = (α + β)/αβ. α³ + β³ = (α + β)³ − 3αβ(α + β). (α − β)² = (α + β)² − 4αβ. For the zeroes of x² − 5x + 6 (sum 5, product 6): α² + β² = 25 − 12 = 13; 1/α + 1/β = 5/6; check with α = 2, β = 3: 4 + 9 = 13 and 1/2 + 1/3 = 5/6.

In the examination, after finding zeroes by factorisation, always write the two relations with the coefficients substituted and show that both sides agree; this verification is a separate mark.

The relations of the previous section can be run backwards. If a quadratic polynomial has zeroes α and β, then it is k(x − α)(x − β) = k[x² − (α + β)x + αβ] for any non-zero constant k. Taking k = 1 gives the simplest form:

Required polynomial: x² − (sum of zeroes)x + (product of zeroes).

Example 1. Zeroes 3 and −2. Sum = 1, product = −6. Polynomial: x² − x − 6. Check: (x − 3)(x + 2) = x² − x − 6.

Example 2. Sum of zeroes 1/4 and product −1. Polynomial: x² − (1/4)x − 1. To clear fractions multiply by 4: 4x² − x − 4. Both are correct; the second is usually preferred. Note that multiplying by a constant does not change the zeroes.

Example 3. Zeroes √2 and −√2. Sum = 0, product = −2. Polynomial: x² − 2.

Example 4. Sum of zeroes −3 and product 2. Polynomial: x² + 3x + 2 = (x + 1)(x + 2), zeroes −1 and −2, whose sum is −3 and product 2. Correct.

Example 5. Zeroes 1/2 and 1/3. Sum = 5/6, product = 1/6. Polynomial: x² − (5/6)x + 1/6, or multiplying by 6: 6x² − 5x + 1 = (2x − 1)(3x − 1).

Example 6. Sum √2 and product 1/3. Polynomial: x² − √2 x + 1/3, or 3x² − 3√2 x + 1.

Example 7. A polynomial whose zeroes are related to those of a given one. If α and β are the zeroes of x² − 5x + 6, find a polynomial whose zeroes are α + 1 and β + 1. New sum = (α + β) + 2 = 5 + 2 = 7; new product = αβ + (α + β) + 1 = 6 + 5 + 1 = 12. Polynomial: x² − 7x + 12. Check: the original zeroes are 2 and 3, the new ones 3 and 4, and (x − 3)(x − 4) = x² − 7x + 12.

Example 8. If α and β are the zeroes of x² + 3x + 2, find the polynomial whose zeroes are 1/α and 1/β. New sum = (α + β)/αβ = −3/2; new product = 1/(αβ) = 1/2. Polynomial: x² + (3/2)x + 1/2, or 2x² + 3x + 1. Check: original zeroes −1, −2; reciprocals −1, −1/2; (x + 1)(2x + 1) = 2x² + 3x + 1.

Two cautions. The formula is x² minus the sum plus the product; the sign on the sum is the most common error. And when the question says a quadratic polynomial rather than the quadratic polynomial, any non-zero multiple is acceptable, so state that the general answer is k(x² − Sx + P) and give the simplest.

Let α, β and γ (gamma) be the zeroes of the cubic polynomial p(x) = ax³ + bx² + cx + d, a ≠ 0. Then p(x) = a(x − α)(x − β)(x − γ). Expanding the product of the three brackets: (x − α)(x − β)(x − γ) = x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ. Multiplying by a and comparing coefficients with ax³ + bx² + cx + d gives three relations:

α + β + γ = −b/a; αβ + βγ + γα = c/a; αβγ = −d/a.

Notice the alternating signs: minus, plus, minus. The sum of the zeroes and the product of the zeroes carry a negative sign; the sum of the products taken two at a time does not.

Verification 1. p(x) = x³ − 6x² + 11x − 6 with zeroes 1, 2, 3. Check p(1) = 1 − 6 + 11 − 6 = 0, p(2) = 8 − 24 + 22 − 6 = 0, p(3) = 27 − 54 + 33 − 6 = 0. Sum = 6 = −(−6)/1. Sum of products in pairs = 1×2 + 2×3 + 3×1 = 2 + 6 + 3 = 11 = 11/1. Product = 6 = −(−6)/1. All three hold.

Verification 2. p(x) = 3x³ − 5x² − 11x − 3 with zeroes 3, −1, −1/3. p(3) = 81 − 45 − 33 − 3 = 0; p(−1) = −3 − 5 + 11 − 3 = 0; p(−1/3) = 3(−1/27) − 5(1/9) + 11/3 − 3 = −1/9 − 5/9 + 33/9 − 27/9 = 0. Sum = 3 − 1 − 1/3 = 5/3 = −(−5)/3. Pairs: 3(−1) + (−1)(−1/3) + (−1/3)(3) = −3 + 1/3 − 1 = −11/3 = c/a. Product = 3 × (−1) × (−1/3) = 1 = −(−3)/3. All hold.

Verification 3. p(x) = x³ − 4x² + 5x − 2 with zeroes 2, 1, 1. Sum = 4 = −(−4). Pairs: 2 + 1 + 2 = 5. Product = 2 = −(−2). All hold; note the repeated zero is counted twice in each relation.

Forming a cubic from its zeroes. A cubic with zeroes α, β, γ is k[x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ]. Find a cubic with zeroes 2, −1, 3. Sum = 4; pairs = −2 − 3 + 6 = 1; product = −6. Polynomial: x³ − 4x² + x + 6. Check by expanding (x − 2)(x + 1)(x − 3) = (x² − x − 2)(x − 3) = x³ − 4x² + x + 6.

Find a cubic whose sum of zeroes, sum of products of zeroes in pairs and product of zeroes are 2, −7, −14 respectively. Polynomial: x³ − 2x² − 7x + 14.

Using the relations. If the zeroes of x³ − 3x² + x + 1 are a − b, a, a + b, find a and b. Sum: 3a = 3, so a = 1. Product: (1 − b)(1)(1 + b) = 1 − b² = −1, so b² = 2, b = ±√2. The zeroes are 1 − √2, 1, 1 + √2.

Examination questions on cubics are of two kinds: verify the three relations for given zeroes (show all three, including the substitution check that each number really is a zero), and form a cubic from given data. Write the three relations clearly with their signs before substituting.

Just as integers can be divided with a quotient and remainder, so can polynomials. The division algorithm for polynomials states: if p(x) and g(x) are polynomials with g(x) ≠ 0, then there exist unique polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or the degree of r(x) is less than the degree of g(x). Here p(x) is the dividend, g(x) the divisor, q(x) the quotient and r(x) the remainder. This is the polynomial version of Euclid's lemma from the first chapter.

Long division procedure. Arrange both polynomials in descending powers, filling missing powers with zero coefficients. Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient. Multiply the whole divisor by this term and subtract from the dividend. Repeat with the new dividend until its degree is less than the degree of the divisor.

Example 1. Divide 3x³ + x² + 2x + 5 by 1 + 2x + x². Write the divisor as x² + 2x + 1. First term: 3x³ ÷ x² = 3x. Multiply: 3x(x² + 2x + 1) = 3x³ + 6x² + 3x. Subtract: (3x³ + x² + 2x + 5) − (3x³ + 6x² + 3x) = −5x² − x + 5. Next term: −5x² ÷ x² = −5. Multiply: −5(x² + 2x + 1) = −5x² − 10x − 5. Subtract: (−5x² − x + 5) − (−5x² − 10x − 5) = 9x + 10. The degree of 9x + 10 is 1, less than 2, so stop. Quotient q(x) = 3x − 5, remainder r(x) = 9x + 10. Verify: (x² + 2x + 1)(3x − 5) + 9x + 10 = 3x³ − 5x² + 6x² − 10x + 3x − 5 + 9x + 10 = 3x³ + x² + 2x + 5. Correct.

Example 2. Divide x⁴ − 3x² + 4x + 5 by x² + 1 − x. Divisor: x² − x + 1. Dividend with the missing x³ term: x⁴ + 0x³ − 3x² + 4x + 5. Step 1: x⁴ ÷ x² = x²; subtract x²(x² − x + 1) = x⁴ − x³ + x² to get x³ − 4x² + 4x + 5. Step 2: x³ ÷ x² = x; subtract x³ − x² + x to get −3x² + 3x + 5. Step 3: −3x² ÷ x² = −3; subtract −3x² + 3x − 3 to get 8. Quotient x² + x − 3, remainder 8.

Example 3. Check whether x² + 3x + 1 is a factor of 3x⁴ + 5x³ − 7x² + 2x + 2. Divide: first term 3x², subtract 3x⁴ + 9x³ + 3x² to get −4x³ − 10x² + 2x + 2; next term −4x, subtract −4x³ − 12x² − 4x to get 2x² + 6x + 2; next term 2, subtract 2x² + 6x + 2 to get 0. Remainder zero, so it is a factor, and the quotient is 3x² − 4x + 2.

Degree bookkeeping. deg p = deg g + deg q whenever the division is possible with deg p ≥ deg g. If deg p < deg g, the quotient is 0 and the remainder is p(x) itself. The remainder has degree strictly less than the divisor; if the divisor is quadratic, the remainder is at most linear.

Write the division neatly in columns, aligning like powers, and always verify with the identity p = gq + r; the verification is part of the answer in the examination.

The division algorithm has a powerful application: if some zeroes of a polynomial are known, the corresponding linear factors can be multiplied together and divided out, leaving a polynomial of lower degree whose zeroes are the remaining ones.

Principle. If α is a zero of p(x), then (x − α) is a factor and p(x) = (x − α) q(x). If α and β are both zeroes, then (x − α)(x − β) is a factor and p(x) = (x − α)(x − β) q(x). The remaining zeroes are the zeroes of q(x).

Example 1. Find all zeroes of 2x⁴ − 3x³ − 3x² + 6x − 2, given that two of its zeroes are √2 and −√2. Since √2 and −√2 are zeroes, (x − √2)(x + √2) = x² − 2 is a factor. Divide: 2x⁴ ÷ x² = 2x²; subtract 2x⁴ − 4x² to get −3x³ + x² + 6x − 2. Next −3x³ ÷ x² = −3x; subtract −3x³ + 6x to get x² − 2. Next x² ÷ x² = 1; subtract x² − 2 to get 0. Quotient 2x² − 3x + 1 = (2x − 1)(x − 1). So the other zeroes are 1/2 and 1. All four zeroes: √2, −√2, 1/2, 1.

Example 2. Find all zeroes of x³ − 3x² − x + 3, given that 1 is a zero. Divide by (x − 1): quotient x² − 2x − 3 = (x − 3)(x + 1). Zeroes: 1, 3, −1. Check: 1 + 3 − 1 = 3 = −(−3)/1, matching the sum relation.

Example 3. Find all zeroes of x⁴ − 6x³ − 26x² + 138x − 35, given that two zeroes are 2 + √3 and 2 − √3. The factor from these two zeroes: sum = 4, product = (2)² − (√3)² = 1, so the factor is x² − 4x + 1. Divide: x⁴ ÷ x² = x², subtract x⁴ − 4x³ + x² to get −2x³ − 27x² + 138x − 35; −2x³ ÷ x² = −2x, subtract −2x³ + 8x² − 2x to get −35x² + 140x − 35; −35x² ÷ x² = −35, subtract −35x² + 140x − 35 to get 0. Quotient x² − 2x − 35 = (x − 7)(x + 5). Zeroes: 2 + √3, 2 − √3, 7, −5.

Example 4. Finding what to subtract. What must be subtracted from x⁴ + 2x³ − 13x² − 12x + 21 so that the result is exactly divisible by x² − 4x + 3? Divide and find the remainder: the quotient works out to x² + 6x + 8 and the remainder to 2x − 3. Subtracting the remainder 2x − 3 makes the division exact.

Example 5. Finding a polynomial from its quotient and remainder. On dividing x³ − 3x² + x + 2 by a polynomial g(x), the quotient is x − 2 and the remainder −2x + 4. Find g(x). From p = gq + r, g(x) = (p − r)/q = (x³ − 3x² + x + 2 + 2x − 4)/(x − 2) = (x³ − 3x² + 3x − 2)/(x − 2). Dividing gives x² − x + 1 with zero remainder, so g(x) = x² − x + 1.

When two given zeroes are conjugate surds a ± √b, build their quadratic factor from sum 2a and product a² − b rather than multiplying brackets with surds, which invites errors. After finding all zeroes, check with the sum-of-zeroes relation as a final safeguard.

This chapter answered one question in several ways: what are the zeroes of a polynomial? Algebraically, they are the numbers k with p(k) = 0, found by factorisation or division. Geometrically, they are where the graph meets the x-axis. Structurally, they are tied to the coefficients by the sum-and-product relations.

What to memorise. Degree and the names linear, quadratic, cubic. Zero of ax + b is −b/a. A quadratic has 0, 1 or 2 real zeroes according as its parabola misses, touches or cuts the x-axis. α + β = −b/a, αβ = c/a. A cubic's three relations with alternating signs. The polynomial from zeroes: x² − Sx + P. The division algorithm p = gq + r with deg r < deg g.

One-mark questions. State the degree; find p(2) for a given p(x); write the zero of 3x + 5; from a given graph, state the number of zeroes; write a quadratic whose zeroes are 2 and −3; state the sum of the zeroes of 2x² − 5x + 3 without finding them.

Two-mark questions. Find the zeroes of a quadratic by factorisation and verify the relations. Form a quadratic with given sum and product. Check whether a given number is a zero. Find k if a given number is a zero of a polynomial with parameter k.

Four-mark questions. Draw the graph of a quadratic on graph paper and find its zeroes. Verify the three relations for a cubic with given zeroes. Divide one polynomial by another and verify the algorithm. Find all zeroes of a degree-4 polynomial given two of them. Find what must be added or subtracted to make a division exact.

Common mistakes. Sign error in −b/a. Forgetting brackets when substituting a negative number. Writing the zero of 3x − 4 as 4 instead of 4/3. Omitting the zero coefficient for a missing power when dividing. Not writing the verification p = gq + r. Counting the y-intercept as a zero when reading a graph. Forgetting that multiplying a polynomial by a constant leaves the zeroes unchanged.

Revision routine. Factorise five quadratics by splitting the middle term. Verify the sum-product relations for each. Form the polynomial back from the zeroes. Do two long divisions and verify. Draw one parabola and one cubic from a table of values.

Looking ahead. The next chapter, Pair of Linear Equations, uses the graph of linear polynomials; Quadratic Equations turns the search for zeroes into solving ax² + bx + c = 0 by formula and introduces the discriminant, which is the algebraic form of the three-cases picture; Progressions uses the sum-product relations again; and the whole of Intermediate algebra builds on polynomials.