Chapter 2 — Polynomials

An algebraic expression in one variable x is a polynomial if it is a sum of terms of the form (constant) × xⁿ, where each exponent n is a whole number 0, 1, 2, 3, ... . So 4x + 2, 2y² − 3y + 4, 5x³ − 4x² + x − √2 and 7 are polynomials. Expressions with a variable in a denominator, under a root or with a negative or fractional exponent are not: 1/(x + 1), √x + 3, x⁻² + 1 and 3/x are not polynomials. The constants multiplying the powers are the coefficients; in 5x³ − 4x² + x − √2 the coefficients of x³, x², x and x⁰ are 5, −4, 1 and −√2. A term with no variable, like −√2 here, is the constant term.

The highest power of the variable in a polynomial is its degree. The degree of 4x + 2 is 1; of 2y² − 3y + 4 it is 2; of 5x³ − 4x² + x − √2 it is 3. A non-zero constant such as 7 is a polynomial of degree 0. The zero polynomial, 0, is not assigned a degree. Polynomials are named by their degree. A polynomial of degree 1 is linear, and its general form is ax + b with a ≠ 0. A polynomial of degree 2 is quadratic (from the Latin quadratus, square), with general form ax² + bx + c, a ≠ 0, where a, b, c are real numbers. A polynomial of degree 3 is cubic, with general form ax³ + bx² + cx + d, a ≠ 0. Higher degrees exist but this chapter goes up to cubics.

Polynomials are written with the variable as an argument: p(x) = x² − 3x − 4, q(y) = y³ − 1, r(t) = t² + 2t − 3. This notation lets us speak of the value of the polynomial at a number: p(2) is what you get by putting x = 2 in p(x). For p(x) = x² − 3x − 4, p(2) = 4 − 6 − 4 = −6, p(−1) = 1 + 3 − 4 = 0, and p(4) = 16 − 12 − 4 = 0.

Polynomials are also classified by the number of terms: a monomial has one term (5x²), a binomial two (x + 3), a trinomial three (x² + x + 1). This is a different classification from degree: 5x² is a monomial of degree 2, and x + 3 is a binomial of degree 1.

The examination begins with recognition: given a list of expressions, say which are polynomials and give the degree and type. Check each exponent is a whole number, find the largest one, and name the degree. Remember that the coefficient a of the highest power must be non-zero for the general forms; 0x² + 3x + 1 is really linear, not quadratic.

A real number k is called a zero of the polynomial p(x) if p(k) = 0, that is, if substituting k for x makes the polynomial's value zero. For p(x) = x² − 3x − 4 we found p(−1) = 0 and p(4) = 0, so −1 and 4 are zeroes of p(x). The zeroes are the numbers that 'kill' the polynomial. They are also called roots when we speak of the equation p(x) = 0, but in this chapter we use 'zeroes of the polynomial'.

To find the zero of a linear polynomial ax + b, solve ax + b = 0, giving x = −b/a. So the zero of 2x + 3 is −3/2, the zero of 5x − 10 is 2, the zero of x + 7 is −7. A linear polynomial has exactly one zero, and it is the negative of the constant term divided by the coefficient of x. Notice that the zero is expressed in terms of the coefficients; this is the first instance of the relation between zeroes and coefficients that runs through the chapter.

To find the zeroes of a quadratic polynomial ax² + bx + c, factorise it into two linear factors and set each to zero. For x² − 3x − 4 = (x − 4)(x + 1), the zeroes are 4 and −1. For x² + 7x + 10 = (x + 2)(x + 5), the zeroes are −2 and −5. The method of splitting the middle term from Class 9 is used: for 6x² − 7x − 3, find two numbers whose product is 6 × (−3) = −18 and sum is −7; they are −9 and 2, so 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3), zeroes −1/3 and 3/2.

A quadratic polynomial can have two zeroes, one zero (when the two factors are equal, as in x² − 4x + 4 = (x − 2)², where 2 is a repeated zero), or no real zero (as in x² + 1, which is never zero for real x because x² ≥ 0). So a quadratic has at most two zeroes. Similarly a cubic has at most three zeroes, and in general a polynomial of degree n has at most n zeroes. This is a key fact: the degree bounds the number of zeroes.

Checking whether a number is a zero is only substitution. Is 1/2 a zero of 2x² + x − 1? Compute 2(1/4) + 1/2 − 1 = 1/2 + 1/2 − 1 = 0; yes. Is −2 a zero of x³ + 8? (−2)³ + 8 = −8 + 8 = 0; yes. Is 3 a zero of x² − 2x − 1? 9 − 6 − 1 = 2 ≠ 0; no. In examinations, always show the substitution and the arithmetic.

A polynomial p(x) can be drawn: for each x we plot the point (x, p(x)) and the points join into a curve, the graph of y = p(x). The graph makes the zeroes visible: since a zero k satisfies p(k) = 0, the point (k, 0) is on the graph, and it lies on the x-axis. So the zeroes of a polynomial are the x-coordinates of the points where its graph meets the x-axis. Counting those points counts the zeroes.

The graph of a linear polynomial y = ax + b is a straight line. It meets the x-axis at exactly one point, (−b/a, 0), which is why a linear polynomial has exactly one zero. For y = 2x + 3, plot (−2, −1) and (2, 7); the line through them crosses the x-axis at (−3/2, 0). The line rises to the right if a > 0 and falls if a < 0.

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) and has a lowest point; if a < 0 it opens downwards (like a cap) and has a highest point. This turning point is the vertex, and the parabola is symmetric about the vertical line through it. For y = x² − 3x − 4, a table of values gives (−2, 6), (−1, 0), (0, −4), (1, −6), (2, −6), (3, −4), (4, 0), (5, 6); plotting shows an upward parabola crossing the x-axis at −1 and 4, the two zeroes, with its vertex at (1.5, −6.25).

The parabola can meet the x-axis in three ways. Case 1: it cuts the axis at two distinct points A and A', and the polynomial has two distinct zeroes (x² − 3x − 4). Case 2: it touches the axis at exactly one point, the vertex sitting on the axis, and the polynomial has one repeated zero (x² − 4x + 4 = (x − 2)² touches at (2, 0)). Case 3: it lies entirely above or entirely below the axis, never meeting it, and the polynomial has no real zero (x² + 1 lies above; −x² − 1 lies below). This is the geometric reason a quadratic has at most two zeroes.

The graph of a cubic polynomial is a curve with an S-like shape that may cross the x-axis one, two or three times; y = x³ − 4x = x(x − 2)(x + 2) crosses at −2, 0 and 2, while y = x³ crosses only at 0. A cubic always crosses the axis at least once, because for large positive x it goes one way and for large negative x the other way.

The examination shows graphs and asks for the number of zeroes: count the crossings (a touch counts as one zero). It may also ask you to draw the graph of a given quadratic and read its zeroes; use a table of at least six values around the vertex, plot carefully, and mark the intersections with the x-axis.

Take the quadratic p(x) = 2x² − 8x + 6. Factorising, 2x² − 8x + 6 = 2(x² − 4x + 3) = 2(x − 1)(x − 3), so the zeroes are 1 and 3. Now compare: the sum of the zeroes is 1 + 3 = 4, and −(coefficient of x)/(coefficient of x²) = −(−8)/2 = 4. The product of the zeroes is 1 × 3 = 3, and (constant term)/(coefficient of x²) = 6/2 = 3. Both match. This is not luck; it holds for every quadratic.

Theorem. If α and β are the zeroes of the quadratic polynomial p(x) = ax² + bx + c, a ≠ 0, then α + β = −b/a and αβ = c/a. In words: sum of zeroes = −(coefficient of x)/(coefficient of x²), product of zeroes = constant term/(coefficient of x²).

Why it is true. Since α and β are zeroes, (x − α) and (x − β) are factors of p(x), and because p(x) has degree 2 with leading coefficient a, p(x) = a(x − α)(x − β). Expanding, a(x − α)(x − β) = a[x² − (α + β)x + αβ] = ax² − a(α + β)x + aαβ. Comparing with ax² + bx + c, coefficient by coefficient: −a(α + β) = b, so α + β = −b/a; and aαβ = c, so αβ = c/a.

Verifying in problems. Find the zeroes of x² + 7x + 10 and verify the relationship. Factorise: x² + 7x + 10 = (x + 2)(x + 5), zeroes −2 and −5. Sum = −7 = −(7)/1 = −b/a. Product = 10 = 10/1 = c/a. Verified. For 3x² − x − 4 = (3x − 4)(x + 1), zeroes 4/3 and −1; sum = 1/3 = −(−1)/3; product = −4/3 = (−4)/3. For x² − 2 with zeroes √2 and −√2, sum = 0 = −0/1 (b = 0), product = −2 = −2/1.

Using the relations without finding zeroes. If α and β are the zeroes of x² − 5x + 6, then without factorising we know α + β = 5 and αβ = 6, so α² + β² = (α + β)² − 2αβ = 25 − 12 = 13, and 1/α + 1/β = (α + β)/(αβ) = 5/6, and (α − β)² = (α + β)² − 4αβ = 25 − 24 = 1. If one zero of x² − 8x + k is three times the other, let them be α and 3α: sum 4α = 8 gives α = 2, and product 3α² = k gives k = 12. If the zeroes of x² + (a + 1)x + b are 2 and −3, then 2 + (−3) = −(a + 1) gives a = 0, and 2 × (−3) = b gives b = −6. These manipulations are the heart of the chapter's examination questions.

Note the sign in the sum: it is minus b over a. Students lose marks by writing b/a. A good check: for x² − 5x + 6 the zeroes 2 and 3 are positive and their sum is positive, so the coefficient of x must be negative, which it is.

The relations of the last section work in reverse: if we know the sum S and product P of the zeroes of a quadratic, we can write the quadratic down. Since ax² + bx + c = a[x² − (α + β)x + αβ], every quadratic with zeroes α and β is a constant multiple of x² − (sum of zeroes)x + (product of zeroes). Taking the constant as 1 gives the simplest polynomial; any non-zero multiple k[x² − Sx + P] has the same zeroes.

Given the zeroes. Find a quadratic polynomial whose zeroes are 3 and −2. Sum S = 1, product P = −6. Polynomial: x² − x − 6. Check by factorising: (x − 3)(x + 2) = x² − x − 6. Zeroes √2 and −√2: S = 0, P = −2, polynomial x² − 2. Zeroes 2 + √3 and 2 − √3: S = 4, P = (2 + √3)(2 − √3) = 4 − 3 = 1, polynomial x² − 4x + 1. Zeroes 1/2 and −1/3: S = 1/6, P = −1/6, polynomial x² − x/6 − 1/6, or multiplying by 6 to clear fractions, 6x² − x − 1. Both are correct; the second is neater.

Given the sum and product. Find a quadratic polynomial with sum of zeroes 1/4 and product −1. Polynomial: x² − (1/4)x − 1, or 4x² − x − 4. Sum √2, product 1/3: x² − √2 x + 1/3, or 3x² − 3√2 x + 1. Sum 0, product √5: x² + √5. Sum 1, product 1: x² − x + 1 (this one has no real zeroes, but the polynomial still exists). Sum −1/4, product 1/4: x² + x/4 + 1/4, or 4x² + x + 1.

Zeroes related to another polynomial's zeroes. This is a favourite examination pattern. If α and β are the zeroes of x² − 5x + 6, find the polynomial whose zeroes are 2α and 2β. From the given polynomial α + β = 5 and αβ = 6. New sum = 2α + 2β = 2(α + β) = 10, new product = 4αβ = 24. Required polynomial: x² − 10x + 24. Find the polynomial whose zeroes are 1/α and 1/β: new sum = (α + β)/(αβ) = 5/6, new product = 1/(αβ) = 1/6, polynomial x² − (5/6)x + 1/6, or 6x² − 5x + 1. Find the polynomial whose zeroes are α + 1 and β + 1: new sum = (α + β) + 2 = 7, new product = αβ + (α + β) + 1 = 6 + 5 + 1 = 12, polynomial x² − 7x + 12. Find the polynomial whose zeroes are α² and β²: new sum = α² + β² = 25 − 12 = 13, new product = (αβ)² = 36, polynomial x² − 13x + 36.

In every case the method is the same three lines: compute the new sum, compute the new product, write x² − Sx + P. Do not try to find α and β individually unless the question asks for them; the relations do the work. And always state at the end 'or any non-zero multiple of this'.

A cubic polynomial ax³ + bx² + cx + d, a ≠ 0, can have up to three zeroes, and there are three relations connecting them with the coefficients. Theorem. If α, β and γ are the zeroes of ax³ + bx² + cx + d, then α + β + γ = −b/a, αβ + βγ + γα = c/a, and αβγ = −d/a. In words: the sum of the zeroes is −(coefficient of x²)/(coefficient of x³); the sum of the products of the zeroes taken two at a time is (coefficient of x)/(coefficient of x³); the product of the zeroes is −(constant term)/(coefficient of x³). Notice the alternating signs: minus, plus, minus.

Why. As for quadratics, a cubic with zeroes α, β, γ and leading coefficient a is a(x − α)(x − β)(x − γ). Multiply out: (x − α)(x − β) = x² − (α + β)x + αβ; multiplying by (x − γ) gives x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ. So ax³ + bx² + cx + d = a x³ − a(α + β + γ)x² + a(αβ + βγ + γα)x − aαβγ, and comparing coefficients gives the three relations.

Verifying. Verify that 3, −1 and −1/3 are the zeroes of p(x) = 3x³ − 5x² − 11x − 3, and check the relations. First, p(3) = 81 − 45 − 33 − 3 = 0; p(−1) = −3 − 5 + 11 − 3 = 0; p(−1/3) = 3(−1/27) − 5(1/9) − 11(−1/3) − 3 = −1/9 − 5/9 + 11/3 − 3 = −6/9 + 33/9 − 27/9 = 0. All three are zeroes. Here a = 3, b = −5, c = −11, d = −3. Sum: 3 + (−1) + (−1/3) = 5/3 = −(−5)/3 = −b/a. Sum of products in 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 = −d/a. All verified.

A second example: for x³ − 4x² + 5x − 2, check 2, 1, 1. p(2) = 8 − 16 + 10 − 2 = 0, p(1) = 1 − 4 + 5 − 2 = 0. Sum 2 + 1 + 1 = 4 = −(−4)/1; pairs 2 + 1 + 2 = 5 = 5/1; product 2 = −(−2)/1. Verified; note that 1 is a repeated zero.

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, −3 and 1: sum = 0, pairs = −6 − 3 + 2 = −7, product = −6; polynomial x³ − 0x² − 7x + 6 = x³ − 7x + 6. Find a cubic whose sum of zeroes, sum of pairwise products and product are 2, −7 and −14: x³ − 2x² − 7x + 14. Cubic with zeroes 0, 1, −1: sum 0, pairs −1, product 0: x³ − x.

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: (a − b)a(a + b) = a(a² − b²) = −1, so 1 − b² = −1, b² = 2, b = ±√2. The zeroes are 1 − √2, 1, 1 + √2. If two zeroes of x³ − 4x² + x + 6 are 3 and −1, the third zero γ satisfies 3 + (−1) + γ = 4, so γ = 2; check product 3 × (−1) × 2 = −6 = −6/1. Such questions reward knowing the three relations exactly, signs included.

Euclid's lemma for integers says a = bq + r with r smaller than b. Polynomials obey the same law with 'smaller' meaning 'of lower degree'. Division algorithm. If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then there exist 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. In words, dividend = divisor × quotient + remainder. If r(x) = 0, g(x) is a factor of p(x).

How to divide. The process is long division. Step 1: write both polynomials in descending powers of x, filling any missing power with a zero coefficient. Step 2: divide the first term of the dividend by the first term of the divisor to get the first term of the quotient. Step 3: multiply the whole divisor by this term and subtract from the dividend. Step 4: treat the result as the new dividend and repeat until the degree of the remainder is less than the degree of the divisor.

Worked example. Divide 3x³ + x² + 2x + 5 by 1 + 2x + x². Arrange 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. Check: (x² + 2x + 1)(3x − 5) + (9x + 10) = 3x³ − 5x² + 6x² − 10x + 3x − 5 + 9x + 10 = 3x³ + x² + 2x + 5. Correct.

Second example. Divide x⁴ − 3x² + 4x + 5 by x² + 1 − x. Write the divisor as x² − x + 1 and the dividend as x⁴ + 0x³ − 3x² + 4x + 5. x⁴ ÷ x² = x²; x²(x² − x + 1) = x⁴ − x³ + x²; subtract to get x³ − 4x² + 4x + 5. x³ ÷ x² = x; x(x² − x + 1) = x³ − x² + x; subtract to get −3x² + 3x + 5. −3x² ÷ x² = −3; −3(x² − x + 1) = −3x² + 3x − 3; subtract to get 8. Quotient x² + x − 3, remainder 8.

Checking a factor. Is t² − 3 a factor of 2t⁴ + 3t³ − 2t² − 9t − 12? Divide: quotient 2t² + 3t + 4, remainder 0. So yes. Is x² + 3x + 1 a factor of 3x⁴ + 5x³ − 7x² + 2x + 2? Divide: quotient 3x² − 4x + 2, remainder 0. Yes.

Always check the answer by multiplying back; the check is quick and the examiner expects the equation dividend = divisor × quotient + remainder written out. Keep the columns aligned by powers and be careful with signs when subtracting.

If k is a zero of p(x), then (x − k) is a factor of p(x); this is the factor theorem of Class 9. So when some zeroes of a polynomial are known, we can multiply the corresponding factors together, divide p(x) by their product, and the quotient contains the remaining zeroes. This is the most examined application of the division algorithm.

Worked example 1. Find all the zeroes of 2x⁴ − 3x³ − 3x² + 6x − 2, given that two of them are √2 and −√2. Since √2 and −√2 are zeroes, (x − √2)(x + √2) = x² − 2 is a factor. Divide 2x⁴ − 3x³ − 3x² + 6x − 2 by x² − 2: 2x⁴ ÷ x² = 2x², and 2x²(x² − 2) = 2x⁴ − 4x²; subtracting gives −3x³ + x² + 6x − 2. Then −3x³ ÷ x² = −3x, and −3x(x² − 2) = −3x³ + 6x; subtracting gives x² − 2. Then x² ÷ x² = 1, and 1(x² − 2) = x² − 2; subtracting gives 0. So the quotient is 2x² − 3x + 1 and the remainder is 0. Now factorise the quotient: 2x² − 3x + 1 = 2x² − 2x − x + 1 = 2x(x − 1) − 1(x − 1) = (2x − 1)(x − 1), with zeroes 1/2 and 1. Hence all four zeroes are √2, −√2, 1/2 and 1.

Worked example 2. Find all zeroes of x⁴ − 6x³ − 26x² + 138x − 35 given that two zeroes are 2 ± √3. The factor is (x − 2 − √3)(x − 2 + √3) = (x − 2)² − 3 = x² − 4x + 1. Dividing x⁴ − 6x³ − 26x² + 138x − 35 by x² − 4x + 1 gives the quotient x² − 2x − 35 and remainder 0. Factorise: x² − 2x − 35 = (x − 7)(x + 5), zeroes 7 and −5. All zeroes: 2 + √3, 2 − √3, 7, −5.

Worked example 3. Find all zeroes of 3x⁴ + 6x³ − 2x² − 10x − 5 if two of them are ±√(5/3). The factor is x² − 5/3, or, clearing the fraction, 3x² − 5. Divide: 3x⁴ ÷ 3x² = x²; x²(3x² − 5) = 3x⁴ − 5x²; subtracting gives 6x³ + 3x² − 10x − 5. Then 6x³ ÷ 3x² = 2x; 2x(3x² − 5) = 6x³ − 10x; subtracting gives 3x² − 5. Then 1 × (3x² − 5); remainder 0. Quotient x² + 2x + 1 = (x + 1)², giving the repeated zero −1. All zeroes: √(5/3), −√(5/3), −1, −1.

When one zero of a cubic is known. Given that −1 is a zero of x³ − 4x² + x + 6, first confirm it: (−1)³ − 4(1) + (−1) + 6 = −1 − 4 − 1 + 6 = 0. Then divide by x + 1: x³ ÷ x = x², x²(x + 1) = x³ + x², subtract to get −5x² + x + 6; −5x² ÷ x = −5x, −5x(x + 1) = −5x² − 5x, subtract to get 6x + 6; 6x ÷ x = 6, 6(x + 1) = 6x + 6, remainder 0. The quotient x² − 5x + 6 = (x − 2)(x − 3), so the zeroes are −1, 2 and 3. Had the substitution not given zero, the division would have left a non-zero remainder, which is the check that catches a wrongly assumed zero.

The method in summary: form the factor from the known zeroes (product of (x − k) for each), divide, confirm the remainder is zero, factorise the quotient. When the known zeroes are ±√a, the factor is simply x² − a. When they are p ± √q, the factor is (x − p)² − q.

The equation p(x) = g(x) q(x) + r(x) has four parts, and if any three are known the fourth can be found. This gives a family of questions that examiners like.

Finding the dividend. On dividing p(x) by x² − 2x + 1 the quotient is x − 2 and the remainder is 4x + 3. Find p(x). p(x) = (x² − 2x + 1)(x − 2) + (4x + 3) = x³ − 2x² − 2x² + 4x + x − 2 + 4x + 3 = x³ − 4x² + 9x + 1. No division is needed; only multiplication and collection of terms.

Finding the divisor. On dividing x³ − 3x² + x + 2 by a polynomial g(x), the quotient and remainder are x − 2 and −2x + 4 respectively. Find g(x). From p(x) = g(x) q(x) + r(x), g(x) q(x) = p(x) − r(x) = x³ − 3x² + x + 2 − (−2x + 4) = x³ − 3x² + 3x − 2. So g(x) = (x³ − 3x² + 3x − 2) ÷ (x − 2). Divide: x³ ÷ x = x², x²(x − 2) = x³ − 2x², subtract to get −x² + 3x − 2; −x² ÷ x = −x, −x(x − 2) = −x² + 2x, subtract to get x − 2; x ÷ x = 1, 1(x − 2) = x − 2, remainder 0. So g(x) = x² − x + 1.

Finding unknown coefficients for exact division. If x⁴ − 6x³ + 16x² − 25x + 10 is divided by x² − 2x + k the remainder is x + a. Find k and a. Perform the division with k as a letter: x⁴ ÷ x² = x², x²(x² − 2x + k) = x⁴ − 2x³ + kx², subtract to get −4x³ + (16 − k)x² − 25x + 10. −4x³ ÷ x² = −4x, −4x(x² − 2x + k) = −4x³ + 8x² − 4kx, subtract to get (8 − k)x² + (4k − 25)x + 10. (8 − k)x² ÷ x² = 8 − k, (8 − k)(x² − 2x + k) = (8 − k)x² − 2(8 − k)x + k(8 − k), subtract to get [(4k − 25) + 2(8 − k)]x + [10 − k(8 − k)] = (2k − 9)x + (10 − 8k + k²). This remainder must equal x + a, so 2k − 9 = 1, giving k = 5, and a = 10 − 40 + 25 = −5.

What must be added or subtracted. What must be subtracted from x⁴ + 2x³ − 13x² − 12x + 21 so that the result is exactly divisible by x² − 4x + 3? Divide: the remainder comes out as 2x − 3. Subtracting the remainder makes the division exact, so subtract 2x − 3. What must be added to x³ − 3x² + 6x − 15 so that it is divisible by x − 3? Divide: remainder 3 (or use the remainder theorem: p(3) = 27 − 27 + 18 − 15 = 3). Add −3, that is, subtract 3.

Degree bookkeeping. When deg p(x) = deg g(x) + deg q(x), the degrees add: dividing a degree-5 polynomial by a degree-2 polynomial gives a degree-3 quotient. When deg p < deg g, the quotient is 0 and the remainder is p(x) itself. Questions may ask you to give examples of divisions satisfying a condition on degrees. For deg p = deg q, take a constant divisor: divide 2x² + 4x + 6 by 2 to get q = x² + 2x + 3, r = 0. For deg q = deg r, take p = x³ + x + 1 and g = x²: q = x and r = x + 1, both of degree 1. For deg r = 0, take p = x³ + 1 and g = x + 1: q = x² − x + 1, r = 0, or p = x² + 1 and g = x: q = x, r = 1. Build such examples by choosing g and q first and then computing p = gq + r.

Most problems in this chapter end with factorising a quadratic, so the methods deserve a place of their own. There are four to master.

Common factor. Take out anything common first: 3x² − 12x = 3x(x − 4), zeroes 0 and 4. 2x² − 8 = 2(x² − 4) = 2(x − 2)(x + 2). Always look for this before anything else.

Difference of squares. a² − b² = (a − b)(a + b). So x² − 9 = (x − 3)(x + 3); 4x² − 25 = (2x − 5)(2x + 5); x² − 5 = (x − √5)(x + √5); 3x² − 7 = (√3 x − √7)(√3 x + √7). This is how zeroes like ±√5 arise.

Perfect squares. a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)². x² + 6x + 9 = (x + 3)², repeated zero −3; 4s² − 4s + 1 = (2s − 1)², repeated zero 1/2; x² − 2√2 x + 2 = (x − √2)², repeated zero √2. Recognise them by checking that the middle term is twice the product of the square roots of the outer terms.

Splitting the middle term. For ax² + bx + c, find two numbers whose product is ac and whose sum is b, split bx into two terms with those coefficients, and factor by grouping. Example: 6x² − 3 − 7x; arrange as 6x² − 7x − 3; ac = −18, b = −7; numbers −9 and +2; 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3); zeroes −1/3 and 3/2. Example: x² − 15 − 2x... arrange as x² − 2x − 15; ac = −15, b = −2; numbers −5 and 3; (x − 5)(x + 3); zeroes 5, −3. Example: 4u² + 8u; common factor 4u(u + 2); zeroes 0, −2. Example: t² − 15 = (t − √15)(t + √15). Example: 3x² − x − 4; ac = −12, b = −1; numbers −4 and 3; 3x² − 4x + 3x − 4 = x(3x − 4) + 1(3x − 4) = (3x − 4)(x + 1); zeroes 4/3, −1.

When the coefficients are surds. √3 x² + 10x + 7√3: ac = √3 × 7√3 = 21, b = 10; numbers 7 and 3; √3 x² + 7x + 3x + 7√3 = x(√3 x + 7) + √3(√3 x + 7) = (x + √3)(√3 x + 7); zeroes −√3 and −7/√3 = −7√3/3. 2√3 x² − 5x + √3: ac = 6, b = −5; numbers −3 and −2; 2√3 x² − 3x − 2x + √3 = √3 x(2x − √3) − 1(2x − √3) = (√3 x − 1)(2x − √3); zeroes 1/√3 and √3/2. Verify each by the sum and product relations: for 2√3 x² − 5x + √3, sum = 1/√3 + √3/2 = (2 + 3)/(2√3) = 5/(2√3) = −b/a, product = 1/2 = √3/(2√3) = c/a.

Whichever method you use, finish by checking the zeroes against −b/a and c/a; it takes ten seconds and catches almost every slip.

Polynomials describe many real quantities, and a few standard situations show how zeroes, the vertex and the coefficients acquire meaning.

Area and dimensions. A rectangular garden has length 3 metres more than its width w. Its area is A(w) = w(w + 3) = w² + 3w, a quadratic in w. Its zeroes are 0 and −3, the widths at which the area vanishes; only w > 0 makes sense physically, and the graph is the right half of an upward parabola. If the area is 40 m², then w² + 3w − 40 = 0 = (w + 8)(w − 5), so w = 5 and the length is 8 m. The negative zero is rejected.

Projectile height. A ball thrown upward has height h(t) = 20t − 5t² metres after t seconds. h(t) = 5t(4 − t), a downward parabola (a = −5 < 0) with zeroes t = 0 (launch) and t = 4 (landing). The vertex is at t = −b/(2a) = −20/(−10) = 2 seconds, where h(2) = 40 − 20 = 20 metres, the greatest height. The symmetry of the parabola says the ball takes as long to rise as to fall.

Profit. A shop's daily profit for selling x items is P(x) = −x² + 60x − 500. Zeroes: x² − 60x + 500 = (x − 10)(x − 50), so x = 10 and x = 50 are the break-even points; between them the profit is positive; the maximum is at x = 30 with P(30) = −900 + 1800 − 500 = 400.

Reading a drawn graph. Examination papers show the graph of a polynomial and ask for its zeroes and, sometimes, its expression. If a parabola crosses at (−2, 0) and (3, 0) and passes through (0, −6), then p(x) = k(x + 2)(x − 3) with p(0) = −6k = −6, so k = 1 and p(x) = x² − x − 6. If a parabola touches at (1, 0) and passes through (0, 2), then p(x) = k(x − 1)² with k = 2, p(x) = 2x² − 4x + 2. If a cubic crosses at −1, 0 and 2 and passes through (1, −2), then p(x) = k x(x + 1)(x − 2), and p(1) = k(1)(2)(−1) = −2k = −2, so k = 1, p(x) = x³ − x² − 2x.

Symmetry and the vertex. For any quadratic with zeroes α and β, the axis of symmetry is x = (α + β)/2 = −b/(2a), midway between the zeroes. For x² − 3x − 4 with zeroes −1 and 4, the axis is x = 1.5. Knowing this halves the plotting work when drawing a graph: compute values on one side of the axis and mirror them.

The number of zeroes and the degree in life. A quantity governed by a quadratic can be zero at most twice: a ball is at ground level at most twice in one flight; a profit function crosses zero at most twice. A cubic can cross three times. These facts, obvious from the graphs, are the physical face of the theorem that a polynomial of degree n has at most n zeroes.

The Class 10 paper draws on this chapter for every mark value. Here are the patterns and the traps.

One-mark. The degree of a given polynomial; the number of zeroes from a graph; the zero of a linear polynomial; the sum or product of zeroes of a given quadratic read directly as −b/a and c/a; whether a given expression is a polynomial; the quadratic with sum 0 and product −1 (x² − 1); the value of k if 2 is a zero of x² − kx + 6 (4 − 2k + 6 = 0 gives k = 5). Answer in one line but with the reason.

Two-mark. Find the zeroes of a quadratic and verify the relations; form a quadratic from a given sum and product; divide a polynomial by a linear or quadratic divisor; check whether g(x) is a factor of p(x).

Four-mark. Find the zeroes of a quadratic with surd coefficients and verify; form the polynomial whose zeroes are functions of the zeroes of a given polynomial; verify the three cubic relations for given zeroes; find all zeroes of a quartic given two of them; find k and a in a division with unknowns.

Eight-mark. Draw the graph of a quadratic and find its zeroes; or a combined problem: divide, find remaining zeroes, and verify the cubic or quartic relations at the end.

Common errors. (1) The sign of the sum: α + β = −b/a, not b/a. For x² + 5x + 6 the zeroes are −2 and −3 and their sum is −5. (2) Forgetting a in the relations when a ≠ 1: for 2x² − 8x + 6 the sum is 8/2 = 4, not 8. (3) In the cubic relations, mixing up which of the three carries a minus: sum is −b/a, pairs +c/a, product −d/a. (4) Not arranging the dividend and divisor in descending powers, or leaving out a zero placeholder for a missing power, which misaligns the long division. (5) Sign errors when subtracting during division; write the subtraction as adding the negative if that is safer. (6) Stopping division too early or too late: stop when the remainder's degree is less than the divisor's. (7) Dividing by (x − √2) and (x + √2) separately instead of by x² − 2, which doubles the work and the chance of error. (8) In 'form a polynomial' questions, writing x² + Sx + P instead of x² − Sx + P. (9) Reading a touching point on a graph as two zeroes or as no zero; it is one repeated zero. (10) Rejecting a surd or negative zero as 'not possible' in a pure algebra question; only word problems with physical constraints reject zeroes.

A final habit: after every zero you find, substitute it back into the polynomial once, and after every division, multiply back once. Those two checks turn a hopeful answer into a certain one, and they are exactly what the examiner's marking scheme rewards as 'verification'.