Chapter 5 — Polynomials Optional

Let α and β be the zeroes of ax² + bx + c. We know α + β = −b/a and αβ = c/a. Any expression that stays the same when α and β are swapped, called a symmetric expression, can be written in terms of these two quantities, and hence found from the coefficients without ever solving for α and β. This is the technique behind an entire family of optional problems.

The standard identities. α² + β² = (α + β)² − 2αβ. α³ + β³ = (α + β)³ − 3αβ(α + β). (α − β)² = (α + β)² − 4αβ. 1/α + 1/β = (α + β)/αβ. α/β + β/α = (α² + β²)/αβ. 1/α² + 1/β² = (α² + β²)/(αβ)². α²β + αβ² = αβ(α + β).

Example 1. If α, β are the zeroes of x² − 5x + 6, find α² + β², α³ + β³ and α/β + β/α. Sum = 5, product = 6. α² + β² = 25 − 12 = 13. α³ + β³ = 125 − 3(6)(5) = 125 − 90 = 35. α/β + β/α = 13/6. Check with α = 2, β = 3: 4 + 9 = 13; 8 + 27 = 35; 2/3 + 3/2 = 13/6.

Example 2. If α, β are the zeroes of 2x² + 3x − 5, find 1/α + 1/β and 1/α² + 1/β². Sum = −3/2, product = −5/2. 1/α + 1/β = (−3/2)/(−5/2) = 3/5. α² + β² = 9/4 + 5 = 29/4. 1/α² + 1/β² = (29/4)/(25/4) = 29/25.

Example 3. If α, β are the zeroes of x² + 7x + 12, find (α − β)² and hence α − β. Sum = −7, product = 12. (α − β)² = 49 − 48 = 1, so α − β = ±1. Indeed the zeroes are −3 and −4.

Example 4. If α, β are the zeroes of x² − x − 1, show that α² + β² = 3 and α⁴ + β⁴ = 7. Sum = 1, product = −1. α² + β² = 1 + 2 = 3. α⁴ + β⁴ = (α² + β²)² − 2(αβ)² = 9 − 2 = 7. Notice the same identity applied to the squares.

Example 5. If α, β are the zeroes of 3x² − 4x + 1, find α²/β + β²/α. This equals (α³ + β³)/αβ. Sum = 4/3, product = 1/3. α³ + β³ = 64/27 − 3(1/3)(4/3) = 64/27 − 4/3 = 64/27 − 36/27 = 28/27. Divide by αβ = 1/3: 28/27 × 3 = 28/9. Check with zeroes 1 and 1/3: 1/(1/3) + (1/9)/1 = 3 + 1/9 = 28/9.

The examination expects the identity to be written first, then the substitution. Marks are lost when a student finds the zeroes and substitutes numerically, because the question specifically says without finding the zeroes; the method, not only the value, is examined.

Many optional problems give a polynomial with an unknown constant k and a condition on its zeroes, and ask for k. The method is to translate the condition into an equation in α + β and αβ, substitute the coefficient expressions, and solve.

Example 1. If the zeroes of x² − (k + 6)x + 2(2k − 1) satisfy α + β = αβ/2, find k. Sum = k + 6, product = 2(2k − 1) = 4k − 2. The condition gives k + 6 = (4k − 2)/2 = 2k − 1, so k = 7. Check: polynomial x² − 13x + 26, sum 13, product 26, and 26/2 = 13.

Example 2. If one zero of 2x² − 8x − m is 5/2, find m and the other zero. Let the other be β. Sum: 5/2 + β = 8/2 = 4, so β = 3/2. Product: (5/2)(3/2) = −m/2, so 15/4 = −m/2 and m = −15/2. Check with p(5/2) = 2(25/4) − 20 − m = 25/2 − 20 + 15/2 = 0.

Example 3. If the zeroes of x² + px + 45 differ by 4 and have (α − β)² = 16... rather: if α − β = 4 with product 45, find p. (α − β)² = (α + β)² − 4αβ, so 16 = p² − 180, p² = 196, p = ±14. The zeroes are then 9 and 5 (p = −14) or −5 and −9 (p = 14).

Example 4. Find k so that the zeroes of kx² + 2x + 3k are equal in magnitude but opposite in sign... note this needs sum zero, which gives −2/k = 0, impossible; so no such k exists. This kind of answer, that no value works, is legitimate and should be stated with the reason.

Example 5. If the sum of the zeroes of kx² + 2x + 3k equals their product, find k. Sum = −2/k, product = 3. So −2/k = 3 gives k = −2/3.

Example 6. If α and β are the zeroes of x² − 6x + k and 3α + 2β = 20, find k. From α + β = 6 we have β = 6 − α; substituting, 3α + 12 − 2α = 20 so α = 8 and β = −2. Then k = αβ = −16. Check: x² − 6x − 16 = (x − 8)(x + 2).

Example 7. If the zeroes of x² − 8x + k are in the ratio 3 : 5, find k. Let the zeroes be 3t and 5t. Sum 8t = 8, t = 1, zeroes 3 and 5, k = 15.

Example 8. If one zero of (a² + 9)x² + 13x + 6a is the reciprocal of the other, find a. Product = 1, so 6a/(a² + 9) = 1, a² − 6a + 9 = 0, (a − 3)² = 0, a = 3.

Always finish by writing the polynomial with the value found and checking that the zeroes really satisfy the condition; a sign slip in −b/a is otherwise undetectable.

Given the zeroes α and β of one quadratic, we are often asked for a quadratic whose zeroes are related to them: 2α and 2β, α + 1 and β + 1, 1/α and 1/β, α² and β², α/β and β/α, or α + β and αβ. The method is always the same: find the new sum and new product in terms of the old sum S and product P, then write x² − (new sum)x + (new product).

Table of transformations. Zeroes kα, kβ: sum kS, product k²P. Zeroes α + t, β + t: sum S + 2t, product P + tS + t². Zeroes 1/α, 1/β: sum S/P, product 1/P. Zeroes α², β²: sum S² − 2P, product P². Zeroes α/β, β/α: sum (S² − 2P)/P, product 1. Zeroes −α, −β: sum −S, product P.

Example 1. α, β are the zeroes of x² − 3x + 2. Find a quadratic whose zeroes are 2α + 1 and 2β + 1. S = 3, P = 2. New sum = 2S + 2 = 8. New product = (2α + 1)(2β + 1) = 4P + 2S + 1 = 8 + 6 + 1 = 15. Polynomial x² − 8x + 15. Check: zeroes 1 and 2 become 3 and 5, and (x − 3)(x − 5) = x² − 8x + 15.

Example 2. α, β are the zeroes of 2x² − 5x + 7. Find a quadratic whose zeroes are 2α + 3β and 3α + 2β. S = 5/2, P = 7/2. New sum = 5α + 5β = 5S = 25/2. New product = (2α + 3β)(3α + 2β) = 6α² + 4αβ + 9αβ + 6β² = 6(α² + β²) + 13αβ = 6(S² − 2P) + 13P = 6(25/4 − 7) + 91/2 = 6(−3/4) + 91/2 = −9/2 + 91/2 = 41. Polynomial x² − (25/2)x + 41, or 2x² − 25x + 82.

Example 3. α, β are the zeroes of x² + 5x + 6. Find a quadratic whose zeroes are 1/α and 1/β. S = −5, P = 6. New sum = S/P = −5/6, new product = 1/6. Polynomial x² + (5/6)x + 1/6, or 6x² + 5x + 1. Note the shortcut: reversing the coefficients of ax² + bx + c gives cx² + bx + a, whose zeroes are the reciprocals.

Example 4. α, β are the zeroes of x² − 2x − 8. Find a quadratic whose zeroes are α² and β². S = 2, P = −8. New sum = 4 + 16 = 20. New product = 64. Polynomial x² − 20x + 64. Check: zeroes 4, −2 give 16 and 4, and (x − 16)(x − 4) = x² − 20x + 64.

Example 5. α, β are the zeroes of x² − 4x + 3. Find a quadratic whose zeroes are α/β and β/α. S = 4, P = 3. New sum = (16 − 6)/3 = 10/3. New product = 1. Polynomial x² − (10/3)x + 1, or 3x² − 10x + 3. Check: zeroes 1 and 3 give 1/3 and 3, and (3x − 1)(x − 3) = 3x² − 10x + 3.

When the original polynomial has simple zeroes, verify by transforming them directly; when it does not, trust the algebra but recompute the new product twice, since that is where arithmetic slips occur.

For a cubic ax³ + bx² + cx + d with zeroes α, β, γ: α + β + γ = −b/a, αβ + βγ + γα = c/a, αβγ = −d/a. Optional problems tell us the zeroes follow a pattern, and the relations then determine them.

Zeroes in arithmetic progression. If the zeroes of x³ − 3x² + x + 1 are a − b, a, a + b, find a and b. Sum: (a − b) + a + (a + b) = 3a = 3, so a = 1. Product: (a − b)(a)(a + b) = a(a² − b²) = 1 − b² = −1, so b² = 2, b = ±√2. The zeroes are 1 − √2, 1, 1 + √2. Check with the middle relation: αβ + βγ + γα = a(a − b) + a(a + b) + (a − b)(a + b) = 2a² + a² − b² = 3a² − b² = 3 − 2 = 1 = c/a. Correct.

Another AP example. The zeroes of x³ − 12x² + 39x − 28 are in AP; find them. Let them be a − d, a, a + d. Sum 3a = 12, a = 4. Product a(a² − d²) = 28, so 16 − d² = 7, d² = 9, d = ±3. Zeroes: 1, 4, 7. Check: 1 + 4 + 7 = 12; 4 + 28 + 7 = 39; 28. All match.

The key insight. When zeroes are in AP, the middle zero is always the mean of the three, that is, −b/(3a). This gives one zero immediately, and dividing by the corresponding linear factor leaves a quadratic for the other two. For x³ − 12x² + 39x − 28, the middle zero is 12/3 = 4, and dividing by (x − 4) gives x² − 8x + 7 = (x − 1)(x − 7).

Two zeroes summing to zero. If two zeroes of x³ − 4x² − 7x + 28 are equal in magnitude and opposite in sign, find all three. Let them be α, −α, γ. Sum: γ = 4. Product: −α² × 4 = −28, so α² = 7, α = ±√7. Zeroes: √7, −√7, 4. Check: (x − 4)(x² − 7) = x³ − 4x² − 7x + 28.

One zero given. If 2 is a zero of x³ − 7x + 6, find the others. Sum of all three is 0 (no x² term), so β + γ = −2. Product is −6, so 2βγ = −6, βγ = −3. Hence β, γ are zeroes of t² + 2t − 3 = (t + 3)(t − 1): −3 and 1. Zeroes 2, 1, −3. This avoids long division.

Zeroes in geometric progression. If the zeroes of x³ − 7x² + 14x − 8 are a/r, a, ar, the product a³ = 8 gives a = 2 as one zero; dividing by (x − 2) leaves x² − 5x + 4 = (x − 1)(x − 4). Zeroes 1, 2, 4, which are indeed in GP.

The recurring strategy: use whichever relation isolates a single unknown, usually the sum for AP and the product for GP, find one zero, then reduce to a quadratic.

A fourth-degree polynomial may have zeroes that come in conjugate pairs a + √b and a − √b. When two such zeroes are given, the quadratic factor they produce is built from their sum and product, and long division by that factor leaves a quadratic for the remaining two zeroes.

Building the factor. For zeroes 2 + √3 and 2 − √3: sum = 4, product = (2 + √3)(2 − √3) = 4 − 3 = 1. Factor: x² − 4x + 1. For zeroes √5 and −√5: sum 0, product −5, factor x² − 5. For zeroes 3 + √2 and 3 − √2: sum 6, product 9 − 2 = 7, factor x² − 6x + 7. For zeroes −1 ± √2: sum −2, product 1 − 2 = −1, factor x² + 2x − 1.

Example 1. Find all zeroes of x⁴ − 6x³ − 26x² + 138x − 35 given that two zeroes are 2 ± √3. Factor: x² − 4x + 1. Division: x⁴ ÷ x² = x², subtract x⁴ − 4x³ + x², remainder so far −2x³ − 27x² + 138x − 35. Next −2x³ ÷ x² = −2x, subtract −2x³ + 8x² − 2x, giving −35x² + 140x − 35. Next −35, subtract −35x² + 140x − 35, giving 0. Quotient x² − 2x − 35 = (x − 7)(x + 5). All zeroes: 2 + √3, 2 − √3, 7, −5. Check sum: 4 + 7 − 5 = 6 = −(−6). Correct.

Example 2. Find all zeroes of 2x⁴ − 9x³ + 5x² + 3x − 1 given that two zeroes are 2 ± √3. Divide by x² − 4x + 1: 2x⁴ ÷ x² = 2x², subtract 2x⁴ − 8x³ + 2x² to get −x³ + 3x² + 3x − 1; −x³ ÷ x² = −x, subtract −x³ + 4x² − x to get −x² + 4x − 1; −1, subtract −x² + 4x − 1 to get 0. Quotient 2x² − x − 1 = (2x + 1)(x − 1). Zeroes: 2 ± √3, −1/2, 1.

Example 3. Find all zeroes of x⁴ + x³ − 9x² − 3x + 18 given that two zeroes are √3 and −√3. Divide by x² − 3: x², subtract x⁴ − 3x² to get x³ − 6x² − 3x + 18; x, subtract x³ − 3x to get −6x² + 18; −6, subtract −6x² + 18 to get 0. Quotient x² + x − 6 = (x + 3)(x − 2). Zeroes: √3, −√3, −3, 2.

Example 4. Find all zeroes of x⁴ − 3x³ − x² + 9x − 6 given that two zeroes are ±√3. Divide by x² − 3: quotient x² − 3x + 2 = (x − 1)(x − 2). Zeroes: √3, −√3, 1, 2.

Why conjugates come together. If a polynomial has rational coefficients and a + √b is a zero with √b irrational, then a − √b is also a zero. This is why questions give the pair, and why the factor built from them has rational coefficients, so that the division works out with no surds at all. If the quotient comes out with surds, an arithmetic error has occurred.

Write the division in columns with a zero placeholder for any missing power, and end with the sum-of-zeroes check against −b/a, which catches most sign errors.

The most demanding optional problem divides a quartic by a quadratic that contains an unknown k and states that the remainder is a given linear expression involving another unknown a. Comparing the remainder obtained by division with the given remainder yields two equations, from which both unknowns are found.

Example 1. If x⁴ − 6x³ + 16x² − 25x + 10 is divided by x² − 2x + k, the remainder is x + a. Find k and a.

Divide step by step, keeping k as a symbol. Step 1: x⁴ ÷ x² = x². Multiply: x⁴ − 2x³ + kx². Subtract: −4x³ + (16 − k)x² − 25x + 10. Step 2: −4x³ ÷ x² = −4x. Multiply: −4x³ + 8x² − 4kx. Subtract: (16 − k − 8)x² + (−25 + 4k)x + 10 = (8 − k)x² + (4k − 25)x + 10. Step 3: (8 − k)x² ÷ x² = 8 − k. Multiply: (8 − k)x² − 2(8 − k)x + k(8 − k). Subtract: [(4k − 25) + 2(8 − k)]x + [10 − k(8 − k)] = (2k − 9)x + (10 − 8k + k²).

So the remainder is (2k − 9)x + (k² − 8k + 10). This must equal x + a. Comparing coefficients of x: 2k − 9 = 1, so k = 5. Comparing constants: a = k² − 8k + 10 = 25 − 40 + 10 = −5. Answer: k = 5, a = −5. The quotient is x² − 4x + 3.

Verification. (x² − 2x + 5)(x² − 4x + 3) + (x − 5) = x⁴ − 4x³ + 3x² − 2x³ + 8x² − 6x + 5x² − 20x + 15 + x − 5 = x⁴ − 6x³ + 16x² − 25x + 10. Correct.

Example 2. If x³ + 2x² + kx + 3 divided by x − 3 leaves remainder 21, find k. Here the divisor is linear, so the remainder theorem applies: p(3) = 27 + 18 + 3k + 3 = 48 + 3k = 21, so k = −9. The quotient is then x² + 5x + 6.

Example 3. Find k so that x² + 2x + k is a factor of 2x⁴ + x³ − 14x² + 5x + 6. Divide symbolically: 2x², subtract 2x⁴ + 4x³ + 2kx², leaving −3x³ + (−14 − 2k)x² + 5x + 6; −3x, subtract −3x³ − 6x² − 3kx, leaving (−8 − 2k)x² + (5 + 3k)x + 6; (−8 − 2k), subtract (−8 − 2k)x² + 2(−8 − 2k)x + k(−8 − 2k), leaving [5 + 3k + 16 + 4k]x + [6 + 8k + 2k²] = (7k + 21)x + (2k² + 8k + 6). For a factor, both must be zero: 7k + 21 = 0 gives k = −3, and 2(9) − 24 + 6 = 0 confirms. So k = −3 and the factor is x² + 2x − 3 = (x + 3)(x − 1); the quotient is 2x² − 3x − 2 = (2x + 1)(x − 2), so the zeroes of the quartic are −3, 1, −1/2, 2.

The discipline needed is to keep every term in k, to bracket carefully when subtracting, and to compare the two coefficients of the remainder separately. Write the general remainder in the form (linear in x) before comparing; this form is what carries the marks.

If p(x) divided by g(x) leaves remainder r(x), then p(x) − r(x) is exactly divisible by g(x), because p − r = gq. So the polynomial that must be subtracted to make the division exact is the remainder itself, and the polynomial that must be added is −r(x), or equivalently g(x) − r(x) when a positive-looking expression of smaller degree than g is wanted.

Example 1. What must be subtracted from x⁴ + 2x³ − 13x² − 12x + 21 so that the result is exactly divisible by x² − 4x + 3? Divide: x², subtract x⁴ − 4x³ + 3x² to get 6x³ − 16x² − 12x + 21; 6x, subtract 6x³ − 24x² + 18x to get 8x² − 30x + 21; 8, subtract 8x² − 32x + 24 to get 2x − 3. Remainder 2x − 3. So 2x − 3 must be subtracted. Check: the result x⁴ + 2x³ − 13x² − 14x + 24 = (x² − 4x + 3)(x² + 6x + 8).

Example 2. What must be added to x⁴ + 2x³ − 2x² + x − 1 so that the result is exactly divisible by x² + 2x − 3? Divide: x², subtract x⁴ + 2x³ − 3x² to get x² + x − 1; 1, subtract x² + 2x − 3 to get −x + 2. Remainder −x + 2. To make the division exact, add −(−x + 2) = x − 2. Check: x⁴ + 2x³ − 2x² + 2x − 3 = (x² + 2x − 3)(x² + 1).

Example 3. What must be subtracted from 4x⁴ − 2x³ − 6x² + x − 5 so that the result is exactly divisible by 2x² + x − 2? Divide: 2x², subtract 4x⁴ + 2x³ − 4x² to get −4x³ − 2x² + x − 5; −2x, subtract −4x³ − 2x² + 4x to get −3x − 5. Since the degree is now 1, less than 2, stop. Remainder −3x − 5. Subtract −3x − 5, that is, add 3x + 5.

Example 4. Finding the divisor. On dividing 2x³ + 4x² + 5x + 7 by g(x), the quotient is 2x and the remainder 7 − 5x. Find g(x). g(x) = (p − r)/q = (2x³ + 4x² + 5x + 7 − 7 + 5x)/(2x) = (2x³ + 4x² + 10x)/(2x) = x² + 2x + 5.

Example 5. Finding the dividend. A polynomial divided by x² − 1 gives quotient x + 3 and remainder 2x − 5. Find the polynomial. p = gq + r = (x² − 1)(x + 3) + 2x − 5 = x³ + 3x² − x − 3 + 2x − 5 = x³ + 3x² + x − 8.

Read the question carefully for the word added or subtracted: subtract the remainder, add its negative. A final expansion of g × q confirms the answer and should be shown.

A graph of y = ax² + bx + c reveals the signs of a, b and c, and the optional exercise asks for this reading. Three observations do the work.

Sign of a. If the parabola opens upwards (a cup), a > 0. If it opens downwards (a cap), a < 0.

Sign of c. The y-intercept is the value at x = 0, which is c. If the curve cuts the y-axis above the origin, c > 0; below, c < 0; through the origin, c = 0.

Sign of b. The vertex is at x = −b/(2a). If the vertex is to the right of the y-axis, −b/(2a) > 0, so b and a have opposite signs. If the vertex is to the left, b and a have the same sign. If the vertex is on the y-axis, b = 0. Equivalently, look at the sum of the zeroes −b/a: both zeroes positive means −b/a > 0.

Example 1. A parabola opens upwards, cuts the y-axis below the origin, and has both zeroes positive. Then a > 0; c < 0; sum of zeroes positive means −b/a > 0, so b < 0. Product of zeroes c/a should be positive for two positive zeroes, but c/a < 0 here: contradiction. So the description is impossible; with c < 0 and a > 0 the zeroes have opposite signs. Such consistency checks are part of the exercise.

Example 2. A parabola opens downwards, y-intercept positive, vertex to the right of the y-axis. a < 0, c > 0, and vertex right means b and a have opposite signs, so b > 0. Example polynomial: −x² + 2x + 3, vertex at x = 1, zeroes −1 and 3.

Example 3. Sketch y = x² − 6x + 8 and confirm the reading. a = 1 > 0 (opens up). c = 8 > 0 (cuts y-axis above origin). Vertex at x = 3 (right of y-axis), so b < 0, consistent with b = −6. Zeroes 2 and 4, both positive, product 8 = c/a, sum 6 = −b/a.

Both zeroes negative. Then sum −b/a < 0 and product c/a > 0. With a > 0 this means b > 0 and c > 0: all coefficients positive. Example: x² + 5x + 6 with zeroes −2, −3. So a quadratic with all positive coefficients (a, b, c > 0) cannot have a positive zero, a neat fact: if x > 0 then ax² + bx + c > 0.

Zeroes of opposite sign. Then product c/a < 0, so a and c have opposite signs; the y-intercept and the direction of opening disagree. Example: x² − x − 6 with zeroes 3 and −2.

Zeroes equal. The vertex touches the x-axis; algebraically b² = 4ac, which the next chapter studies as the discriminant.

Number of zeroes and the vertex. For a > 0: two zeroes if the vertex is below the x-axis, one if on it, none if above. For a < 0 the reverse. So the sign of the y-coordinate of the vertex, c − b²/(4a), combined with the sign of a, decides the number of real zeroes.

These readings appear as short reasoning questions: given a sketch, state the signs of a, b, c with reasons, or given the signs, sketch a possible graph. Justify each sign with one of the three observations.

Some optional problems are proofs: show that a stated relation between the coefficients follows from a condition on the zeroes, or vice versa. They are handled by writing the condition in terms of α + β and αβ and substituting the coefficient forms.

Problem 1. If the zeroes of ax² + bx + c are reciprocals of each other, show that a = c. Reciprocal zeroes mean αβ = 1. But αβ = c/a, so c/a = 1 and c = a.

Problem 2. If one zero of ax² + bx + c is the negative of the other, show that b = 0. Then α + β = 0, so −b/a = 0, so b = 0. The polynomial has the form ax² + c with zeroes ±√(−c/a), which are real only when a and c have opposite signs.

Problem 3. If the zeroes of x² + px + q are α and β, show that the zeroes of x² − px + q are −α and −β. For the second polynomial the sum of zeroes is p and the product q. Now (−α) + (−β) = −(α + β) = −(−p) = p and (−α)(−β) = αβ = q. So −α and −β have the right sum and product; they are its zeroes.

Problem 4. If α, β are the zeroes of ax² + bx + c, show that a(α² + β²) = (b² − 2ac)/a. α² + β² = (α + β)² − 2αβ = b²/a² − 2c/a = (b² − 2ac)/a². Multiplying by a gives (b² − 2ac)/a.

Problem 5. Show that if the zeroes of ax² + bx + c are α and β, then (α − β)² = (b² − 4ac)/a². (α − β)² = (α + β)² − 4αβ = b²/a² − 4c/a = (b² − 4ac)/a². Hence the zeroes are equal exactly when b² − 4ac = 0, and they are real exactly when b² − 4ac ≥ 0, since a square of a real number cannot be negative. This is the algebraic form of the three-cases picture and is the discriminant of the next chapter.

Problem 6. If α, β are the zeroes of x² − p(x + 1) − c, show that (α + 1)(β + 1) = 1 − c. Rewrite the polynomial: x² − px − p − c, so α + β = p and αβ = −p − c. Then (α + 1)(β + 1) = αβ + (α + β) + 1 = −p − c + p + 1 = 1 − c.

Problem 7. If the zeroes of x² + bx + c and x² + cx + b (b ≠ c) have exactly one common zero, find it. Let the common zero be t: t² + bt + c = 0 and t² + ct + b = 0. Subtracting: (b − c)t + (c − b) = 0, so (b − c)(t − 1) = 0, and since b ≠ c, t = 1. Then substituting, 1 + b + c = 0, so b + c = −1 is the condition.

Problem 8. Show that x² + x + 1 has no real zero. The product of zeroes would be 1 and the sum −1, so (α − β)² = 1 − 4 = −3 < 0, impossible for real α, β. Alternatively, x² + x + 1 = (x + 1/2)² + 3/4 ≥ 3/4 > 0.

Each proof takes three or four lines. The marks are for the correct translation of the condition into sum and product and for the clean algebra that follows; there is no need for any numerical work.

The three relations for a cubic support the same kind of symmetric computation as for a quadratic, and the construction of a cubic from information about its zeroes.

Symmetric expressions. With α + β + γ = S₁, αβ + βγ + γα = S₂, αβγ = S₃: α² + β² + γ² = S₁² − 2S₂. 1/α + 1/β + 1/γ = S₂/S₃. 1/(αβ) + 1/(βγ) + 1/(γα) = S₁/S₃. α³ + β³ + γ³ = S₁³ − 3S₁S₂ + 3S₃.

Example 1. For the zeroes of x³ − 6x² + 11x − 6: S₁ = 6, S₂ = 11, S₃ = 6. α² + β² + γ² = 36 − 22 = 14 (check: 1 + 4 + 9 = 14). 1/α + 1/β + 1/γ = 11/6 (check: 1 + 1/2 + 1/3 = 11/6). α³ + β³ + γ³ = 216 − 198 + 18 = 36 (check: 1 + 8 + 27 = 36).

Example 2. For the zeroes of 2x³ + x² − 5x + 2: S₁ = −1/2, S₂ = −5/2, S₃ = −1. α² + β² + γ² = 1/4 + 5 = 21/4. 1/α + 1/β + 1/γ = (−5/2)/(−1) = 5/2. Check: the zeroes are 1, 1/2, −2: 1 + 1/4 + 4 = 21/4; 1 + 2 − 1/2 = 5/2.

Constructing a cubic. With zeroes 2, −1 and 1/2: S₁ = 3/2, S₂ = −2 − 1/2 + 1 = −3/2, S₃ = −1. Polynomial x³ − (3/2)x² − (3/2)x + 1, or 2x³ − 3x² − 3x + 2. Check: (x − 2)(x + 1)(2x − 1) = (x² − x − 2)(2x − 1) = 2x³ − x² − 2x² + x − 4x + 2 = 2x³ − 3x² − 3x + 2.

A cubic with transformed zeroes. If α, β, γ are the zeroes of x³ − 6x² + 11x − 6, find a cubic whose zeroes are 2α, 2β, 2γ. New S₁ = 12, S₂ = 4 × 11 = 44, S₃ = 8 × 6 = 48. Polynomial x³ − 12x² + 44x − 48. Check: zeroes 2, 4, 6: (x − 2)(x − 4)(x − 6) = x³ − 12x² + 44x − 48. In general, scaling zeroes by k scales S₁, S₂, S₃ by k, k², k³.

Reciprocal zeroes. A cubic whose zeroes are the reciprocals of those of ax³ + bx² + cx + d is dx³ + cx² + bx + a, the coefficients reversed. For x³ − 6x² + 11x − 6, the reciprocal polynomial is −6x³ + 11x² − 6x + 1, with zeroes 1, 1/2, 1/3.

Verifying a given set of zeroes. Are 1/2, 1, −2 the zeroes of 2x³ + x² − 5x + 2? Substitute each: p(1/2) = 1/4 + 1/4 − 5/2 + 2 = 0; p(1) = 2 + 1 − 5 + 2 = 0; p(−2) = −16 + 4 + 10 + 2 = 0. Yes. Then confirm the three relations as a further check.

Write S₁, S₂, S₃ with their signs from the polynomial before any computation; the sign of S₃, which is −d/a, is the one most often written wrongly.

The optional exercise includes graph questions beyond the quadratic: reading the number of zeroes of a cubic or quartic from its picture, matching a graph to a polynomial, and sketching a cubic from its zeroes.

Counting zeroes from a picture. The number of zeroes equals the number of distinct points where the curve meets the x-axis, whether crossing or touching. A curve that crosses at −2, touches at 1 and crosses at 3 has three zeroes, −2, 1, 3, and the polynomial has degree at least 4 because the touch counts double: (x + 2)(x − 1)²(x − 3) is the simplest. A curve crossing the axis once and having no other contact could be a cubic such as x³ + x or a linear polynomial; the end behaviour distinguishes them: a cubic has ends going in opposite vertical directions with a bend, a line has none.

Matching a graph to a polynomial. Given the graphs of y = x³ − 4x, y = x³ and y = x³ − x², identify each. x³ − 4x = x(x − 2)(x + 2) crosses at −2, 0, 2. x³ passes through the origin, flat there, and nowhere else. x³ − x² = x²(x − 1) touches at 0 and crosses at 1. So count and classify the contacts to match.

Sketching from zeroes. Sketch y = (x + 1)(x − 2)(x − 4). Zeroes −1, 2, 4. Leading coefficient positive, so the curve comes from the bottom left and leaves to the top right. y-intercept: (1)(−2)(−4) = 8. Between −1 and 2 the curve is above the axis (test x = 0: y = 8), between 2 and 4 it is below (test x = 3: (4)(1)(−1) = −4), and beyond 4 it is above. The sketch has a hump between −1 and 2 and a dip between 2 and 4.

Sign pattern. For a polynomial with distinct real zeroes, the sign of y changes at each zero where the curve crosses, and does not change at a zero where it touches (even multiplicity). This lets you fill in the sign of y on every interval with a single test point.

End behaviour. For degree n with leading coefficient a: if n is even, both ends go the same way (up if a > 0, down if a < 0), like a parabola. If n is odd, the ends go opposite ways (bottom left to top right if a > 0, top left to bottom right if a < 0), like a cubic. So an even-degree polynomial can miss the x-axis altogether, while an odd-degree one must cross it at least once.

Example: how many zeroes? y = x⁴ − 5x² + 4 = (x² − 1)(x² − 4) = (x − 1)(x + 1)(x − 2)(x + 2): four zeroes, ±1, ±2, and the graph is a W shape crossing four times. y = x⁴ + 1: never zero, the graph stays above the axis. y = x⁴ − 2x² + 1 = (x² − 1)²: touches at ±1, two zeroes.

In an examination sketch, mark the zeroes and the y-intercept, indicate the end behaviour with arrows, and show one test point per interval. A sketch is not a plot; accuracy at the zeroes and the correct shape earn the marks.

This closing section maps each type of optional problem to its method, so that an unseen question can be classified in seconds.

The question mentions α and β and an expression in them. Write S = −b/a and P = c/a, express the target through the standard identities, substitute. Do not solve for the zeroes unless the question asks for them.

The question has a parameter k and a condition on the zeroes. Translate the condition into S and P, which contain k, and solve the resulting equation. Ratio conditions use zeroes mt and nt; reciprocal conditions use P = 1; equal-and-opposite uses S = 0; a linear condition such as 3α + 2β = 20 is combined with α + β = S to find α and β, then k from P.

The question asks for a new polynomial whose zeroes are related to the old. Compute the new sum and product from S and P and write x² − (new S)x + (new P). Reciprocals: reverse coefficients. Scaled: S becomes kS, P becomes k²P. Shifted by t: S + 2t and P + tS + t². Squared: S² − 2P and P².

The polynomial is cubic and the zeroes follow a pattern. AP: middle zero = −b/(3a). GP: middle zero = cube root of −d/a. Two equal and opposite: third zero = −b/a. One zero given: use sum and product to get a quadratic for the other two. Then reduce and factorise.

The polynomial is quartic and two surd zeroes are given. Build x² − (sum)x + (product), divide, factorise the quotient, check with the sum of all zeroes.

The divisor contains k and the remainder is given. Divide symbolically, express the remainder as (linear in k)x + (quadratic in k), compare coefficients with the given remainder, solve. If instead the question says the divisor is a factor, set both remainder coefficients to zero.

The question asks what to add or subtract for exact divisibility. Divide, take the remainder r; subtract r, or add −r. Recover a divisor by g = (p − r)/q and a dividend by p = gq + r.

The question shows a graph. a from the opening direction, c from the y-intercept, b from the side of the vertex; the number of zeroes from x-axis contacts; the degree from the number of bends and the end behaviour.

The question is a proof. Express everything through S and P, or through S₁, S₂, S₃ for a cubic, and manipulate; end with the required statement written out in full.

Two safeguards. After every problem, substitute the answer back into the original condition. And write each relation with its sign before substituting: −b/a, c/a; −b/a, c/a, −d/a. Most lost marks on this chapter are sign errors made in a hurry rather than misunderstanding.

Where this leads. The identities for α and β become the tools of the Quadratic Equations chapter; the pattern-zero problems reappear in Progressions; the division algorithm underlies the factor theorem used throughout Intermediate algebra; and the sign reading of graphs is the beginning of curve sketching in calculus.