Questions on roots of polynomials

Question 1

The equation 2z³ + 6z² − 4z + 10 = 0 has roots α, β, γ. Write down α + β + γ and αβγ.

Accepted Answer

Divide by 2 first: z³ + 3z² − 2z + 5 = 0. Then α + β + γ = −3 and αβγ = −5. B1 for the sum of the roots, B1 for the product. The formulae carry a division by the leading coefficient, which is not 1 here.

Question 2

The quadratic z² − 6z + 4 = 0 has roots α and β. Without solving, find α + β, αβ and 1/α + 1/β.

Accepted Answer

α + β = 6 and αβ = 4 straight from the coefficients. Then 1/α + 1/β = (α + β)/(αβ) = 6/4 = 3/2. B1 for the sum and product, M1 for the common denominator, A1 for 3/2.

Question 3

The equation z⁴ + z³ − 7z² − z + 6 = 0 has roots α, β, γ, δ. Write down α + β + γ + δ, αβ + αγ + αδ + βγ + βδ + γδ, and αβγδ.

Accepted Answer

For a quartic the signs continue to alternate: α + β + γ + δ = −1, the pairwise sum is −7, and αβγδ = 6. B1 for the sum, B1 for the pairwise sum, B1 for the product. The triple sum αβγ + αβδ + αγδ + βγδ, if asked for, would be the z-coefficient with its sign flipped again: −(−1) = 1.

Question 4

The quadratic 2z² − 5z − 3 = 0 has roots α and β, with α > β. Find (α − β)² and hence α − β, both exactly.

Accepted Answer

Dividing by 2, α + β = 5/2 and αβ = −3/2. (α − β)² = (α + β)² − 4αβ = 25/4 + 6 = 49/4, so α − β = 7/2, taking the positive root since α > β. M1 for the identity (α + β)² − 4αβ, A1 for 49/4, A1 for 7/2. This identity isolates the difference directly, without going through the sum of squares α² + β².

Question 5

The quadratic z² − 6z + 4 = 0 has roots α and β. Find α² + β², and hence write down a quadratic equation with roots α² and β².

Accepted Answer

α² + β² = (α + β)² − 2αβ = 36 − 8 = 28, and α²β² = (αβ)² = 16. A quadratic with those roots is z² − 28z + 16 = 0, that is z² minus (sum)z plus product. M1 for (α + β)² − 2αβ, A1 for 28, B1 for α²β² = 16, A1 for the equation.

Question 6

Verify that z = 2 is a root of z³ − 4z² + z + 6 = 0, find the other two roots, and check the sum and product against the coefficients.

Accepted Answer

2³ − 16 + 2 + 6 = 0, so z − 2 is a factor and the cubic is (z − 2)(z² − 2z − 3) = (z − 2)(z − 3)(z + 1), roots 2, 3, −1. Sum: 4 = −(−4); product: −6 = −6/1. M1 for substituting z = 2, A1 for the value zero, M1 for factorising the cubic, A1 for the other two roots, B1 for the check.

Question 7

The equation z² − 6z + 4 = 0 has roots α, β. Find a quadratic equation with integer coefficients whose roots are α + 1 and β + 1.

Accepted Answer

Substitute w = z + 1, so z = w − 1: (w − 1)² − 6(w − 1) + 4 = w² − 8w + 11 = 0. Alternatively the new sum is 6 + 2 = 8 and the new product αβ + α + β + 1 = 4 + 6 + 1 = 11. Both routes give z² − 8z + 11 = 0. M1 for the substitution w = z + 1, dM1 for expanding, A1 for the new sum 8, A1 for the new product 11.

Question 8

A cubic with real coefficients has roots that sum to 0 and multiply to −8, and the sum of its pairwise products is −2. Write down the cubic, taking the leading coefficient as 1.

Accepted Answer

Read the coefficients off with alternating signs, as z³ − (sum)z² + (pairs)z − (product). That gives z³ + 0z² − 2z + 8, that is z³ − 2z + 8 = 0. B1 for the alternating signs, M1 for inserting the three values, A1 for the cubic.

Question 9

The cubic z³ − 3z² − 6z + 8 = 0 has roots α, β, γ. Without solving the equation, find the value of (3 + α)(3 + β)(3 + γ).

Accepted Answer

Expanding the product in terms of the elementary symmetric functions: (3 + α)(3 + β)(3 + γ) = 27 + 9(α + β + γ) + 3(αβ + αγ + βγ) + αβγ. M1 for expanding the triple product into the four symmetric terms. From the coefficients, α + β + γ = 3, αβ + αγ + βγ = −6 and αβγ = −8. B1 for the three symmetric functions read off correctly. Substituting these in: 27 + 9(3) + 3(−6) + (−8) = 27 + 27 − 18 − 8. M1 for the substitution. So the value is 28. A1 for 28. The actual roots are 1, −2 and 4, and 4 × 1 × 7 = 28 confirms it.

Question 10

The cubic z³ − 2z² − 5z + 6 = 0 has roots α, β, γ. Without solving the equation, find 1/α² + 1/β² + 1/γ², giving your answer as an exact fraction.

Accepted Answer

From the coefficients, α + β + γ = 2, αβ + αγ + βγ = −5 and αβγ = −6. First find the two reciprocal symmetric functions: 1/α + 1/β + 1/γ = (αβ + αγ + βγ)/(αβγ) = (−5)/(−6) = 5/6, and 1/(αβ) + 1/(αγ) + 1/(βγ) = (α + β + γ)/(αβγ) = 2/(−6) = −1/3. Then square the first and subtract twice the second, exactly as for α² + β² + γ² earlier in this lesson: 1/α² + 1/β² + 1/γ² = (5/6)² − 2(−1/3) = 25/36 + 24/36 = 49/36. M1 for both reciprocal symmetric functions with correct signs, A1 for both values, M1 for the identity (Σ1/α)² − 2Σ1/(αβ), A1 for the substitution, A1 for 49/36.

Question 11

Given that 1 + 2i is a root of z³ − 4z² + 9z − 10 = 0, find all three roots, and check the sum of the roots against the coefficients.

Accepted Answer

The coefficients are real, so the conjugate 1 − 2i is a root too. Sum of roots = 4, and (1 + 2i) + (1 − 2i) = 2, so the third root is 2. The roots are 2, 1 + 2i, 1 − 2i. M1 for stating the conjugate root, M1 for using the sum of roots, A1 for the third root, A1 for all three roots stated. Check: sum = 2 + 1 + 2i + 1 − 2i = 4, which matches −(−4). B1 for the check. Real coefficients are essential here: without that condition 1 − 2i need not be a root at all.

Question 12

The equation z³ − 4z² + z + 6 = 0 has roots α, β and γ. Without solving the equation, find α² + β² + γ² and α³ + β³ + γ³, and hence find a cubic equation with integer coefficients whose roots are α², β² and γ².

Accepted Answer

From the coefficients, Σα = 4, Σαβ = 1 and αβγ = −6. Squaring the first, (Σα)² = Σα² + 2Σαβ, so Σα² = 16 − 2 = 14. For the cubes, use the equation itself. Each root satisfies α³ = 4α² − α − 6, so summing over the three roots gives Σα³ = 4Σα² − Σα − 18 = 56 − 4 − 18 = 34. For the new cubic, its three symmetric functions are Σα² = 14, Σα²β² = (Σαβ)² − 2(Σα)(αβγ) = 1 + 48 = 49, and α²β²γ² = (αβγ)² = 36. So the equation is z³ − 14z² + 49z − 36 = 0. B1 for the three symmetric functions, M1 for squaring the sum, A1 for 14, M1 for using the equation on each root, A1 for 34, M1 for Σα²β², A1 for 49, B1 for the product 36, A1 for the cubic. The actual roots are 2, 3 and −1, whose squares 4, 9 and 1 do satisfy it. Squaring each coefficient of the original cubic is a common and completely invalid shortcut.

Question 13

Given that 1 + i is a root of z⁴ − 7z³ + 18z² − 22z + 12 = 0, find all four roots.

Accepted Answer

The coefficients are real, so 1 − i is a root too, and (z − (1 + i))(z − (1 − i)) = z² − 2z + 2 is a quadratic factor. Dividing the quartic by z² − 2z + 2: the quotient starts z², since z⁴ ÷ z² = z², and z²(z² − 2z + 2) = z⁴ − 2z³ + 2z². Subtracting leaves −5z³ + 16z² − 22z + 12, then −5z³ ÷ z² gives the next term −5z, and −5z(z² − 2z + 2) = −5z³ + 10z² − 10z, leaving 6z² − 12z + 12. Finally 6z² ÷ z² = 6, and 6(z² − 2z + 2) = 6z² − 12z + 12 exactly, so the division is exact with quotient z² − 5z + 6. M1 for the quadratic factor from the conjugate pair, M1 for the polynomial division, A1 for the quotient z² − 5z + 6. Factorising the quotient, z² − 5z + 6 = (z − 2)(z − 3), giving the last two roots. M1 for factorising, A1 for both real roots. The four roots are 1 + i, 1 − i, 2, 3. A1 for stating all four together. As a check, the sum of all four is 7, which matches −(−7).

Question 14

Prove that α³ + β³ + γ³ − 3αβγ = (α + β + γ)(α² + β² + γ² − αβ − βγ − γα), and use it, together with the coefficients of z³ + 2z² − 5z − 6 = 0, to find α³ + β³ + γ³ for its roots α, β, γ.

Accepted Answer

Expand the right-hand side: (α + β + γ)(α² + β² + γ² − αβ − βγ − γα) = α³ + αβ² + αγ² + α²β + β³ + βγ² + α²γ + β²γ + γ³ − α²β − αβ² − αβγ − αβγ − β²γ − βγ² − α²γ − αγ² − αβγ, and every term outside α³, β³, γ³ and −3αβγ cancels in pairs, leaving the left-hand side exactly. M1 for expanding the product, A1 for the identity confirmed. From the coefficients, α + β + γ = −2, αβ + αγ + βγ = −5 and αβγ = −(−6)/1 = 6. B1 for the three symmetric functions. Then α² + β² + γ² = (−2)² − 2(−5) = 14. Substituting into the identity: α³ + β³ + γ³ − 3(6) = (−2)(14 − (−5)) = (−2)(19) = −38, so α³ + β³ + γ³ = −38 + 18 = −20. M1 for Σα² = 14, M1 for substituting into the proved identity, A1 for −20. The actual roots are 2, −1 and −3, and 8 − 1 − 27 = −20 confirms it.

Question 15

The cubic z³ − 7z + 6 = 0 has roots α, β, γ, none of them zero. By substituting z = 1/w, find a cubic equation with integer coefficients whose roots are 1/α, 1/β and 1/γ.

Accepted Answer

No root is zero, since z = 0 gives 6 ≠ 0, so every reciprocal exists. B1 for identifying that no root is zero as the condition the substitution needs. Substitute z = 1/w: 1/w³ − 7/w + 6 = 0. M1 for the substitution z = 1/w, A1 for the equation in w. Multiplying through by w³ clears every fraction: 6w³ − 7w² + 1 = 0. M1 for multiplying by w³, A1 for the cleared cubic. In general the substitution reverses the order of the coefficients whenever no root is zero. B1 for the check: the actual roots are 1, 2 and −3, so the reciprocal roots are 1, 1/2 and −1/3, which sum to 7/6, matching −(−7)/6 = 7/6 from the new cubic.