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Roots of polynomials questions
A polynomial's coefficients determine the symmetric functions of its roots. Their sum, their product and other symmetric combinations can all be found without solving the equation.
7 original questions · 30 marks · the roots of polynomials notes · Further algebra and series
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The equation 2z³ + 6z² − 4z + 10 = 0 has roots α, β, γ. Write down α + β + γ and αβγ.
Worked 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. Skipping the divide is the standard way to lose both marks.The quadratic z² − 6z + 4 = 0 has roots α and β. Without solving, find α + β, αβ and 1/α + 1/β.
Worked 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. Combine over a common denominator and both known quantities appear.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.
Worked 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, with the symmetric functions doing all the work. M1 for (α + β)² − 2αβ, A1 for 28, B1 for α²β² = 16, A1 for the equation.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.
Worked 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. Both relations hold, a worthwhile check on the factorising.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.
Worked 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.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.
Worked 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. The relations run both ways, from roots to coefficients as directly as from coefficients to roots.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.
Worked 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, but the marks here are for the symmetric-function route. Squaring each coefficient of the original cubic is a common and completely invalid shortcut.
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