Preamble
Firstly, let’s make sure that we are absolutely clear about what a POLYNOMIAL is.
In this topic, instead of just solving polynomials, we are going to look at some of the relationships between their solutions, in order to tackle more different problems. We will work our way up from quadratic equations through to quartic equations, but the logic applied will be pretty consistent throughout.
Part 1: Quadratic Equations
Consider the general quadratic equation: ax^{2}+bx+c=0
Without losing generality we can rewrite this as: (1), which is a more useful form, because it only has two unknowns instead of three.
Let’s call the roots (i.e. solutions) of this equation 𝝰 and β. In factorised form, the equation could then be written: (x𝝰)(xβ)=0 (2)
If we compare (1) and (2) above, we get:
, so
.
We have seen this before when dealing with quadratic equation, that the product of the roots is the constant and the negative of the sum of the roots is the coefficient of x. (technically this is known as Vieta’s Theorem, or at least the Quadratic instance of it).
New notation
Let’s start to use a new notation, which will become especially useful when we apply our thinking to polynomials with a higher degree:
Σ𝝰 = 𝝰+β – we will use this for the sum of all of the roots; regardless of how many there are
Σ𝝰β = 𝝰β – this represents the sum of the products of the roots, taken 2 at a time
Applying this new notation to quadratic equations above, we have:
= Σ𝝰 and = Σ𝝰β.
Let’s use the above results to derive Σ𝝰^{2} and Σ(1/𝝰). As with most of our Further Mathematics course, we get greatest benefit by learning both the formulae and the method for deriving them.
We should also know the notation S_{n}. This means Σ𝛼^{n}.
Recursive method
Suppose that we know equation f(x)=2x^{2}+3x2 has roots 𝝰 and β and we want to know Σ𝝰^{3}. We could use a very similar method to the one we used above to find Σ𝝰^{2}. We are going to learn a different method though, that becomes particularly useful when we start dealing with higher powers.
First, let’s multiply the whole equation through by x to give: 2x^{3}+3x^{2}2x=0 (we are doing this because the original equation didn’t have a cubic term and we are interested in cubes).
As we already know that 𝝰 and β are roots, so we know that:
2𝝰^{3}+3𝝰^{2}2𝝰=0 (3) and
2β^{3}+3β^{2}2β=0 (4).
Adding together (3) and (4) gives:
2Σ𝝰^{3}+3Σ𝝰^{2}2Σ𝝰=0, which we can easily solve for Σ𝝰^{3}.
Worked Example
The roots of the equation 3z^{2}4z1=0 are 𝝰 and β. Find the quadratic equation with roots 𝝰+1 and β+1.
Exercise 1
 Each of the following quadratic equations has roots 𝛼 and β. Find the values of 𝛼 + β and 𝛼β.
 x^{2}+ 5x + 9 = 0
 x^{2} – 4x + 8 = 0
 2x^{2} + 3x – 7 = 0
 Given that 3x^{2} + 4x + 12 = 0 has roots 𝛼 and β, find:
 𝛼 + β and 𝛼β
 𝛼^{2} + β^{2}
 x^{2} – (2+p)x + (7+p) = 0 has roots that differ by 1. Find the value of p given that p>0
 If a + b = 3 and a^{2} + b^{2} = 7, find the value of ab and hence write down a quadratic equation with roots a and b.
 If x^{2} + bx + c = 0 has roots 𝛼 and β, prove that:
 If 𝛼 = 3β then
 If 𝛼 = β2, then b^{2} = 4(c+1)
 You are given the quadratic equation px^{2} + qx – 16 = 0, which has roots 𝛼 and β. Given also that and 𝛼β = 8, find the values of p and q.
 The quadratic equation x^{2} + 2x – 6 has roots 𝛼 and β. Find the values of (𝛼 – β)^{2} and
 A quadratic equation has roots 𝛼 and β. Given that and 𝛼^{2} + β^{2} = 12, find two possible quadratic equations that satisfy these values (both of which have 1 as the leading coefficient).
 The quadratic equation 3x^{2} + 2x – 4 = 0 has roots 𝛼 and β. Find the values of S_{1}, S_{2} and S_{1}.
 You are given the quadratic equation 4x^{2} – x + 6 = 0 which has roots 𝛼 and β.
 Find 𝛼^{2} + β^{2}
 Without solving the quadratic equation, state what your answer above tells you about the roots.
Answers to Exercise 1

 𝛼 + β = 5 𝛼β = 9
 𝛼 + β = 4 𝛼β = 8
 𝛼 + β = 3/2 𝛼β = 7/2
 5
 The value of ab = 1.x^{2} + 3x + 1 = 0
 p=2, q=1
 28 and 4/9
 x^{2} + 2x – 4 = 0 and x^{2} – 6x + 12 = 0
 S_{1} = 2/3 , S_{2} = 28/9 , S_{1} = 1/2

 47/16 (or )
 They must be complex roots, because 𝛼^{2} + β^{2} < 0 (squares of real numbers are always positive)
Part 2: Cubic Equations
These follow exactly the same principle as quadratics.
So if the cubic equation ax^{3}+bx^{2}+cx+d=0 has roots 𝝰, β and ɣ, then:
Σ𝝰 =
Σ𝝰β =
Σ𝝰βɣ =
(you can follow the process that we followed above for quadratic equations to check that you are confident with this)
Also, as with quadratics, we can find Σ𝝰^{2} by expanding (𝝰+β+ɣ)^{2}. Let’s do that:
(𝝰+β+ɣ)(𝝰+β+ɣ) = …
Worked Example
Exercise 2
 Each of the following cubic equations has roots 𝛼, β and ɣ. Find, for each case, 𝛼 + β + ɣ and 𝛼βɣ.
 x^{3} + 3x^{2} – 5 = 0
 2x^{3} + 5x^{2} – 6 = 0
 x^{3} + 7x – 9 = 0
 Given that x^{3} – 3x^{2} + 12 = 0 has roots 𝛼, β and ɣ, find the following values:
 𝛼 + β + ɣ
 𝛼β + 𝛼ɣ + βɣ
 𝛼^{2} + β^{2} + ɣ^{2}
 The roots of each of the following cubic equations are 𝛼, β and ɣ. In each case, find the values of S_{2} and S_{1}.
 x^{3} – 2x^{2} + 5 = 0
 3x^{3} + 4x – 1 = 0
 x^{3} + 3x^{2} + 5x – 7 =0
 The cubic equation x^{3} – x + 7 = 0 has roots 𝛼, β and ɣ. Find the values of Σ𝛼 and Σ𝛼^{2}.
 Given that 2x^{3} + 5x^{2} + 1 = 0 has roots 𝛼, β and ɣ and that Sn = 𝛼^{n} + β^{n} + ɣ^{n}, find the values of S_{2} and S_{3}.
 The cubic equation x^{3} + ax^{2} + bx + a = 0 has roots 𝛼, β and ɣ and the constants a and b are real and positive.
 Find in terms of a and b the values of Σ𝛼 and
 Given that , does this cubic equation have complex roots? Give a reason for your answer.
 The cubic equation x^{3} – x + 3 = 0 has roots 𝛼, β and ɣ.
 Using the relations S_{n} = 𝛼^{n} + β^{n} + ɣ^{n}, find the value of S_{4}.
 By considering S_{1} and S_{4}, determine the value of 𝛼^{3}(β+ɣ) + β^{3}(𝛼+ɣ) + ɣ^{3}(𝛼+β)
 A cubic polynomial is given as 2x^{3} – x^{2} + x – 5 = 0, having roots 𝛼, β and ɣ.
 Show that 2S_{n+3} – S_{n+2} + S_{n+1} – 5S_{n} = 0
 Find the value of S_{2}
 The cubic equation px^{3} + qx^{2} + r = 0 has roots 𝛼, β and ɣ. Find in terms of p, q and r:
 S_{1}
 S_{2}
 S_{3}
 The equation x^{3} + px^{2} + qx + r = 0 is such that S_{1} = 0, S_{2} = 2 and S_{1} = 1/5. Find the values of the constants p, q and r.
Answers to Exercise 2

 𝛼 + β + ɣ = 3 𝛼βɣ = 5
 𝛼 + β + ɣ = 5/2 𝛼βɣ = 3
 𝛼 + β + ɣ = 0 𝛼βɣ = 9

 𝛼 + β + ɣ = 3 𝛼β + 𝛼ɣ + βɣ = 0
 9
 0 and 2
 25/4 and 137/8
 p = 0, q = 1 and r = 5
Quartic Equations
With quartic equations we apply the same principles, however with these higher order equations the recurrence technique is much more often required, especially for Σ𝝰^{3}.
If the quartic equation ax^{4}+bx^{3}+cx^{2}+dx+e=0 has roots 𝝰, β, ɣ and 𝛿, then:
Σ𝝰 =
Σ𝝰β =
Σ𝝰βɣ =
Σ𝝰βɣ𝛿 =
When using the recurrence formula you must be careful to make sure that you multiply the constant by n (because you will be adding n “copies” of the equation, each with its own constant)
e.g. If f(x)=x^{4}+3=0, then Σ𝝰^{4}+12=0
Worked Example
The roots of the equation x^{4} – 3x^{2} + 5x – 2 = 0 are 𝛼, β, ɣ and 𝛿. 𝛼^{n} + β^{n} + ɣ^{n} + 𝛿^{n} is denoted by S_{n}. Show that S_{n+4} – 3S_{n+2} + 5S_{n+1} – 2S_{n} = 0, and use this to find S_{1}, S_{2}, S_{3}, S_{4} and S_{5}.
Exercise 3
 For each of the following quartic equations, find the values of Σ𝛼 and Σ𝛼β:
 x^{4} – 2x^{3} + 5x^{2} + 7 = 0
 2x^{4} + 5x^{3} – 3x + 4 = 0
 3x^{4} – 2x^{2} + 9x – 11 = 0
 The quartic equation 5x^{4} – 3x^{3} + x – 13 = 0 has roots 𝛼, β, ɣ and 𝛿. Find:
 Σ𝛼 and Σ𝛼^{2}
 A quartic equation is given as x^{4} + x + 2 = 0. It has roots 𝛼, β, ɣ and 𝛿. State the values of S_{1} and S_{1} and find the value of S_{2}.
 The quartic equation 2x^{4} + x^{3} – x + 7 = 0 has roots 𝛼, β, ɣ and 𝛿. Given that $latex S_3 = \frac{11}{8} and using S_{n}, find the value of S_{4}.
 You are given that x4 – x3 + x + 2 = 0, where the roots are 𝛼, β, ɣ and 𝛿. Find the values of Σ𝛼, Σ𝛼^{2} and . Hence, determine the value of Σ𝛼^{3}.
 The quartic polynomial x^{4} + ax^{2} + bx + 1 = 0 has roots 𝛼, β, ɣ and 𝛿. Given that S_{2} = S_{1}, find S_{3} in terms of a.
 The polynomial 3x^{4} + 2x^{3} + 7x^{2} + 4 = 0 has roots 𝛼, β, ɣ and 𝛿, where S_{n} = 𝛼^{n} + β^{n} + ɣ^{n} + 𝛿^{n}.
 Find the values of S_{1} and S_{2}.
 Find the values of S_{3} and S_{4}.
 Are there any complex roots? Give a reason for your answer.
 For the polynomial x^{4}+ ax^{3} + bx^{2} + c = 0, with roots 𝛼, β, ɣ and 𝛿, it is given that 𝛼 + β + ɣ + 𝛿 = 2, 𝛼βɣ𝛿 = 1 and 𝛼^{2} + β^{2} + ɣ^{2} + 𝛿^{2} = 0. Find the value of the coefficients a, b and c.
 The roots of the quartic x^{4} – 2x^{3} + x^{2} – 4 = 0 are 𝛼, β, ɣ and 𝛿. Show that S_{4} = 9S_{3}.
Answers to Exercise 3

 Σ𝛼 = 2, Σ𝛼β = 5
 Σ𝛼 = 5/2, Σ𝛼β = 0
 Σ𝛼 = 0, Σ𝛼β = 2/3
 0, 1/2, 0
 239/16
 1 and 1, 1/2 and 2
 6a

 2/3 and 38/9
 118/27 and 130/81
 There are complex roots (because S_{2}<0)
 a = 2, b = 2, c = 1
 Proof
Subsitutions
If we want a polynomial f(y) with roots that are double the roots of a given polynomial f(x), we let y=2x, so , and substitute this into the original equation to form a new polynomial f(y).
This can be useful when finding S_{n} = Σ𝝰^{n} for larger values of n, such as 4 or 6. By changing the polynomial to a polynomial whose roots are the squares of the former polynomial, we halve the required value of n.
Worked Example 1
Worked Example 2
Exercise 4
Note: To get a quadratic equation in question 2a you need to do some algebraic manipulation after substituting
 The quadratic equation x^{2} + 5x + 3 = 0 has roots 𝛼 and β. Find the quadratic equation with roots 3𝛼 and 3β.
 The quadratic equation 2x^{2} – 4x + 7 = 0 has roots 𝛼 and β.
 Find the quadratic equation with roots 𝛼^{2} and β^{2}.
 Find the quadratic equation with roots 2𝛼 – 3 and 2β – 3.
 Given that 3x^{2} – 2x + 9 = 0 has roots 𝛼 and β, find the quadratic equation with roots
 The quadratic equation x^{2} – 4x + 9 = 0 has roots 𝛼 and β. Find the quadratic that has roots and .
 Given that 2x^{3} – 5x + 1 = 0 has roots 𝛼, β and ɣ, find the cubic equation with roots 𝛼^{2}, β^{2} and ɣ^{2}. Hence find the value of S_{4}.
 The cubic equation x^{3} + 3x^{2} – 1 = 0 has roots 𝛼, β and ɣ. Show that the cubic equation with roots , and is y^{3} – 3y^{2} – 9y + 3 = 0. Hence, determine the value of:
 A quartic equation 2x^{4} – x^{3} – 6 = 0 has roots 𝛼, β, ɣ and 𝛿. Show that the quartic equation with roots 𝛼^{3}, β^{3}, ɣ^{3} and 𝛿^{3} is 8y^{4} – y^{3} – 18y^{2} – 108y – 216 = 0. Hence find the values of S_{6} and S_{3}.
 The cubic equation x^{3} – x + 4 = 0 has roots 𝛼, β and ɣ. Find the cubic equation that has roots 𝛼^{2}, β^{2} and ɣ^{2}. Hence, or otherwise, determine the values of S_{6}, S_{8} and S_{10}.
Answers to Exercise 4
 x^{2} + 15x + 27 = 0
 9x^{2} – 20x + 14 = 0
 9y^{2} – 4y + 1 = 0
 4y^{3} – 20y^{2} + 25y – 1 = 0, 25/2
 Proof
 Proof 289/64 1/2
 y^{3} – 2y^{2} + y – 16 = 0 50 130 242
End of Roots of Polynomials Chapter Mixed Exercise
Answers to Roots of Polynomials Chapter Mixed Exercise
Relevant Past Exam Questions:
2017: s1q7: Q A, s3q1 Q A, w1q4 Q A;
2016: s1q1 Q A, s3q8 Q A, w1q2 Q A;
2015: s1q4, s3q1, w1q5;
2014: s1q1, w1q’either’;
2013: s1q3, s3q2, w1q2, w3q5;
2012: s1q2, s3q8, w1q’either’, w3q7;
2011: s1q2, s3q3, w1q1, w3q3 (inc. matrices);
2010: s1q6, s3q10, wq7;
2009: sq1, wq5;
2008: sq5, wq’or’;
2007: sq7, wq4;
2006: sq’either’, wq4;
2005: sq4, wq5;
2004: q’or’, wq3;
2003: sq5, wq6;
2002: wq2