9231. Further Pure 1. Roots of Polynomials

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: ax²+bx+c=0

Without losing generality we can rewrite this as: $x^2 + \frac{b}{a} x + \frac{c}{a}$ (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: (x-𝝰)(x-β)=0 (2)

If we compare (1) and (2) above, we get:

$x^2 + \frac{b}{a} x + \frac{c}{a} = (x - \alpha )(x - \beta )$ , so

$x^2 + \frac{b}{a} x + \frac{c}{a} = x^2 - ( \alpha + \beta ) + \alpha \beta$ .

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:

$- \frac{b}{a}$ = Σ𝝰 and $\frac{c}{a}$ = Σ𝝰β.

Let’s use the above results to derive Σ𝝰² 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 Σ𝛼ⁿ.

Recursive method

Suppose that we know equation f(x)=2x²+3x-2 has roots 𝝰 and β and we want to know Σ𝝰³. We could use a very similar method to the one we used above to find Σ𝝰². 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: 2x³+3x²-2x=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:

2𝝰³+3𝝰²-2𝝰=0 (3) and

2β³+3β²-2β=0 (4).

Adding together (3) and (4) gives:

2Σ𝝰³+3Σ𝝰²-2Σ𝝰=0, which we can easily solve for Σ𝝰³.

Worked Example

The roots of the equation 3z²-4z-1=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.
- x²+ 5x + 9 = 0
- x² – 4x + 8 = 0
- 2x² + 3x – 7 = 0
Given that 3x² + 4x + 12 = 0 has roots 𝛼 and β, find:
- 𝛼 + β and 𝛼β
- 𝛼² + β²
x² – (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 a² + b² = 7, find the value of ab and hence write down a quadratic equation with roots a and b.
If x² + bx + c = 0 has roots 𝛼 and β, prove that:
- If 𝛼 = 3β then $b^2 = \frac{16}{3}c$
- If 𝛼 = β-2, then b² = 4(c+1)
You are given the quadratic equation px² + qx – 16 = 0, which has roots 𝛼 and β. Given also that $\alpha + \beta = - \frac{1}{2}$ and 𝛼β = -8, find the values of p and q.
The quadratic equation x² + 2x – 6 has roots 𝛼 and β. Find the values of (𝛼 – β)² and $\frac{1}{ \alpha^2} + \frac{1}{ \beta^2}$
A quadratic equation has roots 𝛼 and β. Given that $\frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{1}{2}$ and 𝛼² + β² = 12, find two possible quadratic equations that satisfy these values (both of which have 1 as the leading coefficient).
The quadratic equation 3x² + 2x – 4 = 0 has roots 𝛼 and β. Find the values of S₁, S₂ and S_-1.
You are given the quadratic equation 4x² – x + 6 = 0 which has roots 𝛼 and β.
- Find 𝛼² + β²
- 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
- 𝛼 + β = -4/3. 𝛼β = 4
- -56/9 (or $-6 \frac{2}{9}$ )
5
The value of ab = 1.x² + 3x + 1 = 0
- Proof
- Proof
p=2, q=1
28 and 4/9
x² + 2x – 4 = 0 and x² – 6x + 12 = 0
S₁ = -2/3 , S₂ = 28/9 , S_-1 = 1/2
- -47/16 (or $-2 \frac{15}{16}$ )
- They must be complex roots, because 𝛼² + β² < 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 ax³+bx²+cx+d=0 has roots 𝝰, β and ɣ, then:

Σ𝝰 = $- \frac{b}{a}$

Σ𝝰β = $\frac{c}{a}$

Σ𝝰βɣ = $- \frac{d}{a}$

(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 Σ𝝰² 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.
- x³ + 3x² – 5 = 0
- 2x³ + 5x² – 6 = 0
- x³ + 7x – 9 = 0
Given that x³ – 3x² + 12 = 0 has roots 𝛼, β and ɣ, find the following values:
- 𝛼 + β + ɣ
- 𝛼β + 𝛼ɣ + βɣ
- 𝛼² + β² + ɣ²
The roots of each of the following cubic equations are 𝛼, β and ɣ. In each case, find the values of S₂ and S_-1.
- x³ – 2x² + 5 = 0
- 3x³ + 4x – 1 = 0
- x³ + 3x² + 5x – 7 =0
The cubic equation x³ – x + 7 = 0 has roots 𝛼, β and ɣ. Find the values of Σ𝛼 and Σ𝛼².
Given that 2x³ + 5x² + 1 = 0 has roots 𝛼, β and ɣ and that Sn = 𝛼ⁿ + βⁿ + ɣⁿ, find the values of S₂ and S₃.
The cubic equation x³ + ax² + 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 $\Sigma \frac{1}{ \alpha }$
- Given that $\Sigma \alpha = \Sigma \frac{1}{ \alpha }$ , does this cubic equation have complex roots? Give a reason for your answer.
The cubic equation x³ – x + 3 = 0 has roots 𝛼, β and ɣ.
- Using the relations S_n = 𝛼ⁿ + βⁿ + ɣⁿ, find the value of S₄.
- By considering S₁ and S₄, determine the value of 𝛼³(β+ɣ) + β³(𝛼+ɣ) + ɣ³(𝛼+β)
A cubic polynomial is given as 2x³ – x² + 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 px³ + qx² + r = 0 has roots 𝛼, β and ɣ. Find in terms of p, q and r:
- S₁
- S₂
- S₃
The equation x³ + px² + qx + r = 0 is such that S₁ = 0, S₂ = -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
- -a and -b/a
- Yes, because Σ𝛼² = -a²
- 2
- -2
- Proof
- -9/25
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 Σ𝝰³.

If the quartic equation ax⁴+bx³+cx²+dx+e=0 has roots 𝝰, β, ɣ and 𝛿, then:

Σ𝝰 = $- \frac{b}{a}$

Σ𝝰β = $\frac{c}{a}$

Σ𝝰βɣ = $- \frac{d}{a}$

Σ𝝰βɣ𝛿 = $\frac{e}{a}$

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)=x⁴+3=0, then Σ𝝰⁴+12=0

Worked Example

The roots of the equation x⁴ – 3x² + 5x – 2 = 0 are 𝛼, β, ɣ and 𝛿. 𝛼ⁿ + βⁿ + ɣⁿ + 𝛿ⁿ is denoted by S_n. Show that S_n+4 – 3S_n+2 + 5S_n+1 – 2S_n = 0, and use this to find S₁, S₂, S₃, S₄ and S₅.

Exercise 3

For each of the following quartic equations, find the values of Σ𝛼 and Σ𝛼β:
- x⁴ – 2x³ + 5x² + 7 = 0
- 2x⁴ + 5x³ – 3x + 4 = 0
- 3x⁴ – 2x² + 9x – 11 = 0
The quartic equation 5x⁴ – 3x³ + x – 13 = 0 has roots 𝛼, β, ɣ and 𝛿. Find:
- Σ𝛼 and Σ𝛼²
- $\Sigma \frac{1}{ \alpha }$
A quartic equation is given as x⁴ + x + 2 = 0. It has roots 𝛼, β, ɣ and 𝛿. State the values of S₁ and S_-1 and find the value of S₂.
The quartic equation 2x⁴ + x³ – 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 S₄.
You are given that x4 – x3 + x + 2 = 0, where the roots are 𝛼, β, ɣ and 𝛿. Find the values of Σ𝛼, Σ𝛼² and $\Sigma \frac{1}{ \alpha }$ . Hence, determine the value of Σ𝛼³.
The quartic polynomial x⁴ + ax² + bx + 1 = 0 has roots 𝛼, β, ɣ and 𝛿. Given that S₂ = S_-1, find S₃ in terms of a.
The polynomial 3x⁴ + 2x³ + 7x² + 4 = 0 has roots 𝛼, β, ɣ and 𝛿, where S_n = 𝛼ⁿ + βⁿ + ɣⁿ + 𝛿ⁿ.
- Find the values of S₁ and S₂.
- Find the values of S₃ and S₄.
- Are there any complex roots? Give a reason for your answer.
For the polynomial x⁴+ ax³ + bx² + c = 0, with roots 𝛼, β, ɣ and 𝛿, it is given that 𝛼 + β + ɣ + 𝛿 = 2, 𝛼βɣ𝛿 = 1 and 𝛼² + β² + ɣ² + 𝛿² = 0. Find the value of the coefficients a, b and c.
The roots of the quartic x⁴ – 2x³ + x² – 4 = 0 are 𝛼, β, ɣ and 𝛿. Show that S₄ = 9S₃.

Answers to Exercise 3

- Σ𝛼 = 2, Σ𝛼β = 5
- Σ𝛼 = -5/2, Σ𝛼β = 0
- Σ𝛼 = 0, Σ𝛼β = -2/3
- Σ𝛼 = 3/5, Σ𝛼² = 9/25
- 1/13
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 S₂<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 $x = \frac{y}{2}$ , and substitute this into the original equation to form a new polynomial f(y).

This can be useful when finding S_n = Σ𝝰ⁿ 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 x² + 5x + 3 = 0 has roots 𝛼 and β. Find the quadratic equation with roots 3𝛼 and 3β.
The quadratic equation 2x² – 4x + 7 = 0 has roots 𝛼 and β.
- Find the quadratic equation with roots 𝛼² and β².
- Find the quadratic equation with roots 2𝛼 – 3 and 2β – 3.
Given that 3x² – 2x + 9 = 0 has roots 𝛼 and β, find the quadratic equation with roots $\frac{ \alpha + 1}{ \alpha } and \frac{ \beta + 1}{ \beta }$
The quadratic equation x² – 4x + 9 = 0 has roots 𝛼 and β. Find the quadratic that has roots $\frac{1}{ \alpha }$ and $\frac{1}{ \beta }$ .
Given that 2x³ – 5x + 1 = 0 has roots 𝛼, β and ɣ, find the cubic equation with roots 𝛼², β² and ɣ². Hence find the value of S₄.
The cubic equation x³ + 3x² – 1 = 0 has roots 𝛼, β and ɣ. Show that the cubic equation with roots , and is y³ – 3y² – 9y + 3 = 0. Hence, determine the value of:
- $\frac{( \alpha + 2)( \beta + 2)( \gamma + 2)}{ \alpha \beta \gamma}$
- $\frac{ \alpha }{ \alpha + 2 } + \frac{ \beta }{ \beta + 2 } + \frac{ \gamma }{ \gamma + 2 }$
A quartic equation 2x⁴ – x³ – 6 = 0 has roots 𝛼, β, ɣ and 𝛿. Show that the quartic equation with roots 𝛼³, β³, ɣ³ and 𝛿³ is 8y⁴ – y³ – 18y² – 108y – 216 = 0. Hence find the values of S₆ and S_-3.
The cubic equation x³ – x + 4 = 0 has roots 𝛼, β and ɣ. Find the cubic equation that has roots 𝛼², β² and ɣ². Hence, or otherwise, determine the values of S₆, S₈ and S₁₀.