# 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.

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 Sn. 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

3+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

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
2. Given that 3x2 + 4x + 12 = 0 has roots 𝛼 and β, find:
• 𝛼 + β and 𝛼β
• 𝛼2 + β2
3. x2 – (2+p)x + (7+p) = 0 has roots that differ by 1. Find the value of p given that p>0
4. 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.
5. 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)
6. 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.
7. The quadratic equation x2 + 2x – 6 has roots 𝛼 and β. Find the values of (𝛼 – β)2 and $\frac{1}{ \alpha^2} + \frac{1}{ \beta^2}$
8. 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).
9. The quadratic equation 3x2 + 2x – 4 = 0 has roots 𝛼 and β. Find the values of S1, S2 and S-1.
10. You are given the quadratic equation 4x2 – x + 6 = 0 which has roots 𝛼 and β.
• Find 𝛼2 + β2

1. 5
2. The value of ab = 1.x2 + 3x + 1 = 0
3. p=2, q=1
4. 28 and 4/9
5. x2 + 2x – 4 = 0 and x2 – 6x + 12 = 0
6. S1 = -2/3 , S2 = 28/9 , S-1 = 1/2

Part 2: Cubic Equations

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

1. 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
2. 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
3. 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
4. The cubic equation x3 – x + 7 = 0 has roots 𝛼, β and ɣ. Find the values of Σ𝛼 and Σ𝛼2.
5. 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.
6. 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.
7. The cubic equation x3 – x + 3 = 0 has roots 𝛼, β and ɣ.
• Using the relations Sn = 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)
8. A cubic polynomial is given as 2x3 – x2 + x – 5 = 0, having roots 𝛼, β and ɣ.
• Show that 2Sn+3 – Sn+2 + Sn+1 – 5Sn = 0
• Find the value of S-2
9. The cubic equation px3 + qx2 + r = 0 has roots 𝛼, β and ɣ. Find in terms of p, q and r:
• S1
• S2
• S3
10. 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.

• 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
1. 0 and 2
2. 25/4 and -137/8
3. 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 Sn. Show that Sn+4 – 3Sn+2 + 5Sn+1 – 2Sn = 0, and use this to find S1, S2, S3, S4 and S5.

Exercise 3

1. 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
2. The quartic equation 5x4 – 3x3 + x – 13 = 0 has roots 𝛼, β, ɣ and 𝛿. Find:
• Σ𝛼 and Σ𝛼2
• $\Sigma \frac{1}{ \alpha }$
3. 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.
4. 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 Sn, find the value of S4.
5. 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.
6. 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.
7. The polynomial 3x4 + 2x3 + 7x2 + 4 = 0 has roots 𝛼, β, ɣ and 𝛿, where Sn = 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.
8. 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.
9. The roots of the quartic x4 – 2x3 + x2 – 4 = 0 are 𝛼, β, ɣ and 𝛿. Show that S4 = 9S3.

1. 0, -1/2, 0
2. -239/16
3. 1 and 1, -1/2 and -2
4. -6a
5. a = -2, b = 2, c = 1
6. 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 Sn = 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

1. The quadratic equation x2 + 5x + 3 = 0 has roots 𝛼 and β. Find the quadratic equation with roots 3𝛼 and 3β.
2. 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.
3. Given that 3x2 – 2x + 9 = 0 has roots 𝛼 and β, find the quadratic equation with roots $\frac{ \alpha + 1}{ \alpha } and \frac{ \beta + 1}{ \beta }$
4. The quadratic equation x2 – 4x + 9 = 0 has roots 𝛼 and β. Find the quadratic that has roots $\frac{1}{ \alpha }$ and $\frac{1}{ \beta }$.
5. 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.
6. The cubic equation x3 + 3x2 – 1 = 0 has roots 𝛼, β and ɣ. Show that the cubic equation with roots $\frac{ \alpha + 2}{ \alpha}$, $\frac{ \beta + 2}{ \beta}$ and $\frac{ \gamma + 2}{ \gamma}$ is y3 – 3y2 – 9y + 3 = 0. Hence, determine the value of:
• $\frac{( \alpha + 2)( \beta + 2)( \gamma + 2)}{ \alpha \beta \gamma}$
• $\frac{ \alpha }{ \alpha + 2 } + \frac{ \beta }{ \beta + 2 } + \frac{ \gamma }{ \gamma + 2 }$
7. 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.
8. 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.

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