1. Polynomials: The relationship between roots and coefficients.
By Mr Porter

2. Asssumed Knowledge: Quadratic: ax2 + bx + c = 0
If α, β are the roots of the quadtraic, ax2 + bx + c = 0, then
Cubic: ax3 + bx2 + cx + d = 0
If α, β, γ are the roots of a cubic, ax3 + bx2 + cx + d = 0, then
Quartic: ax4 + bx3 + cx2 +dx + e = 0
If α, β, γ, δ are the roots of the quartic, ax4 + bx3 + cx2 +dx + e = 0 , then
Students need to learn and be able to write these relationships out quickly.

3. Example 1:
Given P(x) = x3 + 5x + 1 has roots α, β, γ, write down the exact value of:
If α, β, γ are the roots of a cubic, ax3 + bx2 + cx + d = 0, then: a = 1, b = 0, c = 5, d = 1
You need to find all 3 relationships.
Even if they are not asked to be found.
To add fraction, convert to a common denominator.
Check the factorisation by expanding and simplifying the brackets.

4. Example 2:
Given P(x) = 4x3 – 5x2 + x – 6 has roots α, β, γ, write down the exact value of:
If α, β, γ are the roots of a cubic, ax3 + bx2 + cx + d = 0, then: a = 4, b = -5, c = 1, d = -6
You need to find all 3 relationships.
Even if they are not asked to be found.
We know that α, β, γ is a solutions.
4α3 – 5α2 + α – 6 = 0
4β3 – 5β2 + β – 6 = 0
4γ3 – 5γ2 + γ – 6 = 0
Rearrange all three equations of the cubics =
4α3 = 5α2 – α + 6
4β3 = 5β2 – β + 6
4γ3 = 5γ2 – γ + 6
Check the factorisation by expanding the brackets.
ADD
4(α3 + β3 + γ3) = 5(α2 + β2 + γ2) –(α + β + γ) + 18

5. Example 3:
Find the cubic equation whose roots are twice those of the equation 2x3 – 3x2 + 1 = 0.
let equation 2x3 – 3x2 + 1 = 0 have roots α, β, γ
The new cubic equation must have have roots m = 2α, n = 2β, p = 2γ
rearrange
Now, when x = α,
2α3 – 3α2 + 1 = 0 is true.
Multiply by ‘8’
Divide by ‘2’
The required equations is P(x) = x3 – 3x2 + 4

6. Example 4(a):
If a root of the cubic x3 – 4x2 + 2x + 4 = 0 is equal to the sum of the other two roots, find the roots.
I will show 2 methods of solving this question
Let the roots be α, β, γ and γ =α + β
From α + β = 2  β = 2 – α
Substituting into
αβ(α + β) = -4
α(2 – α )(2) = -4
simplify
So,
α2 – 2α – 2 = 0
Quadratic, ∆ = b2 – 4ac = 12 ( ≠n2)
Simplifying,
So, the roots α, β, γ are
We have 2 UNKNOWNS and 3 equations. Solve simultaneously.

7. Example 4(b):
If a root of the cubic x3 – 4x2 + 2x + 4 = 0 is equal to the sum of the other two roots, find the roots.
Let the roots be α, β, γ and γ =α + β
x2 – 2x – 2 = 0
Quadratic, ∆ = b2 – 4ac = 12 ( ≠n2)
So, replace α + β with γ
So, the roots α, β, γ are
This give us
x3 – 4x2 + 2x + 4 = (x – 2) Q(x)
Using polynomial division
This give us
x3 – 4x2 + 2x + 4 = (x – 2)(x2 – 2x – 2)
This looks easier, but it depends on which substitution you see!
α + β = γ or γ = α + β

8. Example 5:
If a root of the cubic x3 + 2x2 – 12x – 16 = 0 is equal to the product of the other two roots, find the roots.
Let the roots be α, β, γ and γ =αβ
So, the polynomials can be written as
x3 + 2x2 – 12x – 16 = (x + 4) Q(x)
x3 + 2x2 – 12x – 16 = (x + 4)(x2 – 2x – 4)
Now,
x2 – 2x – 4 = 0
Quadratic, ∆ = b2 – 4ac = 20 ( ≠n2)
Using γ =αβ in αβγ = 16
(αβ)2 = 16
Test for a factor P(a) = 0
γ = -4 is one of the roots.
So, the roots α, β, γ are

9. Example 6:
Solve the cubic 2x3 – 11x2 + 12x + 9 = 0 given that two of its roots are equal.
Let the roots be α, β, γ and γ =α. [α, α, β]
Possible solutions for α
Check P(a) = 0, for a factor
We have 2 UNKNOWNS and 3 equations. Solve simultaneously.
So, α = 3 is a solution.
Substituting into
simplify
So, 2x3 – 11x2 + 12x + 9 = 0 when
Factorise
Also, once we found α = 3 was a root,
we could have use polynomial division to find the other 2 roots.