### Polynomial Roots and Coefficients

There are clear relationships with the roots of a polynomial and the coefficients of the latter.

x2 - 3x - 10 = 0,
The roots are, x = 5 or x = -2.
Let's look at the coefficients now:
b/a = -3/1 = -3; c/a = -10/1 = -10
From the roots,
5 + (-2) = 3; 5 x -2 =-10
So, if the roots of ax2 + bx + c = 0, are α and β,
α + β = -b/a and αβ = c/a

In general, if ax2 + bx + c= a(x - α)(x - β),
Sum of Roots, α + β = -b/a and Product of Roots, αβ = c/a
Proof
ax2 + bx + c= a(x - α)(x - β) = a(x2 - (α + β)x + αβ) = ax2 - a(α + β)x + aαβ
Making the coefficients of x and the constants on both sides equal,
b = -a(α + β) => (α + β) = -b/a
c = aαβ => αβ = c/a

E.g.1
If x2 - 4x - 12 = 0, find the values of α + β, αβ (α+β)/αβ and α2 + β2, without solving the equation.

α + β = -(-4/1) = 4; αβ = -12/1 = -12
1/α + 1/β = (α+β)/αβ = 4/-12 = -1/3
(α+β)2 = α2 + β2 + 2αβ
α2 + β2 = (α+β)2 - 2αβ
= 16 - 2 x (-12) = 16 + 24 = 40.

E.g.2
If 2x2 - 5x - 12 = 0, find the values of α + β, αβ and (α+β)/αβ, without solving the equation.

α + β = -(-5/2) = 2.5; αβ = -12/2 = -6
1/α + 1/β = (α+β)/αβ = -2.5/6 = -5/12

E.g.3
If px2 + qx + r = 0, find the values of p, q and r, if α = -2/3 and β= 5/6.

Since the roots are given, x = -2/3 or x = 5/6
So, (x + 2/3)(x - 5/6) = 0
x2 - x/6 - 10/18 = 0
18x2 - 3x - 10 = 0
p = 18; q = -3; r = -10

#### Roots of Cubic Polynomials

Let ax3 + bx2 + cx + d = a(x - α)(x - β)(x - γ)
= a[(x2 - (α + β)x + αβ)(x - γ)]
= ax3 - a(α + β + γ)x2 + a(αβ + αγ + βγ)x - aαβγ
α + β + γ = -b/a; αβ + αγ + βγ = c/a; αβγ = -d/a

In general, if ax3 + bx2 + cx + d = 0,
Sum of Roots, α + β + γ = -b/a; Sum of products of pairs of roots,αβ + αγ + βγ = c/a; Product of Roots, αβγ = -d/a

E.g.1
If α β and γ are the roots of the polynomial, 2x3 - 5x2 -9x + 18, find α + β + γ, αβ + αγ + βγ, αβγ and 1/α + 1/β + 1/γ, without finding the roots.
α + β + γ = -b/a = -(-5/2) = 2.5
αβ + αγ + βγ = c/a = -9/2 = -4.5
αβγ = -d/a = -(18/2) = -9
1/α + 1/β + 1/γ = (αβ + αγ + βγ)/(αβγ) = -4.5/-9 = 1/2

If the roots of the cubic polynomial are complex, they are conjugates. The following question is based on a combination of complex and real roots.

E.g.2
If the roots of ax3 + bx2 +cx + d are α = 2 + i, β = 2 - i and γ = 3, find the values of a, b, c and d.
ax3 + bx2 +cx + d = (x - 2 - i)(x -2 + i)(x - 3)
= (x2 - 2x + ix -2x + 4 - 2i -ix + 2i + 1)(x - 3)
= x3 - 7x2 + 17x - 15
= x3 - 7x2 + 17x - 15
Making the coefficients on both sides equal,
a = 1; b = -7; c = 17; d = -15

#### Roots of Quartic Polynomials

Let ax4 + bx2 + cx2 + dx + e = a(x - α)(x - β)(x - γ)(x - δ)
= a[(x4 - (α + β + γ + δ)x3 + (αβ + αγ + αδ + βγ + βδ + γδ)x2 - (αβγ + αβδ + βδγ)x + (αβγδ)]
= ax4 - a(α + β + γ + δ)x3 + a(αβ + αγ + αδ + βγ + βδ + γδ)x2 - a(αβγ + αβδ + βδγ)x + aαβγδ
α + β + γ + δ = -b/a; αβ + αγ + αδ + βγ + βδ + γδ = c/a; αβγ + αβδ + βδγ + αδγ = -d/a; αβγδ = e/a

In general, if ax4 + bx3 + c2 + dx + e = 0,
Sum of Roots, α + β + γ + δ = -b/a; Sum of products of pairs of roots,αβ + αγ + αδ + βγ + βδ + γδ = c/a; Sum of Product of Triplets of Roots = αβγ + αβδ + βδγ +αδγ = -d/a ; Product of Roots, αβγδ = e/a

E.g.1
If x4 - x3 - 7x2 + x + 6 = 0, find α + β + γ + δ, αβγδ and 1/α + 1/β + 1/γ + 1/δ.
α + β + γ + δ = -b/a = 1
αβγδ = e/a = 6
1/α + 1/β + 1/γ + 1/δ = (αγ + αδ + βγ + βδ + γδ)/(αβγδ) = (-c/a)/(e/a)
= -7/6

E.g.2
If x4 - 5x3 + cx2 + dx + e = 0 and the two complex roots of the function are 2 ±i3. Find the other roots and the values of c and d.
Let α = 2 + i3, β = 2 - i3 and the other roots be γ and δ.
Since α + β + γ + δ = -b/a => 2 + i3 + 2 -i3 + γ + δ = 5
γ + δ = 1
αβγδ = e/a => αβγδ = -78
(2 + i3)(2 - i3)γδ = -78
13γδ = -78 => γδ = -6
Solving the two equations, we get γ = 3; δ =-2
αβ + αγ + αδ + βγ + βδ + γδ = c/a
13 + (6 + i9) + (-4 -i6) + (6 - 9i) + (-4 + 6i) - 6 = c/1 => c = 11
αβγ + αβδ + βδγ + αδγ = -d/a
-26 + 39 + (-12 - i18) + (-12 + i18) = -d/a
-11 = d/1 => d = 11

#### Linear Transformations of Roots

Suppose the roots of a quadratic polynomial are α and β. If their values are changed to, say, α + 1 or β + 1, then, there is going to be a corresponding change in the original polynomial - and a corresponding transformation of the graph too. In this section, the transformation of the graph and the corresponding formation of the new polynomial are explored in detail.

E.g.1
x2 - x - 6 = 0
α + β = -b/a = 1
αβ = c/a = -6
α = 3; β = -2
Now let's transform the two roots - a translation by 1 in the negative x-axis: α + 1 = 4; β + 1 = -1; So, the new roots are, x = 4 and x = -1
Now let's form the new polynomial:
New α = 4; new β = -1;
New α + β = 3; new αβ = -4
Since α + β = -b/a and αβ = c/a, if a =1, the new polynomial is derived as follows:
In order to minimize the confusion, it is recommended to use a new variable for the new polynomial, instead of x. Using w as the new variable, the new polynomial takes the following form:
w2 - 3w - 4 = 0

E.g.2
x2 - x - 6 = 0
α + β = -b/a = 1
αβ = c/a = -6
α = 3; β = -2
Now let's transform the two roots - stretching by a factor 1/2 parallel to x-axis: α/2 = 3/2; 2β = -1; So, the new roots are, x = 3/2 and x = -1
Now let's form the new polynomial:
New α = 3/2; new β = -1;
New α + β = 1/2; new αβ = -3/2
Since α + β = -b/a and αβ = c/a, the new polynomial is derived as follows:
In order to minimize the confusion, it is recommended to use a new variable for the new polynomial, instead of x. Using w as the new variable, the new polynomial takes the following form:
w2 - 1/2 w - 3/2 = 0
4w2 -2w - 6 = 0

E.g.3
x2 - x - 6 = 0
α + β = -b/a = 1
αβ = c/a = -6
α = 3; β = -2
Now let's transform the two roots - reflecting in the y-axis: -1 x α = -3; -1 x β = 2; So, the new roots are, x = -3 and x = 2
Now let's form the new polynomial:
New α = -3; new β = 2;
New α + β = -1; new αβ = -6
Since α + β = -b/a and αβ = c/a, the new polynomial is derived as follows:
In order to minimize the confusion, it is recommended to use a new variable for the new polynomial, instead of x. Using w as the new variable, the new polynomial takes the following form:
w2 + w - 6 = 0

You may practise the above interactively with the following applet; just move the slider and see the corresponding change in the polynomial and the graph. Please select one transformation at a time to avoid the text and graphics being blurred.