ALGEBRA EQUATIONS - SOLVING POLYNOMIAL EQUATIONS OF HIGHER DEGREE WITH EXAMPLES

Please study introduction
given to equations and
Algebra Equations in Math Equations,
if you have not already done so.
We studied about equations
of first degree in one and
two variables and word problems
on them in
Linear Equations.
We also covered equations of
second degree, their solution,
nature of roots, relationship
between roots and coefficients,
word problems in
Quadratic Equations.
In Polynomials,
we covered definition, degree,
zeros of polynomials,
remainder theorem, factor theorem.
Applications of factor theorem
and Synthetic Division to find
the factors of polynomials
were covered in
Algebra Factoring.
Please study all these
before proceeding further.
They are prerequisites here.

Here, we study Polynomial
Equations of higher degree.
In Algebra Equations,
we cover Polynomial Equations.

Fundamental Theorem of Algebra :

Every polynomial equation f(x) = 0 of degree
n ≥ 1 has atleast one root real or complex.

Based on the Fundamental Theorem
of Algebra and mathematical induction,
we can prove that

Every Polynomial equation of degree n ≥ 1 has only n roots real or imaginary and no more.

Relation between the roots and coefficients of an nth degree equation :
Algebra Equations

Consider the nth degree equation
xⁿ + p₁x^{(n - 1)} + p₂x^{(n - 2)} + p₃x^{(n - 3)} + .......... + p_n = 0
Let α₁, α₂, α₃, α₄,..........α_n be the roots of the equation.
Then,
s₁ = sum of the roots
= Σ α₁ = (-1)¹ p₁
s₂ = sum of the products of the roots taken two at a time
= Σ α₁α₂ = (-1)² p₂
s₃ = sum of the products of the roots taken three at a time
= Σ α₁α₂α₃ = (-1)³ p₃
s₄ = sum of the products of the roots taken four at a time
= Σ α₁α₂α₃α₄ = (-1)⁴ p₄
...................................................
....................................................
s_n = sum of the products of the roots taken all at a time
= α₁α₂α₃α.₄..........α_n = (-1)ⁿ p_n

NOTE 1. The above relations are true only when the coefficient of xⁿ is unity. In case it is not unity, the algebra equation should be divided with the coefficient of xⁿ to make it unity before writing the above relations between the roots and coefficients.

2. The number of relations we have is same as the degree of the equation. So there are n relations to find n roots. Though, n relations are sufficient to find n unknowns, it is practically difficult to find the roots with the help of the relations only.

Usually, we are given something more about the roots. If that additional data is not given, we can solve some equations by trial and error, using remainder theorem or factor theorem, as we did in Algebra Factoring.

3. For n = 3, we get a third degree or cubic algebra equation
x³ + p₁x² + p₂x + p₃ = 0
Then
s₁ = α₁ + α₂ + α₃ = -p₁
s₂ = α₁α₂ + α₁α₃ + α₂α₃ = p₂
s₃ = -p₃

4. For n = 4, we get a fourth degree or biquadratic algebra equation
x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0
Then
s₁ = α₁ + α₂ + α₃ + α₄ = -p₁
s₂ = α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄ = p₂
s₃ = α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄ = -p₃
s₄ = α₁α₂α₃α₄ = p₄

Revision of knowledge of Progressions :
Arithmetic Progression (A.P.) : Quantities are said to be in A.P., if the difference of any term and its preceding term is a constant.
Harmonic Progression (H.P.) : Quantities are said to be in H.P., if their reciprocals are in A.P.
Geometric Progression (G.P.) : Quantities are said to be in G.P., if the ratio of any term and its preceding term is a constant.

For a cubic equation, when the roots are
(i) in A.P., they are taken as α - β, α, α + β
(ii) in H.P., they are taken as 1⁄(α - β), 1⁄α, 1⁄(α + β)
(iii) in G.P., they are taken as α⁄β, α, αβ

For a biquadratic equation, when the roots are
(i) in A.P., they are taken as α - 3β, α - β, α + β, α + 3β
(ii) in H.P., they are taken as 1⁄(α - 3β), 1⁄(α - β), 1⁄α, 1⁄(α + 3β)
(iii) in G.P., they are taken as α⁄β³, α⁄β, αβ, αβ³

Set of Solved Examples : Cubic and Biquadratic Algebra Equations

Solved Example 1 : Cubic and Biquadratic Algebra Equations

If one root of the Algebra Equation 2x³ + 3x² - 8x + 3 = 0 is double another root, solve the equation.

Solution :
The given Algebra Equation is 2x³ + 3x² - 8x + 3 = 0
Dividing both sides of the equation by 2, we get
x³ + (3⁄2)x² - 4x + 3⁄2 = 0
Comparing this with x³ + p₁x² + p₂x + p₃ = 0, we get
p₁ = 3⁄2; p₂ = -4; p₃ = 3⁄2

By data one root is double the other root.
So if one root is α, then the other root is 2α.
Let the third root be β
We know, sum of roots = -p₁⇒ α + 2α + β = -3⁄2
⇒ 3α + β = -3⁄2 ⇒ 6α + 2β = -3...........(i)

sum of the product of the roots taken two at a time = p₂ = -4
⇒ (α)(2α) + (α)(β) + (2α)(β) = -4
⇒ α(2α + β + 2β) = -4 ⇒ α(2α + 3β) = -4.............(ii)

product of the roots = -p₃ = -3⁄2
⇒ (α)(2α)(β) = -3⁄2 ⇒ 4(α)²(β) = -3..........(iii)

From (i), β = (-3 - 6α)⁄2⇒ 3β = (-9 - 18α)⁄2 = -9⁄2 - 9α
Using this value of 3β in (ii), we get
α(2α + 3β) = -4⇒ α(2α - 9⁄2 - 9α) = -4
⇒ α( -9⁄2 - 7α) = -4⇒ 7α² + (9⁄2)α - 4 = 0
⇒ 14α² + 9α - 8 = 0⇒ 14α² + 16α - 7α - 8 = 0
⇒ 2α(7α + 8) -1(7α + 8) = 0⇒ (7α + 8)(2α - 1) = 0
⇒ α = 1⁄2 or -8⁄7

when α = 1⁄2, β = (-3 - 6α)⁄2 = (-3 - 3)⁄2 = -3
when α = -8⁄7, β = {-3 - 6(-8⁄7)}⁄2 = (-21 + 48)⁄14 = 27⁄14

For α = -8⁄7 and β = 27⁄14, L.H.S. of (iii) = 4(α)²(β) = 4( -8⁄7)²(27⁄14) = (4)(64)(27)⁄(49)(14) ≠ -3
∴ α = -8⁄7 and β = 27⁄14, do not satisfy (iii) and is not a valid solution.
α = 1⁄2, β = -3 satisfy equation (iii) and is a valid solution.

∴ The roots of the given Algebra Equation are α, 2α, β
= 1⁄2, 1, -3. Ans.

Solved Example 2 : Cubic and Biquadratic Algebra Equations

Solve the Algebra Equation 4x³ + 20x² - 23x + 6 = 0 two of the roots being equal..

Solution :
The given Algebra Equation equation is 4x³ + 20x² - 23x + 6 = 0
Dividing both sides of the equation by 4, we get
x³ + 5x² - (23⁄4)x + 3⁄2 = 0
Comparing this with x³ + p₁x² + p₂x + p₃ = 0, we get
p₁ = 5; p₂ = - (23⁄4); p₃ = 3⁄2

By data two roots are equal.
So if one root is α, then the other root is α.
Let the third root be β
We know, sum of roots = -p₁⇒ α + α + β = -5
⇒ 2α + β = -5 ...........(i)

sum of the product of the roots taken two at a time = p₂ = - (23⁄4)
⇒ (α)(α) + (α)(β) + (α)(β) = -(23⁄4)
⇒ α(α + 2β) = -(23⁄4) ⇒ 4α(α + 2β) = -23.............(ii)

product of the roots = -p₃ = -3⁄2
⇒ (α)(α)(β) = -3⁄2 ⇒ 2(α)²(β) = -3..........(iii)

From (i), β = (-5 - 2α)⇒ 2β = (-10 - 4α)
Using this value of 2β in (ii), we get
4α(α + 2β) = -23⇒ 4α(α - 10 - 4α) = -23
⇒ 4α( -10 - 3α) = -23⇒ 12α² + 40α - 23 = 0
⇒ 12α² + 46α - 6α - 23 = 0
⇒ 2α(6α + 23) -1(6α + 23) = 0⇒ (6α + 23)(2α - 1) = 0
⇒ α = 1⁄2 or -23⁄6

when α = 1⁄2, β = (-5 - 2α) = (-5 - 1) = -6
when α = -23⁄6, β = -5 - 2(-23⁄6) = (-15 + 23)⁄3 = 8⁄3

For α = -23⁄6 and β = 8⁄3, L.H.S. of (iii) = 2(α)²(β) = 2(-23⁄6)²(8⁄3) ≠ -3
∴ α = -23⁄6 and β = 8⁄3, do not satisfy (iii) and is not a valid solution.
α = 1⁄2, β = -6 satisfy equation (iii) and is a valid solution.

∴ the roots of the given Algebra Equation are α, α, β
= 1⁄2, 1⁄2, -6. Ans.

Solved Example 3 : Cubic and Biquadratic Algebra Equations

Solve the Algebra Equation 4x³ - 24x² + 23x + 18 = 0, the roots of which are in A.P.

Solution :
The given Algebra Equation is 4x³ - 24x² + 23x + 18 = 0
Dividing both sides of the equation by 4, we get
x³ - 6x² + (23⁄4)x + 9⁄2 = 0
Comparing this with x³ + p₁x² + p₂x + p₃ = 0, we get
p₁ = -6; p₂ = (23⁄4); p₃ = 9⁄2

By data the roots are in A.P.
Let the roots be α - β, α, α + β
We know, sum of roots = -p₁ ⇒ α - β + α + α + β = 6
⇒ 3α = 6⇒ α = 2.........(i)

sum of the product of the roots taken two at a time = p₂ = (23⁄4)
⇒ (α - β)(α) + (α - β)(α + β) + (α)(α + β) = (23⁄4)
⇒ 3α² - β² = (23⁄4)
Using the value of α = 2 from (i), we get
3(2)² - β² = (23⁄4) ⇒ 48 - 4β² = 23
⇒ 4β² = -23 + 48 = 25 ⇒ 2β = ±5⇒ β = ±5⁄2.......(ii)

product of the roots = -p₃ = -9⁄2
⇒ (α - β )(α)(α + β) = -9⁄2 ⇒ α{(α)² - (β)²} = -9⁄2.......(iii)

Verify α = 2, β = ±5⁄2 satisfy (iii) or not.
L.H.S. of (iii) = 2{2² - (±5⁄2)²} = 2(4 - 25⁄4) = 2(16 - 25)⁄4 = -9⁄2 = R.H.S. of (iii) [satisfied]

∴ the roots are α - β, α, α + β = 2 - (±5⁄2), 2, 2 + (±5⁄2) = -1⁄2 or 9⁄2, 2, 9⁄2 or -1⁄2
= -1⁄2, 2, 9⁄2 (or) 9⁄2, 2, -1⁄2
As you can see, both sets are same.

Thus, the roots of the given Algebra Equation are
-1⁄2, 2, 9⁄2. Ans.

Solved Example 4 : Cubic and Biquadratic Algebra Equations

Solve the Algebra Equation 15x³ - 23x² + 9x - 1 = 0, the roots of which are in H.P.

Solution :
The given Algebra Equation is 15x³ - 23x² + 9x - 1 = 0
Dividing both sides of the equation by 15, we get
x³ - (23⁄15)x² + (9⁄15)x - 1⁄15 = 0
Comparing this with x³ + p₁x² + p₂x + p₃ = 0, we get
p₁ = -23⁄15; p₂ = (9⁄15); p₃ = -1⁄15

By data the roots are in H.P.
Let the roots be 1⁄(α - β), 1⁄α, 1⁄(α + β)
We know, sum of roots = -p₁ ⇒ 1⁄(α - β) + 1⁄α + 1⁄(α + β) = 23⁄15.....(i)
sum of the product of the roots taken two at a time = p₂ = (9⁄15)
⇒ 1⁄{(α - β)(α)} + 1⁄{(α - β)(α + β)} + 1⁄{(α)(α + β)} = (9⁄15)
⇒ {(α + β) + α + (α - β)}⁄{(α - β)(α)(α + β)} = (9⁄15)
⇒ 3α⁄{(α - β)(α)(α + β)} = (9⁄15).......(ii)

product of the roots = -p₃ = 1⁄15
⇒ 1⁄{(α - β )(α)(α + β)} = 1⁄15.......(iii)

(ii) ÷ (iii) gives 3α = (9⁄15)⁄(1⁄15) = 9 ⇒ &alpha = 3
From (iii), (α - β )(α)(α + β) = 15
Using the value of α here, we get
(3 - β )(3)(3 + β) = 15⇒ 9 - β² = 5 ⇒ β² = 4⇒ β = ±2

The values of α and β are found from (ii) and (iii).
Let us verify whether they satisfy (i) or not.

First consider &alpha = 3 and β = +2
L.H.S. of (i) = 1⁄(α - β) + 1⁄α + 1⁄(α + β) = 1⁄(3 - 2) + 1⁄&3 + 1⁄(3 + 2) = 1 + 1⁄&3 + 1⁄5
= (15 + 5 + 3)⁄15 = 23⁄15 = R.H.S. of (i) [satisfied.]
Now consider &alpha = 3 and β = -2 L.H.S. of (i) = 1⁄(α - β) + 1⁄α + 1⁄(α + β) = 1⁄(3 + 2) + 1⁄&3 + 1⁄(3 - 2) = 1⁄5 + 1⁄&3 + 1
= (3 + 5 + 15)⁄15 = 23⁄15 = R.H.S. of (i) [satisfied.]
So, &alpha = 3, β = ±2 satisfy equation (i). ∴ The required roots are 1⁄(α - β), 1⁄α, 1⁄(α + β)
= 1⁄(3 - 2), 1⁄&3, 1⁄(3 + 2) (or) 1⁄(3 + 2), 1⁄&3, 1⁄(3 - 2) = 1, 1⁄&3, 1⁄5 (or) 1⁄5, 1⁄&3, 1
As we can see, both sets are same.

Thus, the roots of the given Algebra Equation are
1, 1⁄&3, 1⁄5. Ans.











Solved Example 5 : Cubic and Biquadratic Algebra Equations

Solve the Algebra Equation 8x³ - 14x² + 7x - 1 = 0, the roots of which are in G.P.

Solution :
The given Algebra Equation is 8x³ - 14x² + 7x - 1 = 0
Dividing both sides of the equation by 8, we get
x³ - (7⁄4)x² + (7⁄8)x - 1⁄8 = 0
Comparing this with x³ + p₁x² + p₂x + p₃ = 0, we get
p₁ = -7⁄4; p₂ = 7⁄8; p₃ = -1⁄8

By data the roots are in G.P.
Let the roots be α⁄β, α, αβ
We know, sum of roots = -p₁ = 7⁄4
⇒ α⁄β + α + αβ = 7⁄4 ⇒ α(1⁄β + 1 + β) = 7⁄4.........(i)

sum of the product of the roots taken two at a time = p₂ = 7⁄8
⇒ (α⁄β)α + (α⁄β)(αβ) + α(αβ) = 7⁄8 ⇒ (α)²(1⁄β + 1 + β) = 7⁄8.......(ii)

product of the roots = -p₃ = 1⁄8
⇒ (α⁄β)(α)(αβ) = 1⁄8 ⇒ α = 1⁄2.........(iii)

(ii) ÷ (i) gives α = 1⁄2 [same as (iii)]

Using the value of α = 1⁄2 in (i), we get
(1⁄2)(1⁄β + 1 + β) = 7⁄4 ⇒ (1⁄β + 1 + β) = 7⁄2
Multiplying both sides with 2β, we get
2 + 2β + 2β² = 7β ⇒ 2β² - 5β + 2 = 0 ⇒ 2β² - 4β - β + 2 = 0
⇒ 2β(β - 2) -1(β - 2) = 0 ⇒ (β - 2)(2β - 1) = 0⇒ β = 2 or 1⁄2

∴ The required roots are α⁄β, α, αβ
= (1⁄2)⁄2, (1⁄2), (1⁄2)2 (or) (1⁄2)⁄(1⁄2), (1⁄2), (1⁄2)(1⁄2) = 1⁄4, 1⁄2, 1 (or) 1, 1⁄2, 1⁄4 As we can see, both sets are same.

Thus, the roots of the given Algebra Equation are
1⁄4, 1⁄2, 1. Ans.

Solved Example 6 : Cubic and Biquadratic Algebra Equations

If the Algebra Equation x⁴ + 4x³ - 2x² - 12x + 9 = 0, has a pair of equal roots, solve the equation.

Solution :
The given Algebra Equation is
x⁴ + 4x³ - 2x² - 12x + 9 = 0
Comparing this with x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0, we get
p₁ = 4; p₂ = -2; p₃ = -12; p₄ = 9

Also we have
s₁ = -p₁ = -4; s₂ = p₂ = -2; s₃ = -p₃ = 12; s₄ = p₄ = 9;

By data two pairs of roots are equal.
∴ Let the roots be α, α, β, β
s₁ = α + α + β + β = 2(α + β) = -4 ⇒ α + β = -2........(i)

s₄ = (α)(α)(β)(β) = (αβ)² = 9⇒ αβ = ±3 ......(ii)

Using (i) and (ii), we get
(α - β)² = (α + β)² - 4αβ = (-2)² - 4(±3) = 4 - 12 or 4 + 12 = -8 or 16
Taking positive value, α - β = ±4.......(iii)
(i) + (iii) gives 2α = 2 or -6⇒ α = 1 or -3
using these in (i), we get
β = -2 - α = -2 -1 or -2 +3 = -3 or 1
These are the same as α values.
∴ the roots are 1, 1, -3, -3.

These roots are found using s₁ and s₄.
Let us verify whether these satisfy s₂ and s₃.
s₂ = (1)(1) + (1)(-3) + (1)(-3) + (1)(-3) + (1)(-3) + (-3)(-3)
= 10 - 12 = -2 [satisfied.]
s₃ = (1)(1)(-3) + (1)(1)(-3) + (1)(-3)(-3) + (1)(-3)(-3) = 18 - 6 = 12 [satisfied.]

Thus the roots of the given Algebra Equation are
1, 1, -3, -3. Ans.

Solved Example 7 : Cubic and Biquadratic Algebra Equations

Solve the Algebra Equation 16x⁴ - 64x³ + 56x² + 16x - 15 = 0, the roots of which are in A.P.

Solution :
The given Algebra Equation is 16x⁴ - 64x³ + 56x² + 16x - 15 = 0
Dividing both sides of the equation by 16, we get
x⁴ - 4x³ + (7⁄2)x² + x - 15⁄16 = 0
Comparing this with x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0, we get
p₁ = -4; p₂ = 7⁄2; p₃ = 1; p₄ = -15⁄16

By data the roots are in A.P.
Let the roots be α - 3β, α - β, α + β, α + 3β
We know, s₁ = -p₁ = 4 ⇒ α - 3β + α - β + α + β + α + 3β = 4 ⇒ 4α = 4⇒ α = 1......(i)

Using the value of α = 1, the roots become 1 - 3β, 1 - β, 1 + β, 1 + 3β

We know, s₂ = p₂ = 7⁄2 ⇒ (1 - 3β)(1 - β) + (1 - 3β)(1 + β) + (1 - 3β)(1 + 3β) + (1 - β)(1 + β) + (1 - β)(1 + 3β) + (1 + β)(1 + 3β) = 7⁄2
⇒ (1 - 4β + 3β²) + (1 - 2β - 3β²) + (1 - 9β²) + (1 - β²) + (1 + 2β - 3β²) + (1 + 4β + 3β²) = 7⁄2
⇒ 6 - 10β² = 7⁄2 ⇒ 12 - 20β² = 7
⇒ 20β² - 5 = 0 ⇒ β² = 1⁄4 ⇒ β = ±(1⁄2)......(ii)

When β = +(1⁄2), the roots become -1⁄2, 1⁄2, 3⁄2, 5⁄2.
When β = -(1⁄2), the roots become 5⁄2, 3⁄2, 1⁄2, -1⁄2.
As we can see, both sets are same.

Let us verify whether the set of roots satisfy values of s₃ and s₄ We know, s₃ = -p₃ = -1 and s₄ = p₄ = -15⁄16 But, s₃ = (-1⁄2)(1⁄2)(3⁄2) + (-1⁄2)(1⁄2)(5⁄2) + (-1⁄2)(3⁄2)(5⁄2) + (1⁄2)(3⁄2)(5⁄2)
= (-3⁄8) + (-5⁄8) + (-15⁄8) + (15⁄8) = -1 [satisfied.] s₄ = (-1⁄2)(1⁄2)(3⁄2)(5⁄2) = -15⁄16 [satisfied.]

∴ The roots of the given Algebra Equation are
-1⁄2, 1⁄2, 3⁄2, 5⁄2. Ans.

Solved Example 8 : Cubic and Biquadratic Algebra Equations

Solve the Algebra Equation 6x⁴ - 29x³ + 40x² - 7x - 12 = 0, if the product of the two of its roots is equal to 2.

Solution :
The given Algebra Equation is
6x⁴ - 29x³ + 40x² - 7x - 12 = 0
Dividing both sides of the equation by 6, we get
x⁴ - (29⁄6)x³ + (20⁄3)x² - (7⁄6)x - 2 = 0
Comparing this with x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0, we get
p₁ = - (29⁄6); p₂ = (20⁄3); p₃ = -(7⁄6); p₄ = -2

Also we have s₁ = -p₁ = (29⁄6); s₂ = p₂ = (20⁄3); s₃ = -p₃ = (7⁄6); s₄ = p₄ = -2;

By data, product of two of its roots = 2.
But, s₄ = product of all the four roots = -2
⇒ (2)(product of the other two roots) = -2
⇒ product of the other two roots = -1.
∴ Let the roots be α, 2⁄α, β, -1⁄β

s₁ = α + 2⁄α + β - 1⁄β = 29⁄6......(i)

s₃ = (α)(2⁄α)(β) + (α)(2⁄α)(-1⁄β) + (α)(β)(-1⁄β) + (2⁄α)(β)(-1⁄β) = 2β - 2⁄β - α - 2⁄α = 7⁄6......(iii)

(i) + (iii) gives 3β - 3⁄β = 29⁄6 + 7⁄6 = 6 ⇒ β - 1⁄β = 2⇒ β² - 1 = 2β
⇒ β² - 2β - 1 = 0⇒ (β - 1)² - 1 - 1 = 0
(β - 1)² = 2⇒ (β - 1) = ±√2 ⇒ β = 1 ± √2 = 1 + √2, 1 - √2.
It is clear that one value is the negative reciprocal of the other.

Using these in (i), we get α + 2⁄α + 1 + √2 + 1 - √2 = 29⁄6
⇒ α + 2⁄α = 29⁄6 - 2 = 17⁄6 ⇒ 6α² + 12 = 17α
⇒ 6α² - 17α + 12 = 0 ⇒ 6α² - 9α - 8α + 12 = 0
⇒ 3α(2α - 3) - 4(2α - 3) = 0 ⇒ (2α - 3)(3α - 4) = 0 ⇒ α = 3⁄2, 4⁄3
It is clear that one value is two times the reciprocal of the other.

∴ The roots are 3⁄2, 4⁄3, 1 + √2, 1 - √2

These values are found using s₁, s₃, s₄
Let us verify whether they satisfy s₂.
s₂ = (3⁄2)(4⁄3) + (3⁄2)(1 + √2) + (3⁄2)(1 - √2) + (4⁄3)(1 + √2) + (4⁄3)(1 - √2) + (1 + √2)(1 - √2)
= 2 + 3⁄2 + 3√2⁄2 + 3⁄2 - 3√2⁄2 + 4⁄3 + 4√2⁄3 + 4⁄3 - 4√2⁄3 + 1 - 2
= 20⁄3 [satisfied.]

Thus the roots of the given Algebra Equation are
3⁄2, 4⁄3, 1 + √2, 1 - √2. Ans.

Solved Example 9 : Cubic and Biquadratic Algebra Equations

Solve the Algebra Equation x³ - 5x² - 4x + 20 = 0 Solution :
The given algebra eqaution is x³ - 5x² - 4x + 20 = 0

Here, no additional condition is given.
Let f(x) = x³ - 5x² - 4x + 20
Such type of equations were factorised in Algebra Factoring Please study there.
We will follow the same procedure.
constant term = 20 has factors 1, 2, 4, 5, 10, 20 -1, -2, -4, -5, -10, -20.
Let us verify whether f(a) = 0, where a = one of those factors.
f(1) and f(-1) are not zero.
f(2) = 2³ - 5(2)² - 4(2) + 20 = 8 - 20 - 8 + 20 = 0
⇒ (x - 2) is a factor of f(x).
⇒ x = 2 is a root of f(x) = 0.
one root is found as 2. Let the other roots be α and β
s₁ = 2 + α + β = -p₁ = 5 ⇒ α + β = 3......(i)

s₂ = (2)(α) + (2)(β) + (α)(β) = p₂ = -4 ⇒ 2(α + β) + (α)(β) = -4.....(ii)
Using (i) in (ii), we get
2(3) + (αβ) = -4 ⇒ αβ = -10.....(iii)

Using (i) and (iii), we can find (α - β)
(α - β)² = (α + β)² - 4αβ = 3² - 4(-10) = 9 + 40 = 49
⇒ (α - β) = ±7 ......(iv)
(i) + (iv) gives 2α = 10 or -4 ⇒ α = 5 or -2.
using these in (i), we get
β = 3 - 5 or 3 + 2 = -2 or 5.
These are same as values of α
⇒ the roots are 2, 5, -2.
Let us verify whether these satisfy s₃ or not.
s₃ = (2)(5)(-2) = -20 = -p₃ [satisfied.]

Thus the roots of the given Algebra Equation are
2, 5, -2. Ans.

Solved Example 10 : Cubic and Biquadratic Algebra Equations

Solve the algebra equation x⁴ - 16x³ + 86x² - 176x + 105 = 0

Solution :
The given algebra equation is x⁴ - 16x³ + 86x² - 176x + 105 = 0

Let f(x) = x⁴ - 16x³ + 86x² - 176x + 105
Constant term = 105 has factors 1, 3, 5, 7, -1, -3, -5, -7. Let us verify whether f(a) = 0, where a = one of those factors.
f(1) = 1 -16 + 86 -176 + 105 = 192 - 192 = 0
⇒ (x - 1) is a factor of f(x).
⇒ x = 1 is a root of f(x) = 0.
You can use synthetic division here, as explained in Algebra Factoring (link given above.)

+1   |   +1   -16    +86    -176    +105
     |   +0   +1     -15    +71     -105
     |---------------------------------------------
     |   +1   -15    +71    -105      0

s₂ = (3)(α) + (3)(β) + (α)(β) = p₂ = 71 ⇒ 3(α + β) + (α)(β) = 71.....(ii)
Using (i) in (ii), we get
3(12) + (αβ) = 71 ⇒ αβ = 35.....(iii)

Using (i) and (iii), we can find (α - β)
(α - β)² = (α + β)² - 4αβ = 12² - 4(35) = 144 - 140 = 4
⇒ (α - β) = ±2 ......(iv)
(i) + (iv) gives 2α = 14 or 10 ⇒ α = 7 or 5.
using these in (i), we get
β = 12 - 7 or 12 - 5 = 5 or 7.
These are same as values of α
⇒ the roots are 3, 5, 7.
Let us verify whether these satisfy s₃ or not.
s₃ = (3)(5)(7) = 105 = -p₃ [satisfied.] ⇒ roots of g(x) = 0 are 3, 5, 7.

Thus, roots of of the given Algebra Equation are
1, 3, 5, 7. Ans.

Exercise : Cubic and Biquadratic Algebra Equations

1. Solve the Algebra Equation 8x³ - 36x² - 18x + 81 = 0, the roots of which are in A.P.
2. Solve the Algebra Equation 6x³ - 11x² + 6x - 1 = 0, the roots of which are in H.P.
3. Solve the Algebra Equation 54x³ - 39x² - 26x + 16 = 0, the roots of which are in G.P.
4. If one root of the Algebra Equation 24x³ - 14x² - 63x + 45 = 0 is double another root, solve the equation.
5. Solve the Algebra Equation 18x³ + 81x² +121x + 60 = 0, given a root is equal to half the sum of the remaining roots.
6. Solve the Algebra Equation 9x³ - 15x² + 7x - 1 = 0 two of the roots being equal..
7. Solve the Algebra Equation x⁴ - 2x³ - 21x² + 22x + 40 = 0, the roots of which are in A.P.
8. Solve the Algebra Equation 8x⁴ - 2x³ - 27x² + 6x + 9 = 0, if the sum of two of its roots is 0.
9. Solve the Algebra Equation x³ - 5x² - 2x + 24 = 0
10. Solve the Algebra Equation x⁴ - 3x³ - 15x² + 19x + 30 = 0

For Answers see at the bottom of the page.

Answers to Exercise : Cubic and Biquadratic Algebra Equations

(1) -3⁄2, 3⁄2, 9⁄2 (2) 1, 1⁄2, 1⁄3 (3) 1⁄2, -2⁄3, 8⁄9
(4) 3⁄4, 3⁄2, -5⁄3 (5) -3⁄2, -4⁄3, -5⁄3 (6) 1⁄3, 1⁄3, 1
(7) -4, -1, 2, 5 (8) -1⁄2, 3⁄4, √3, -√3 (9) 3, 4, -2 (10) -3, -1, 2, 5