SOLVING POLYNOMIALS OF HIGHER DEGREE WITH EXAMPLES

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Please study
Method of Solving Polynomials
if you have not already done so.
There we gave introduction to Algebra
Equations of higher degree and discussed
the method of solving them.

We developed the relations between
the roots and coefficients of the
equation and gave Formulas for the same.
That knowledge is a prerequisite here.
So please study them before proceeding further.

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Solving Polynomials

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Example 1 : Solving Polynomials

Solve the following problem on Solving Polynomials.

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

Solution to Example 1 Solving Polynomials:
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.

Example 1 Solving Polynomials is thus solved :

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Example 2 : Solving Polynomials

Solve the following problem on Solving Polynomials.

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

Solution to Example 2 Solving Polynomials:
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

⇒ f(x) ÷ (x - 1) = x3 - 15x2 + 71x - 105
Let g(x) = x3 - 15x2 + 71x - 105
g(3) = 33 - 15(3)2 + 71(3) -105 = 27 - 135 + 213 - 105 = 0
⇒ (x - 3) is a factor of g(x).
x = 3 is a root of g(x) = 0.
Let α and β be the other roots of g(x) = 0
s1 = 3 + α + β = -p1 = 15⇒ α + β = 12......(i)

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, Solving Polynomials, we get
roots of of the given Algebra Equation are
1, 3, 5, 7. Ans.

Example 2 Solving Polynomials is thus solved :

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Exercise : Solving Polynomials

Solve the following problems on Solving Polynomials.

  1. Solve the Algebra Equation 8x4 - 2x3 - 27x2 + 6x + 9 = 0, if the sum of two of its roots is 0.
  2. Solve the Algebra Equation x4 - 3x3 - 15x2 + 19x + 30 = 0

For Answers see at the bottom of the page.

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Answers to Exercise : Solving Polynomials

Answers to problems on Solving Polynomials.

  1. -1⁄2, 3⁄4, √3, -√3
  2. -3, -1, 2, 5