POLYNOMIAL EQUATION - SOLVING EQUATIONS OF HIGHER DEGREE WITH EXAMPLES

Please study Method of Solving Polynomial Equation 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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Solve the Polynomial Equation 4x^{3} - 24x^{2} + 23x + 18 = 0, the roots of which are in A.P.

Solution to Example 1 of Polynomial Equation : The given Algebra Equation is 4x^{3} - 24x^{2} + 23x + 18 = 0 Dividing both sides of the equation by 4, we get x^{3} - 6x^{2} + (23⁄4)x + 9⁄2 = 0 Comparing this with x^{3} + p_{1}x^{2} + p_{2}x + p_{3} = 0, we get p_{1} = -6; p_{2} = (23⁄4); p_{3} = 9⁄2

By data the roots are in A.P. Let the roots be α - β, α, α + β We know, sum of roots = -p_{1} ⇒ α - β + α + α + β = 6 ⇒ 3α = 6⇒ α = 2.........(i)

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

∴ 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.

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Solution to Example 2 of Polynomial Equation : The given algebra eqaution is x^{3} - 5x^{2} - 4x + 20 = 0

Here, no additional condition is given. Let f(x) = x^{3} - 5x^{2} - 4x + 20 Such type of equations were factorized 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) = 2^{3} - 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 s_{1} = 2 + α + β = -p_{1} = 5⇒ α + β = 3......(i)

s_{2} = (2)(α) + (2)(β) + (α)(β) = p_{2} = -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= 3^{2} - 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 s_{3} or not. s_{3} = (2)(5)(-2) = -20 = -p_{3} [satisfied.]

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

Example 2 of Polynomial Equation is thus solved.

Exercise : Polynomial Equation

Problems on Polynomial Equation :

Solve the Polynomial Equation 54x^{3} - 39x^{2} - 26x + 16 = 0, the roots of which are in G.P.

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