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
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.)
⇒ 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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