Solving Polynomial Equations with Different Powers :
Here we are going to see some example problems of solving polynomials.
Question 1 :
Solve the following equations
(i) sin^{2} x − 5sin x + 4 = 0
Solution :
Let y = sin x
y^{2} − 5 y + 4 = 0
(y - 1) (y - 4) = 0
y = 1 and y = 4
sin x = 1 General solution = nπ ± (-1)^{n }y y = π/2 Here n = 2π = 2nπ + (π/2) |
sin x = 4 No solution. |
(ii) 12x^{3} + 8x = 29x^{2} − 4
Solution :
12x^{3} + 8x = 29x^{2} − 4
12x^{3} - 29x^{2 }+ 8x + 4 = 0
(x - 2) is a factor.The other factors are 12x^{2} - 5x - 2
12x^{2} - 5x - 2 = 0
12x^{2} - 8x + 3x - 2 = 0
4x(3x - 2) + 1(3x - 2) = 0
(4x + 1) (3x - 2) = 0
So, the factors are (x - 2) (4x + 1) (3x - 2)
x - 2 = 0, 4x + 1 = 0, 3x - 2 = 0
x = 2, x = -1/4, x = 2/3
Hence 2, -1/4 and 2/3 are roots of the given polynomial.
Let a_{n} x^{n} + ........ + a_{1}x + a0 with a_{n} ≠ 0 and, be a polynomial with integer coefficients. If p/q, with ( p, q) =1, is a root of the polynomial, then p is a factor of a_{0} and q is a factor of a_{n} .
Question 2 :
Examine for the rational roots of
(i) 2x^{3} − x^{2} − 1 = 0
Solution :
Factor of 1 are ± 1 = p
Factor of 2 are ±1 and ±2 = q
p/q = 1/1, 1/2, -1/1, -1/2
If the given equation will have rational solution, then it must be one of the above given solutions.
1 is the rational number and it is the factor of the given polynomial. 1/2 is not the factor of the given polynomial.
Hence the given polynomial has rational roots.
(ii) x^{8} − 3x + 1 = 0
Solution :
Factor of 1 are ± 1 (the constant) = p
Factor of 1 are ±1 (coefficient of x^{8}) = q
p/q = 1, -1
If the given equation will have rational solution, then it must be one of the above given solutions.
Hence the given polynomial will not have rational roots.
After having gone through the stuff given above, we hope that the students would have understood, "Solving Polynomial Equations with Different Powers".
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