SOLVING POLYNOMIAL EQUATIONS WITH DIFFERENT POWERS

Question 1 :

Solve the following equations

(i) sin² x − 5sin x + 4 = 0

Solution :

Let y = sin x

y² − 5 y + 4 = 0

(y - 1) (y - 4)  =  0

y  =  1 and y  =  4

(ii) 12x³ + 8x = 29x² − 4

Solution :

12x³ + 8x = 29x² − 4

12x³ - 29x² + 8x + 4  =  0

(x - 2) is a factor.The other factors are 12x² - 5x - 2

12x² - 5x - 2  =  0

12x² - 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 xⁿ + ........ + a₁x + 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 a₀ and q is a factor of a_n .

Question  2 :

Examine for the rational roots of

(i) 2x³ − x² − 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) x⁸ − 3x + 1 = 0

Solution :

Factor of 1 are ± 1 (the constant) = p

Factor of 1 are ±1 (coefficient of x⁸) = 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.

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