SOLVING PARTLY FACTORED POLYNOMIAL EQUATIONS

Question 1 :

Solve : (i) (x − 5)(x − 7)(x + 6)(x + 4) = 504

Solution :

(x − 5)(x − 7)(x + 6)(x + 4) = 504

(x − 5)(x + 4) (x − 7)(x + 6) = 504

(x² - x - 20) (x² - x - 42)  =  504

x² - x  =  t

(t - 20)(t - 42)  =  504

t² - 20t - 42t + 840 - 504  =  0

t² - 62t + 336  =  0

t² - 56t - 6t + 336  =  0

t(t - 56) - 6(t - 56)  =  0

(t - 6) (t - 56)  =  0

t  =  6 and t = 56

Hence 3, -2, 8 and -7 are roots of the given polynomial.

(ii) (x − 4)(x − 7)(x − 2)(x +1) = 16

Solution :

(x − 4)(x − 7)(x − 2)(x +1) = 16

(x − 4)(x − 2)(x − 7)(x +1) = 16

(x² - 6x + 8) (x² - 6x - 7)  =  16

x² - 6x  =  t

(t + 8)(t - 7)  =  16

t² - 7t + 8t - 56 - 16  =  0

t² + t - 72  =  0

t² + 9t - 8t - 72  =  0

t(t + 9) - 8(t + 9)  =  0

(t + 9) (t - 8)  =  0

t  =  -9 and t = 8

Hence 3, 3, 3 + 2√17 and 3 - 2√17 are roots of the given polynomial.

Question 2 :

Solve : (2x −1)(x + 3)(x − 2)(2x + 3) + 20 = 0

Solution :

(2x −1)(x + 3)(x − 2)(2x + 3) + 20 = 0

(2x −1)(2x + 3)(x + 3)(x − 2) + 20 = 0

(4x² + 6x - 2x - 3)(x² - 2x + 3x - 6) + 20 = 0

(4x² + 4x - 3)(x² + x - 6) + 20 = 0

(4(x² + x) - 3)(x² + x - 6) + 20 = 0

Let x² + x  = t 

(4t - 3)(t - 6) + 20 = 0

4t² - 24t - 3t + 18 + 20  =  0

4t² - 27t + 38  =  0

4t² - 19t - 8t + 38  =  0

t(4t - 19) - 2(4t - 19)  =  0

(t - 2) (4t - 19)  =  0

t  =  2 and t  =  19/4

Hence 1, -2, (-1 + 2√5)/2 and (-1 - 2√5)/2 are roots of the given polynomial.

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