FIND THE OTHER ROOTS OF THE POLYNOMIAL EQUATION OF DEGREE 6

Question 1 :

Find all zeros of the polynomial x6 − 3x5 − 5x4 + 22x3 − 39x2 − 39x + 135, if it is known that 1 + 2i an3 are two of its zeros.

Solution :

The four roots are 1 + 2i, 1 - 2i, 3 and -3.

Quadratic equation whose roots are 1 + 2i and 1 - 2i.

Sum of roots  =  2

Product of roots  =  1 + 4  =  5

x2 - 2x + 5

Sum of roots   =  0

Product of roots  =  -3

(x2 - 3) is a factor.

(x2 - 2x + 5) (x2 - 3)  =  x4 - 3x2 - 2x3 + 6x + 5x2 - 15

  =  x4 - 2x3 + 2x2 + 6x - 15

x2 - x - 9  =  0

x = (-b ± √b2 - 4ac)/2a

x = (1 ± √1 + 36)/2

x = (1 ± √37)/2

Hence the zeroes are 1 + 2i, 1 - 2i, 3, -3, (1 ± √37)/2.

Question 2 :

Solve the cubic equation

(i) 2x3 − 9x2 +10x = 3

Solution :

2x3 − 9x2 +10x - 3  =  0

(x -1) is a factor

  =  2x2 - 7x + 3  

  =  2x2 - 1x - 6x + 3  

  =  x(2x - 1) - 3(2x - 1)

  =  (x - 3) (2x - 1)

(x - 1) (x - 3) (2x - 1)  =  0

x - 1  =  0, x - 3  =  0,  2x - 1  =  0

x  =  1, x  =  3, x  =  1/2

Hence the solutions are 1, 3 and 1/2.

(ii)  8x3 − 2x2 − 7x + 3 = 0.

Solution :

(x + 1) is a factor.

The other factors are 8x2 - 10x + 3

  =  8x2 - 12x + 2x + 3

  =  4x(2x - 3) -1(2x - 3)

  =  (4x - 1) (2x - 3)

By equating them to zero, we get 

x + 1  =  0, 4x - 1  =  0,  2x - 3  =  0

x  =  -1, x  =  1/4,  x  =  3/2

Hence the solutions are -1, 1/4 and 3/2.

Question 3 :

Solve the equation x4 −14x2 + 45 = 0

Solution :

Let x2  =  t

t2 − 14t + 45 = 0

t2 − 9t - 5t + 45 = 0

t(t - 9) - 5(t - 9)  =  0

(t - 5) (t - 9)  =  0

t  =  5 and t  =  9

x=  5, x2  =  9

x  =  ±√5, x = ±3

Hence the solutions are √5, -√5, 3 and -3.

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