hello do this simple poly question for me (1 Viewer)

[kooltrainer]

1)Find A and B given that x^3 + 3x + 2i = (x-A)^2 (X-B)

as simple as that. But i need help.. :apig:

2) Find all roots of the equation 18x^3 + 3x^2 - 28x + 12=0 if 2 of the roots are equal
could u use sum of roots for this question??

 

[leoyh]

expand the RHS and rearrange so that you can equate coefficients and u should be able to get the answers

 

[3unitz]

Find A and B given that x^3 + 3x + 2i = (x-A)^2 (X-B)

as simple as that. But i need help.. :apig:

f(x) = x^3 + 3x + 2i
f(-i) = 0 & f'(-i) = 0

.'. A = -i

x^3 + 3x + 2i = (x + i)^2(x - B)
sub x = 0,

B = 2i

 

[ronnknee]

LHS
= x^3 + 3x + 2i

RHS
= (x - A)^2 . (x - B)
= x^3 - (2A + B)x^2 + (A^2 + 2AB)x - A^2.B

By equating the coefficients of
x^2:
2A + B = 0
B = -2A

Constant:
-BA^2 = 2i
-(-2A)A^2 = 2i
A^3 = i
A^3 = (-i^2)i [note: -i^2 = 1]
A^3 = -i^3
A = -i

Since B = -2A and A = -i,
B = 2i

 

[kooltrainer]

zomg.. how come sum of roots wont work for question 1 =( .. can sum1 show me..

 

[ronnknee]

Question 2 (without using sum of roots)
P(x) = 18x^3 + 3x^2 - 28x + 12 = 0
Since there is a double root, P(x) = P'(x) = 0
P'(x)
= 54x^2 + 6x - 28
= 27x^2 + 3x - 14
= 0
(now use quadratic formula to find the zeroes of P'(x))
x = 2/3, -7/9

Therefore either 2/3 or -7/9 can be the double root

Test:
P(-7/9) does not equal to 0
P(2/3) = 0

Therefore 2/3 is the double root

P(x) = (Ax - B)(3x - 2)^2
By comparing the coefficients of:
x^3:
18 = 9A
A = 2

Constant term:
-12 = 4B
B = -3

Therefore P(x) = (2x + 3)(3x - 2)^2