Question

A polynomial equation $2x^{3}+px+108=0$ has a double root (a root with multiplicity 2). Find the value of the coefficient p.

Answer 1

Answer:

Hope it's correct thank you

Answer 2

Given :  A polynomial equation 2x³ + px + 108 = 0  has a double root (a root with multiplicity 2).

To Find : the value of the coefficient p.

Solution:

Let say root with multiplicity 2  is a   and other root is b

so roots are

a , a , b

Sum of roots = a + a + b = 0    ( as coefficient of x²  is 0)

=> b = - 2a

Product of roots = -108/2 = -54

a.a.(-2a) = -54

=> a =   3

and b = -2a = -6

Sum of product of pair of roots

aa + ab  + ab  =  p/2

=> 3*3 + 3 * (-6) + 3(-6) = p/2

=> 9 - 18 - 18 = p/2

=> p = -54

the value of the coefficient p. is - 54

or after finding a = 3

just substitute a = 3  in   2x³ + px + 108 = 0

2(3)³ + 3p + 108 = 0

=>  54 + 3p + 108 = 0

=> 3p = -162

=> p = -54

Additional info :

2x³ -54x + 108  = 2(x³ - 27x  + 54)  = 2(x - 3)²(x + 6)

Learn More:

Find a quadratic polynomial whose zeroes are 1/2 , 4 - Brainly.in

brainly.in/question/16158030

Find the quadratic polynomial whose zeroes are log 1000, log0.01*0.1

brainly.in/question/18047168

brainly.in/question/18973744