Determine the polynomial P=x^7+a*x^5+b*x^3+3 if P'(1)=0 !

Expert Answers
hala718 eNotes educator| Certified Educator

P = x^7 + ax^5 + bx^3 + 3

p'(1) ==> P(1) = 0

==> P(1) = 1 + a + b + 3 =0

==> a+b+4 = 0 ........(1)

==> p'(x) = 7x^6 + 5ax^4 + 3bx^2

==> p'(1) = 7+5a + 3b = 0

==>  5a+3b + 7 = 0........(2)

From (1) and (2):

==> 2a - 5 = 0

==> a = 5/2

==> b = -4-a = -4 - 5/2 = -13/2

==> P(x) = x^7 + (5/2)x^5 - (13/2)x^3 + 3

 

giorgiana1976 | Student

If x=1 is a multiple root, it is the root of the the first derivative of the polynomial, also it is the root of the polynomial.

So, that means that P(1)=0 and P'(1)=0

P(1)=1+a+b+3=0

a+b+4=0

P'(x)=7x^6+5ax^4+3bx^2

P'(1)=5a+3b+7=0

a+b=-4, 5a+3b=-7

b=-a-4, 5a + 3(-a-4)=-7, 5a - 3a-12+7=0, 2a=5, a=5/2

b=-5/2-4=-13/2, b=-13/2

P(x) = x^7 + (5*x^5)/2 - (13*x^3)/2 + 3

or

P(x) = 2*x^7 + 5*x^5 - 13*x^3 + 6

neela | Student

To determine P=x^7+a*x^5+b*x^3+3 if P'(1)=0 !

Solution:

P=x^7+a*x^5+b*x^3+3 if P'(1)=0 !

Therefore p(x) =x^7+a*x^5+b*x^3+3 if P'(1)=0 ! = 1

P'(x) = 7x^6+5ax^4+3bx^2+0, as  (x^n)' = nx^(n-1).

p'(1)  = 0! = 1,

So 7 x^6+5ax^4+3bx^2 =1 at x =1 becomes

7*1^6+5a1^4 +3b1^2 = 1

7+5a+3b = 1. Or

5a+3b = 1-7 = -6

5a+3b = -6............(1)

So a = -6-3b = - 2.(3+b)

Therefore for any value of a and  b which  obey

a = -2(3+b) , p'(1) = 0! for the given function.

To detrmine both a and b  only one condition is given. We need 2 conditions to detrmine a and b.

 

 

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