Roots of multiplicity - Polynomials (1 Viewer)

[K4M1N3]

Q. The Polynomial equation P(x) = 0 has a double root at x = a. By writing:

Where Q(x) is a polynomial

Show that P'(a) = 0

I was just wondering since P(x) has only a double root at x = a......Doesn't that mean that the polynomial Q(x) does NOT have a root at x = a?
i.e.

 

[khorne]

Q. The Polynomial equation P(x) = 0 has a double root at x = a. By writing:

Where Q(x) is a polynomial

Show that P'(a) = 0

I was just wondering since P(x) has only a double root at x = a......Doesn't that mean that the polynomial Q(x) does NOT have a root at x = a?
i.e.

Yes, Q(a) is not 0, and x=a is not a root. Instead, the question just says, differentiate it

such as: P'(x) = 2(x-a)Q'(x)

Thus, x=a is a root of P'(x) too.

 

[cutemouse]

Yes, Q(a) is not 0, and x=a is not a root. Instead, the question just says, differentiate it

such as: P'(x) = 2(x-a)Q'(x)

Thus, x=a is a root of P'(x) too.

Product rule?

 

[XTsquared]

Yes, Q(a) is not 0, and x=a is not a root. Instead, the question just says, differentiate it

such as: P'(x) = 2(x-a)Q'(x)

Thus, x=a is a root of P'(x) too.

P'(x) = 2(x-a)Q(x) + ((x-a)^2)Q'(x)

 

[khorne]

i was lazy and did it partially, it changes nothing

 

[XTsquared]

i was lazy and did it partially, it changes nothing

Lol, you used correct grammar etc?

 

[cutemouse]

i was lazy and did it partially, it changes nothing

It's still wrong...