if P(x) = x^4+2x^3-12x^2-40x+c has a triple zero, find c and factorise P(x) over the real field.

Answer 1

To find the value of c, we first need to understand what it means for a polynomial to have a triple zero.

A polynomial has a triple zero at a value a if it can be factored as (x-a)^3. This means that the factor (x-a) occurs three times in the factorization.

So for our polynomial P(x) = x^4 + 2x^3 - 12x^2 - 40x + c to have a triple zero, it must be in the form (x-a)^3.

Expanding this out, we get:

(x-a)^3 = x^3 - 3ax^2 + 3a^2x - a^3

If we compare this to our polynomial P(x), we can equate coefficients to find the value of a:

a^3 = c
-3a = -40
3a^2 = -12
a = 4

Therefore, c = a^3 = 4^3 = 64.

Now that we have found the value of c, we can factorize P(x) = x^4 + 2x^3 - 12x^2 - 40x + 64 over the real field:

P(x) = (x - 4)^3 = (x - 4)(x - 4)(x - 4) = (x - 4)^3

Therefore, the factored form of P(x) over the real field is (x - 4)^3.