The polynomial function has zeroes as given by x=0
In general, given zeroes of a polynomial function, a, b, c, and d, we can write the function as the multiplication of the factors
(x-a), (x-b), (x- c), and (x-d) then the degrees are 4
simply:
f(x)=(x-a) (x-b) (x- c) (x-d)
In this case, we can show that each of a, b, c, and d are zeroes of the function:
f(a)=(a-a) (a-b) (a-c) (a-d)
= (0) (a-b) (a-c) (a-d)
=0
f(b)=(b-a) (b-b) (b-c) (b-d)
=(b-a) (0) (b-c) (b-d)
=0
f(c)=(c-a) (c-b) (c-c) (c-d)
=(c-a) (c-b) (0) (c-d)
=0
f(d)=(d-a) (d-b) (d-c) (d-d)
=(d-a) (d-b) (d-c) (0)
=0
since the value of the function at x=a, x=b, c, and d are equal to 0, then the function
f(x)=(x-a) (x-b) (x- c) (x-d) has zeroes at a, b, c and d
with the generalized form, we can substitute for the given zeroes,
x=-5, x=-2,0and 5 where a=-5, b=-2, c=0 and x=5
f(x)=(x-(-5)) (x-(-2)) (x-0) (x-5)
simplifying gives:
f(x)=(x+5) (x+2) x(x-5)
from here, we can put it in standard polynomial form by foiling the right side:
f(x)=x^3+2x^2-25x-50
To double-check the answer, just plug in the given zeroes, and ensure the value of the function at those points is equal to 0.
f (-5) =x^3+2x^2-25x-50
=(-5)^3+2(-5)^2-2(-5)-50
f (-5) =0
f (-2) =(-2)^3+2(-2)^2-2(-2)-50
f (-2) =0
f (0) =0
f (5) =(5)^3+2(5)^2-2(5)-50
=0
Hence, the function has zeroes as given by x=0.
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