How do you find a polynomial function that has zeros #x=-5, 1, 2# and degree n=4?

1 Answer
George C.
Aug 10, 2018

#f(x) = (x+5)^2(x-1)(x-2) = x^4+7x^3-3x^2-55x+50#

Explanation:

If a polynomial function #f(x)# has zeros #x = -5#, #x=1# and #x=2#, then it has factors #(x+5)#, #(x-1)# and #(x-2)#.

If these were its only factors, then it would be a cubic.

It is not clear from the question whether #-5#, #1# and #2# are supposed to be the only zeros. If so then one of them must be of multiplicity #2#.

In any case, a suitable quartic would be:

#f(x) = (x+5)^2(x-1)(x-2) = x^4+7x^3-3x^2-55x+50#

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