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How do you write a polynomial with zeros -6, 3, 5 and leading coefficient 1?

1 Answer
Jan 21, 2018

Answer:

#x^3-2x^2-33x+90#

Explanation:

This has #3# roots, so it will be a polynomial of degree #3#. If we factor a 3rd degree polynomial we get:

#(x+a)(x+b)(x+c)# where a , b and c are constants:

We can find these constants using the known roots. If the roots are:

#alpha=-6#

#beta=3#

#gamma=5#

Since the constant term in each factor will have an opposite sign to the root, we have:

#(x-alpha)(x-beta)(x-gamma)#

Substituting #alpha , beta and gamma# for root values we have:

#(x-(-6))(x-(3))(x-(5))=(x+6)(x-3)(x-5)#

The coefficient of the first term is the product of the coefficients of the x terms in each factor, since these are all #1#, the coefficient of the first term will be #1#

#(x+6)(x-3)(x-5)=x^3-2x^2-33x+90#