How do you write a polynomial function of least degree with integral coefficients that has the given zeros -3, -1/3, 5?

1 Answer
Shell
Oct 16, 2016

#f(x)=3x^3-5x^2-47x-15#

Explanation:

If the zero is c, the factor is (x-c).

So for zeros of #-3,-1/3, 5#, the factors are

#(x+3)(x+1/3)(x-5)#

Let's take a look at the factor #(x+color(blue)1/color(red)3)#. Using the factor in this form will not result in integer coefficients because #1/3# is not an integer.

Move the #color(red)3# in front of the x and leave the #color(blue)1# in place: #(color(red)3x+color(blue)1)#.

When set equal to zero and solved, both
#(x+1/3)=0# and #(3x+1)=0# result in #x=-1/3#.

#f(x)=(x+3)(3x+1)(x-5)#

Multiply the first two factors.

#f(x)=(3x^2+10x+3)(x-5)#

Multiply/distribute again.

#f(x)=3x^3+10x^2+3x-15x^2-50x-15#

Combine like terms.

#f(x)=3x^3-5x^2-47x-15#

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