One factor of F(x)=x^3-14x^2+61x-84 is (x-7) what are the zeros of the function
+128408 Since (x - 7) is a factor then 7 is a zero
Using a little synthetic division to find the remaining polynomial we have that
7 [ 1 -14 61 -84 ]
7 -49 84
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1 -7 12 0
The remaining polynomial is
x^2 -7x + 12
Set to 0 and factor and we have that
(x - 3) (x - 4) = 0
Setting each factor to 0 and solving for x gives the other two roots (zeroes) of x = 3 and x = 4
