How do you find the rest of the zeros given one of the zero #c=2/3# and the function #f(x)=3x^3+4x^2-x-2#?

1 Answer
George C.
Dec 3, 2016

The zeros of #f(x)# are #2/3# (with multiplicity #1#) and #-1# with multiplicity #2#

Explanation:

The given zero corresponds to a linear factor #(3x-2)#, so #f(x)# will divide evenly as:

#3x^3+4x^2-x-2 = (3x-2)(x^2+2x+1)#

We can recognise the remaining quadratic factor as the perfect square trinomial:

#x^2+2x+1 = (x+1)^2#

So the other zeros are both #-1#

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