Determine all rational zeros of the polynomial P(x)=x3+4x2−3x−18, and write the polynomial in factored form.
The leading coefficient of P is 1, so all the rational zeros are integers:
They are divisors of the constant term −18. Thus, the possible candidates are
±1,±2,±3,±6,±9,±18
Using Synthetic Division
We find that 1 and 3 are not zeros but that 2 is a zero and that P factors as
x3+4x2−3x−18=(x−2)(x2+6x+9)
We now factor x2+6x+9 using the perfect square formula. So
x3+4x2−3x−18=(x−2)(x+3)2x3+4x2−3x−18=(x−2)(x+3)(x+3)
Therefore, the zeros of P are 2 and −3.
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