Determine all rational zeros of the polynomial P(x)=x5−4x4−3x3+22x2−4x−24, 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 −24. Thus, the possible candidates are ±1,±2,±3,±4,±6,±8,±12,±24.
Using Synthetic Division
We find that 1 is not a zeros but that 2 is a zero and P factors as
x5−4x4−3x3+22x2−4x−24=(x−2)(x4−2x3−7x2+8x+12)
We now factor the quotient x4−2x3−7x2+8x+12. Its possible zeros are the divisors of 12, namely
±1,±2,±3,±4,±6,±12
Using Synthetic Division
We find that −1 is a zero and P factors as
x5−4x4−3x3+22x2−4x−24=(x−2)(x+1)(x3−3x2−4x+12)
We now factor the quotient x3−3x2−4x+12. Its possible zeros are the divisors of 12, namely
±1,±2,±3,±4,±6,±12
Using Synthetic Division
We find that 2 is a zero and P factors as
x5−4x4−3x3+22x2−4x−24=(x−2)(x+1)(x−2)(x2−x−6)
We know factor the quotient x2−x−6 using trial and error. We get
x5−4x4−3x3+22x2−4x−24=(x−2)(x+1)(x−2)(x−3)(x+2)
The zeros of P are 2,3,−1 and 2.
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