Determine all rational zeros of the polynomial P(x)=2x3+7x2+4x−4, and write the polynomial in factored form.
The leading coefficient of P is 2 and the factors of 2 are ±1,±2. They are the divisors of the constant term −4 and the factors of −4 are ±1,±2,±4. The possible rational zeros are ±1,±2,±4,±12
Using Synthetic Division
We find that 1 and 2 are not zeros but that 12 is a zero and that P factors as
2x3+7x2+4x−4=(x−12)(2x2+8x+8)
We now factor the quotient 2x2+8x+8 using trial and error. We get,
2x3+7x2+4x−4=(x−12)(2x+4)(x+2)2x3+7x2+4x−4=2(x−12)(x+2)(x+2)
The zeros of P are 12 and −2.
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