Determine all rational zeros of the polynomial P(x)=2x6−3x5−13x4+29x3−27x2+32x−12, and write the polynomial in factored form.
The leading coefficient of P is 2 and its factors ara ±1,±2. They are divisors of the constant term −12 and its factors are ±1,±2,±3,±4,±6,±12. Thus, the possible rational zeros are ±1,±2,±3,±4,±6,±12,±12,±32.
Using Synthetic Division
We find that 1 is not a zeros but that 2 is a zero and P factors as
2x6−3x5−13x4+29x3−27x2+32x−12=(x−2)(2x5+x4−11x3+7x2−13x+6)
We now factor the quotient 2x5+x4−11x3+7x2−13x+6. Its possible zeros are the divisors of 6, namely
±1,±2,±3,±6,±12,±32
Using Synthetic Division
We find that −1 is not a zero but that 2 is a zero and P factors as
2x6−3x5−13x4+29x3−27x2+32x−12=(x−2)(x−2)(2x4+5x3−x2+5x−3)
We now factor the quotient 2x4+5x3−x2+5x−3. Its possible zeros are the divisors of −3, namely
±1,±3,±12,±32
Using Synthetic Division
We find that 3 and 32 are not zeros but that 12 is a zero and that P factors as
2x6−3x5−13x4+29x3−27x2+32x−12=(x−2)(x−2)(x−12)(2x3+6x2+2x+6)
We now factor the quotient 2x3+6x2+2x+69. Its possible zeros are the divisors of 6, namely
±1,±2,±3,±6,±12,±32
We find that −2 is not a zero but that −3 is a zero and that P factors as
2x6−3x5−13x4+29x3−27x2+32x−12=(x−2)(x−2)(x−12)(x+3)(2x2+2)2x6−3x5−13x4+29x3−27x2+32x−12=2(x−2)(x−2)(x−12)(x+3)(x2+1)
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