Wednesday, April 12, 2017

College Algebra, Chapter 4, 4.4, Section 4.4, Problem 44

Determine all rational zeros of the polynomial P(x)=2x63x513x4+29x327x2+32x12, 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

2x63x513x4+29x327x2+32x12=(x2)(2x5+x411x3+7x213x+6)

We now factor the quotient 2x5+x411x3+7x213x+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

2x63x513x4+29x327x2+32x12=(x2)(x2)(2x4+5x3x2+5x3)

We now factor the quotient 2x4+5x3x2+5x3. 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

2x63x513x4+29x327x2+32x12=(x2)(x2)(x12)(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


2x63x513x4+29x327x2+32x12=(x2)(x2)(x12)(x+3)(2x2+2)2x63x513x4+29x327x2+32x12=2(x2)(x2)(x12)(x+3)(x2+1)

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