Monday, January 19, 2015

College Algebra, Chapter 4, 4.4, Section 4.4, Problem 42

Determine all rational zeros of the polynomial P(x)=x54x43x3+22x24x24, 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

x54x43x3+22x24x24=(x2)(x42x37x2+8x+12)

We now factor the quotient x42x37x2+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

x54x43x3+22x24x24=(x2)(x+1)(x33x24x+12)

We now factor the quotient x33x24x+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

x54x43x3+22x24x24=(x2)(x+1)(x2)(x2x6)

We know factor the quotient x2x6 using trial and error. We get

x54x43x3+22x24x24=(x2)(x+1)(x2)(x3)(x+2)

The zeros of P are 2,3,1 and 2.

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