How do you find all the zeros of f(x)=12x3+31x2−17x−6?

1 Answer
George C.
Aug 3, 2016

f(x) has zeros: 23, −14, −3

Explanation:

f(x)=12x3+31x2−17x−6

By the rational root theorem, any rational zeros of f(x) are expressible in the form pq for integers p,q with p a divisor of the constant term −6 and q a divisor of the coefficient 12 of the leading term.

So the only possible rational zeros are:

±112,±16,±14,±13,±12,±23,±34,±1,±43,±32,±2,±3,±6

That's rather a lot of possibilities to check, so let's search by approximately binary chop:

f(0)=−6

f(1)=12+31−17−7=19

f(12)=32+314−172−6=6+31−34−244=−214

f(34)=12(2764)+31(916)−17(34)−6=81+279−204−9616=154

f(23)=12(827)+31(49)−17(23)−6=32+124−102−549=0

So x=23 is a zero and (3x−2) is a factor:

12x3+31x2−17x−6=(3x−2)(4x2+13x+3)

To factor 4x2+13x+2 we can use an AC method: Look for a pair of factors of AC=4⋅3=12 with sum B=13.

The pair 12,1 works. Use this pair to split the middle term and factor by grouping:

4x2+13x+3=4x2+12x+x+3=4x(x+3)+1(x+3)=(4x+1)(x+3)

So the remaining zeros are x=−14 and x=−3

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