How do you find the roots of 12x3+31x2−17x−6=0?

1 Answer
George C.
Jan 8, 2017

Use the rational roots theorem, then factor to find:

x=23 or x=−14 or x=−3

Explanation:

f(x)=12x3+31x2−17x−6

By the rational roots 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.

That means that the only possible rational zeros are:

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

Notice that the signs of the coefficients of f(x) are in the pattern ++−−. By Descartes' Rule of Signs, since there is one change of sign, this cubic has exactly one positive Real zero.

So let's look among the positive possibilities:

f(1)=12+31−17−6=20

f(12)=12(18)+31(14)−17(12)−6=6+31−34−244=−214

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

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

12x3+31x2−17x−6=(3x−2)(4x2+13x+3)

To factor the remaining quadratic, 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)

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

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

So the other two zeros are:

x=−14 and x=−3

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