How do you find all rational roots for 3x^4 - 2x^3 + 8x^2 - 6x - 3 = 03x4−2x3+8x2−6x−3=0?

1 Answer
George C.
May 30, 2016

The rational root theorem can point out the possibilities, but in this particular case there's a shortcut to the rational zeros:

x=1x=1 and x=-1/3x=−13

Explanation:

f(x) = 3x^4-2x^3+8x^2-6x-3f(x)=3x4−2x3+8x2−6x−3

By the rational root theorem, any rational zero of f(x)f(x) is expressible in the form p/qpq for integers p, qp,q with pp a divior of the constant term -3−3 and qq a divisor of the coefficient 33 of the leading term.

So the only possible rational zeros are:

+-1/3±13, +-1±1, +-3±3

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Actually we can proceed more directly to find all the zeros as follows:

Note that the sum of the coefficients of f(x)f(x) is zero. That is:

3-2+8-6-3 = 03−2+8−6−3=0

Hence f(1) = 0f(1)=0 and (x-1)(x−1) is a factor:

3x^4-2x^3+8x^2-6x-3 = (x-1)(3x^3+x^2+9x+3)3x4−2x3+8x2−6x−3=(x−1)(3x3+x2+9x+3)

The remaining cubic can be factored by grouping as follows:

3x^3+x^2+9x+33x3+x2+9x+3

= (3x^3+x^2)+(9x+3)=(3x3+x2)+(9x+3)

=x^2(3x+1)+3(3x+1)=x2(3x+1)+3(3x+1)

=(x^2+3)(3x+1)=(x2+3)(3x+1)

So another rational zero is x=-1/3x=−13

The remaining quadratic has no Real zeros, rational or irrational since x^2+3 >= 3 > 0x2+3≥3>0 for any Real value of xx.

It does have Complex zeros x = +-sqrt(3)ix=±√3i

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