How do you use the rational root theorem to find the roots of 3x^4+2x^3-9x^2-12x-4=03x4+2x3−9x2−12x−4=0?

1 Answer
George C.
Dec 7, 2015

Use the rational root theorem to identify the possible roots, try some and narrow down the search to find roots -2/3−23, -1−1, -1−1 and 22.

Explanation:

f(x) = 3x^4+2x^3-9x^2-12x-4f(x)=3x4+2x3−9x2−12x−4

By the rational root theorem any rational roots of f(x)=0f(x)=0 are expressible in lowest terms as p/qpq for some integers pp and qq, where pp is a divisor of the constant term 44 and qq is a divisor of the coefficient 33 of the leading term.

So the possible rational roots are:

+-1/3±13, +-2/3±23, +-1±1, +-4/3±43, +-2±2, +-4±4

Try:

f(1/3) = 1/27+2/27-1-4-4 = 1/9-9 = -80/9f(13)=127+227−1−4−4=19−9=−809

f(-1/3) = 1/27-2/27-1+4-4 = -28/27f(−13)=127−227−1+4−4=−2827

f(2/3) = 16/27+16/27-4-8-4 = 32/27f(23)=1627+1627−4−8−4=3227

f(-2/3) = 16/27 - 16/27-4+8-4 = 0f(−23)=1627−1627−4+8−4=0

So x=-2/3x=−23 is a root. This 'uses up' one of the possible factors of 22 for pp and the factor of 33 for qq. So the only other possible rational roots are:

+-1±1, +-2±2

Try:

f(1) = 3+2-9-12-4 = -20f(1)=3+2−9−12−4=−20

f(-1) = 3-2-9+12-4 = 0f(−1)=3−2−9+12−4=0

f(2) = 48+16-36-24-4 = 0f(2)=48+16−36−24−4=0

f(-2) = 48-16-36+24-4 = 16f(−2)=48−16−36+24−4=16

So we have identified 33 rational roots out of 44 roots, viz -2/3−23, -1−1 and 22. The fourth root must also be rational and having 'used up' the factor of 22 for pp, it must also be -1−1.

To check, we can multiply out the corresponding factors:

(3x+2)(x+1)(x+1)(x-2)(3x+2)(x+1)(x+1)(x−2)

=(3x^2+5x+2)(x^2-x-2)=(3x2+5x+2)(x2−x−2)

=3x^4+(5-3)x^3-(6+5-2)x^2-(10+2)x-4=3x4+(5−3)x3−(6+5−2)x2−(10+2)x−4

=3x^4+2x^3-9x^2-12x-4=3x4+2x3−9x2−12x−4

Related questions
See all questions in Rational Zeros
Impact of this question
2674 views around the world
You can reuse this answer
Creative Commons License