How do you use the rational root theorem to find the roots of #x^4 - 7x^2 + 12 = 0#?

2 Answers
May 12, 2015

If #p/q# is a root of #x^4-7x^2+12=0# written in lowest terms, then #p# must be a divisor of the constant term #12# and #q# must be a divisor of the coefficient (#1#) of the highest order term #x^4#.

So #q = +-1#. We might as well choose #q = 1#, since we can choose the sign of #p#.

#p# must be a factor of #12#, i.e. #+-12#, #+-6#, #+-4#, #+-3#, #+-2# or #+-1#.

Since #x^4# grows quite fast with larger positive or negative values of #x#, it's probably better to try the possible smaller values of #x# first.

Trying a few substitutions, we quickly find that #x=+-1# are not solutions and #x=+-2# are solutions. When #|x| = 3#, we find that #x^4 - 7x^2 + 12 = 81 - 63 + 12 = 30#. Larger values of #x# will prduce larger values of #x^4-7x^2+12# as the #x^4# term becomes dominant.

So the only rational roots are #x = 2# and #x = -2#.

May 14, 2015

George C gives a great answer to find that #+-2# are zeros of the polynomial. (Roots of the equation.)

If you need to find all (real or complex) roots, there a bit more work to do.

Because, #2# and #-2# are roots, we know that
#x-2# and #x+2# are factors of the polynomial on the left..

Divide by the factors using synthetic division or long division. Or divide by the product of these 2 factors, #x^2-4#

You'll get

#x^2-7x+12=(x^2-4)(x^2-3)=0#

So the roots are:

#2, -2, sqrt3, -sqrt3#