How do you use the rational roots theorem to find all possible zeros of #f(x)=x^4-8x^3+7x-14#?

1 Answer
Aug 9, 2016

Answer:

This #f(x)# has no rational zeros, but we can find the zeros algebraically...

Explanation:

#f(x) = x^4-8x^3+7x-14#

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-14# and #q# a divisor of the coefficient #1# of the leading term. So the only possible rational zeros are:

#+-1, +-2, +-7, +-14#

None of these work, so #f(x)# has no rational zeros.

Can we solve it using other methods?

Substitute #t = (x-2)# to get a quartic with no cubed term...

#x^4-8x^3+7x-14#

#=(x-2)^4-24(x-2)^2-57(x-2)-48#

#=t^4-24t^2-57t-48#

#=(t^2-at+b)(t^2+at+c)#

#=t^4+(b+c-a^2)t^2+(b-c)at+bc#

Equating coefficients and rearranging slightly, we get:

#{ (b+c=a^2-24), (b-c=-57/a), (bc=-48) :}#

Then:

#(a^2-24)^2 = (b+c)^2 = (b-c)^2+4bc = (-57/a)^2-192#

Expanding both ends, we get:

#(a^2)^2-48(a^2)+576 = 3249/((a^2))-192#

Add #192# to both sides, multiply through by #(a^2)# and rearrange slightly to get:

#0 = (a^2)^3-48(a^2)^2+768(a^2)-3249#

#= ((a^2)-16)^3+847#

Hence one Real root gives us:

#a^2 = 16-root(3)(847)#

which is positive.

So we can let #a=sqrt(16-root(3)(847))#

Substituting this into the set of simultaneous equations, we get:

#{ (b+c=a^2-24=(16-root(3)(847))-24 = -8-root(3)(847)), (b - c = -57/a = -57/sqrt(16-root(3)(847))), (bc = -48) :}#

Add the first and second of these and divide by #2# to find:

#b = -4 -1/2 root(3)(847) - 57/(2sqrt(16-root(3)(847)))#

Subtract the second from the first and divide by #2# to find:

#c = -4 -1/2 root(3)(847) + 57/(2sqrt(16-root(3)(847)))#

So we now have two quadratics to solve:

#t^2-(sqrt(16-root(3)(847)))t-(4+1/2 root(3)(847)+57/(2sqrt(16-root(3)(847)))) = 0#

#t^2+(sqrt(16-root(3)(847)))t-(4+1/2 root(3)(847)-57/(2sqrt(16-root(3)(847)))) = 0#

We can solve these using the quadratic formula, then add #2# to get zeros of our original quartic in #x#.

I will leave that as an exercise.