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

##### 1 Answer

This

#### Explanation:

By the rational root theorem, any *rational* zeros of *rational* zeros are:

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

None of these work, so

Can we solve it using other methods?

Substitute

#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

#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

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

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

Subtract the second from the first and divide by

#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

I will leave that as an exercise.