Solve Inequalities With Step

     

To find this, first notice that #x^4 + x^2 + 1 > 0# for all (real) values of #x#. So there are no linear factors, only quadratic ones.

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#x^4 + x^2 + 1 = (ax^2 + bx + c)(dx^2 + ex + f)#

Without bothering to lớn multiply this out fully just yet, notice that the coefficient of #x^4# gives us #ad = 1#. We might as well let #a = 1# & #d = 1#.

... #= (x^2 + bx + c)(x^2 + ex + f)#

Next, the coefficient of #x^3# gives us #b + e = 0#, so #e = -b#.

... #= (x^2 + bx + c)(x^2 -bx + f)#

The constant term gives us #cf = 1#, so either #c = f = 1# or #c = f = -1#. Let"s try #c = f = 1#.

... #= (x^2 + bx + 1)(x^2 - bx + 1)#

lưu ý that the coefficient of #x# will vanish nicely when these are multiplied out.

Finally notice that the coefficient of #x^2# is #(1 - b^2 + 1) = 2 - b^2#, giving us #1 = 2 - b^2#, thus #b^2 = 1#, so #b = 1# or #b = -1#.

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George C.
Nov 15, 2017

#x^4+x^2+1 = (x^2-x+1)(x^2+x+1)#


Explanation:

This really makes a bit more sense in the complex numbers...

First note that:

#(x^2-1)(x^4+x^2+1) = x^6-1#

So zeros of #x^4+x^2+1# are also zeros of #x^6-1#.

What are the zeros of #x^6-1#?

The real zeros are #1# and #-1#, which are zeros of #x^2-1#, the factor we introduced. So the four zeros of #x^4+x^2+1# are the four complex zeros of #x^6-1# apart from #+-1#.

Here are all #6# in the complex plane:

graph((x-1)^2+(y-0)^2-0.002)((x-1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1)^2+(y-0)^2-0.002)((x+1/2)^2+(y+sqrt(3)/2)^2-0.002)((x-1/2)^2+(y+sqrt(3)/2)^2-0.002) = 0 <-2.5, 2.5, -1.25, 1.25>

They khung the vertices of a regular hexagon.

de Moivre"s formula tells us that:

#(cos theta + i sin theta)^n = cos n theta + i sin theta#

where #i# is the imaginary unit, satisfying #i^2=-1#

For instance we find:

#(cos (pi/3) + i sin (pi/3))^6 = cos 2pi + i sin 2pi = 1 + 0 = 1#

That"s the zero #1/2+sqrt(3)/2i# that we see in Q1.

If #a# is a zero of a polynomial then #(x-a)# is a factor.

Hence:

#x^4+x^2+1 = (x-1/2-sqrt(3)/2i)(x-1/2+sqrt(3)/2i)(x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)#

#color(white)(x^4+x^2+1) = (x^2-x+1)(x^2+x+1)#

For example:

#(x-1/2-sqrt(3)/2i)(x-1/2+sqrt(3)/2i)#

#=x^2-((1/2+color(red)(cancel(color(black)(sqrt(3)/2i))))+(1/2-color(red)(cancel(color(black)(sqrt(3)/2i)))))x+(1/2+sqrt(3)/2i)(1/2-sqrt(3/2)i)#