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# find all the real rational zeros

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find all the real rational zeros of x^4+x^3-11x^2+x-12

Oct 13, 2017

#1
+1

First factor this.

x4 + x3 - 11x2 + x - 12     Rearrange.

x3 + x + x4 - 11x2 - 12       Factor  x  out of the first two terms and split the  -11x2 .

x(x2 + 1) + x4 + x2 - 12x2 - 12    Factor  x2 out of  x4 + x2     and  -12  out of  -12x2 - 12 .

x(x2 + 1) + x2(x2 + 1) - 12(x2 + 1)     Factor  (x2 + 1)  out of all the terms.

(x2 + 1)(x2 + x - 12)    Factor the  x2 + x - 12 .

(x2 + 1)(x2 + 4x - 3x - 12)

(x2 + 1)(x(x + 4) - 3(x + 4))

(x2 + 1)(x + 4)(x - 3)

To find out the  x  values that make this equal to zero, set each factor equal to zero and solve for  x .

x2 + 1  =  0           x - 3  =  0           x + 4  =  0

x2  =  -1                 x  = 3                   x  =  -4

not real

Oct 13, 2017
edited by hectictar  Oct 13, 2017
edited by hectictar  Oct 13, 2017

#1
+1

First factor this.

x4 + x3 - 11x2 + x - 12     Rearrange.

x3 + x + x4 - 11x2 - 12       Factor  x  out of the first two terms and split the  -11x2 .

x(x2 + 1) + x4 + x2 - 12x2 - 12    Factor  x2 out of  x4 + x2     and  -12  out of  -12x2 - 12 .

x(x2 + 1) + x2(x2 + 1) - 12(x2 + 1)     Factor  (x2 + 1)  out of all the terms.

(x2 + 1)(x2 + x - 12)    Factor the  x2 + x - 12 .

(x2 + 1)(x2 + 4x - 3x - 12)

(x2 + 1)(x(x + 4) - 3(x + 4))

(x2 + 1)(x + 4)(x - 3)

To find out the  x  values that make this equal to zero, set each factor equal to zero and solve for  x .

x2 + 1  =  0           x - 3  =  0           x + 4  =  0

x2  =  -1                 x  = 3                   x  =  -4

not real

hectictar Oct 13, 2017
edited by hectictar  Oct 13, 2017
edited by hectictar  Oct 13, 2017
#2
+2

Wow, good job Hecticar. Certainy wouldn't haven't gotten that.

AdamTaurus  Oct 13, 2017
#3
+1

I agree with AT.....that's a very nice factoring "trick,"  hectictar.....!!!   Oct 13, 2017
#4
+1

Well thank you! I did have a little hint after I saw what the answer was supposed to be from WolframAlpha though!!!

Oct 13, 2017
#5
+1

LOL!!!!!!.....some people might call that "cheating"....I prefer the term, "research"   Oct 13, 2017