anonymous
  • anonymous
How do you solve for x? x^4 - 4x^3 + 3x^2 + 8x - 16 = 0 I tried using rational root theorem and synthetic division so many times but it won't work. I can't find any zeros. D: Am I simply doing it wrong or what? :(
Algebra
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Try to take the first three terms \(x^4-4x^3 +3x^2\) and factorize.
Jhannybean
  • Jhannybean
what if we .... \[\large x^4 - 4x^3 + 3x^2 + 8x - 16 = 0 \]\[\large x^4 - 4x^3 + 3x^2 + 8x =16\]\[\large x(x^3-4x^2+3x+8)=16\]
anonymous
  • anonymous
I tried those as well but I always reach a dead end. :( Unless, I'm doing it wrong.

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More answers

Jhannybean
  • Jhannybean
Somewhere we're going to have to use the Rational Zero Theorem perhaps?
campbell_st
  • campbell_st
well the quartic has 2 quadratic factors... thats why its hard to solve you are looking at \[(x^2 -3x + 4)(x^2 -x - 4) = 0\] now solve the quadratics for the solutions.
Jhannybean
  • Jhannybean
How did you do that @campbell_st O_o
anonymous
  • anonymous
\(x^4-4x^3 +3x^2=x^2(x - 3)(x-1)\)
campbell_st
  • campbell_st
just practice... and a bit of luck
Luigi0210
  • Luigi0210
aren't we gonna have some imaginary solutions?
Jhannybean
  • Jhannybean
Dx i meant what method of factorization did you use to.... solve this.
campbell_st
  • campbell_st
1 quadratic will the other has unequal irrational solutions.
anonymous
  • anonymous
He used intuition. To factor in 2 terms, the largest orders had to be \(x^2\)
Jhannybean
  • Jhannybean
I see. I'm siding more towards @DanielM_113 's method. Interesting concepts you guys.
campbell_st
  • campbell_st
its just working backwards... and a little trial and error... given the rational root theorem only works for linear roots..
Jhannybean
  • Jhannybean
Ohh I see.
anonymous
  • anonymous
I don't see. D: Can you explain a little more? Sorry.
campbell_st
  • campbell_st
I just see patterns... and work from there.... I saw a lot of 4s 3 = 4 - 1 -4 = -2 x 2 or -1 x 4 -16 = -4 x 4 its just stuff like that...
anonymous
  • anonymous
Oh :O Well, if I do more factoring practice will i be able to be like you... :'D
campbell_st
  • campbell_st
maths is very practical... to more you do the better you get
anonymous
  • anonymous
okay, thanks :D hopefully, i will get this soon.
Jhannybean
  • Jhannybean
And helping others solve problems actually really helps in understanding these concepts,as well as practicing problems :D I've learned so many new techniques i never knew before.
anonymous
  • anonymous
Do you know the Russian multiplication technique?
anonymous
  • anonymous
Nope D:
anonymous
  • anonymous
Russian multiplication @ineptAtMath http://openstudy.com/updates/51ad8bb2e4b06ee3ee25df07

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