Some help with factorizing this polynomial please. I have tried but it is difficult as it factorizes down to a cubic and I can't factorize it further. This is regarding the division of polynomials.
$$P(x) = 4x^4 + 12x^3 - 24x^2 - 32x$$
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23 high reputation user jumps at an easy question, which had no attempts... i think i am starting to observe a pattern here – Lost1 Jan 19 '14 at 10:50
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You could at least have tried to experimentally find at least one simple root of the cubic. – Hagen von Eitzen Jan 19 '14 at 10:56
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HINT:
Taking out $4x $ as common, $4x^4+12x^3-24x^2-32x=4x(x^3+3x^2-6x-8)$
Now, $3x^2-6x=3x(x-2), x^3-8=x^3-2^3$
lab bhattacharjee
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Factor theorem will help.
$$4x^4 + 12x^3 - 24x^2 - 32x=4x(x^3+3x^2-6x-8).$$ Here, letting $f(x)=x^3+3x^2-6x-8,$ we have $$f(-1)=0.$$ This implies that $f(x)$ has a factor $(x-(-1)).$ So, we have $$4x(x+1)(x^2+2x-8).$$ Then, you'll know how to proceed from here.
mathlove
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Thanks this was the best. I got 4x(x + 1)(x + 4)(x - 2) - which is correct. – Simon Jan 19 '14 at 11:19
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You can factor $4 x$ first. Now, by inspection, you could find that $x=-1$ is a solution. Now, reduce from cubic to quadratic and continue.
I am sure you can take from here.
Claude Leibovici
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Maybe this could help: $$4x^4+12x^3-24x^2-32x=(4x^4-24x^2)+(12x^3-3x)=4x^2(x^2-6)+4x^2(-6-\frac8x)$$ Then we can sum the two terms $$4x^2(x^2-12-\frac8x)=4x^2(\frac{x^3-12x-8}x)=4x(x^3-12x-8)=4x^4-48x-32x$$ Hope this would help
Peterix
PunkZebra
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Start by: $$P(x) = 4x^4 + 12x^3 - 24x^2 - 32x=P(x) = 4x(x^3 + 3x^2 - 6x - 8)$$ Then one root is $x=-1$. So divide by $(x+1)$ and get simpler second order polynomial.
In general, if you interested in result rather than the way, much websites like Wolfram Alpha can solve this problems, doing for you the long and sometimes ugly way. All you need to do is to feed the polynomial and get this result saying it is $4 x (x+1) (x-2) (x+4)$.
LeeNeverGup
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