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Factor $4x^3-8x^2-25x+50$ completely

The highest numbers you can take would be $1$, $2$, or $4$. Neither of those apply to all. So let's try the $x$! But the last term $50$ doesn't have an $x$ attached. Anybody want to give a small hint please.

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Can you factor $4x^3-8x^2$? Can you factor $-25x+50$? Then can you see what to do with those factorizations? – Gerry Myerson Jul 16 '12 at 12:09

5 Answers

up vote 9 down vote accepted

Hint: Consider the first two terms, $4x^3-8x^2$, and the last two terms, $-25x+50$, as separate polynomials, and factor each of them. See if you can then put their factorizations together somewhow.

Mouse over the grey box to reveal a spoiler:

$$4x^3-8x^2=4x^2(x-2)\qquad -25x+50=(-25)(x-2)$$ Now use the distributive property of multiplication to give a factorization of $$(4x^3-8x^2)+(-25x+50)$$

However, even the above spoiler is not the complete answer; once you have done this, there is a further factorization that can be done.

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$$(4x^3-8x^2)+(-25x+50)$$ $$4x^2(x-2)-25(x-2)$$ Is this correct? – Austin Broussard Jul 16 '12 at 20:29
What you've written is correct. Are you familiar with the distributive property of multiplication over addition? It may help to write the second line as $$4x^2(x-2)+(-25)(x-2)$$ in order to proceed. – Zev Chonoles Jul 16 '12 at 20:31
The $(x-2)$ are present in both polynomials. – Austin Broussard Jul 16 '12 at 20:36
That's right - what do you get when you combine the two parts, then? – Zev Chonoles Jul 16 '12 at 20:37
$(4x^2-25)(x-2)$ ? I don't know if this has anything to do with the problem, but I noticed that in the first polynomial, all the terms are perfect squares. – Austin Broussard Jul 16 '12 at 20:39
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up vote 0 down vote

HINT: The natural thing to do is attack this polynomial by trial and error with integers. However before doing this you can note that any integer root has to divide $50$ since $$ 4n^3-8n^2-25n+50 \equiv 50 \pmod n $$ This already reduces our search. Further note that when $x$ gets to large that the first term will become dominant. We can see quickly that when $x=\pm 5$ that the first term is $\pm 500$, and too large. SO we are indeed left with cases $\pm 1, \pm 2 \pm 4$. We can also see easily that the constant term will be dominant for $\pm 1$ so there is no need to check that case. Now with only 4 options trial and error is a good option.

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up vote 2 down vote

Hint $ $ Polynomials with coefficients $\rm\: ad, bd, cd,\ldots, ae, be, ce,\ldots\:$ have an obvious factor, e.g.

$$\rm\: (a\,x^2\! +\! b\, x\! +\! c)\,d\,x^3 + (a\,x^2\! +\! b\,x\! +\! c)\,e\ =\ (a\,x^2\! +\! b\, x\! +\! c)\,(d\,x^3\! +\! e)\,$$

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up vote 4 down vote

Another possible approach is to use hit and trial or Rational root theorem (if you prefer doing this more formally) to notice that $2$ is a zero of this polynomial.

Then:

$$ 50-25 x-8 x^2+4 x^3 = 4x^2(x-2) -25(x-2) = (4x^2-25)(x-2)=(2x+5)(2x-5)(x-2) $$

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up vote 5 down vote

Alternatively, it is clear that $x=2$ is a root so $(x-2)$ is a factor. Divide out and you are left with a quadratic to factorise.

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Indeed, direct inspection shows that $2$ is a root. – Lubin Aug 30 '12 at 18:59

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