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How do I solve $\frac {(x^3-4x^2+5x-6)}{(x^2-x-6)=4}$ algebraically?

I tried:

$4(x^2-x-6)=x^3-4x^2+5x-6$

$4x^2-4x-24=x^3-4x^2+5x-6$

$x^3-8x^2+9x+18=0$

I don't know how to solve this algebraically.

Adola
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Thas
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    Use the rational root test to discover a root $r$. Divide the polynomial by $(x-r)$ to get a quadratic. Factor the quadratic to obtain the other two roots. – anon Apr 15 '14 at 17:08
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    You have to be careful multiplying through because $x^2-x-6$ may be equal to zero. – Mark Bennet Apr 15 '14 at 17:15

3 Answers3

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$$x^3-4x^2+5x-6 = (x-3)(x^2-x+2)$$ $$x^2-x-6 = (x-3)(x+2)$$ so the equation becomes: $$\frac{x^2-x+2}{x+2} =4 $$ $$x^2-x+2=4x+8$$ $$x^2-5x-6=0$$ $$(x-6)(x+1)=0$$ $x=6$ or $-1$

TYZ
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Use the fact that: $$\frac{ (x^3-4x^2+5x-6)}{(x^2-x-6)}=(x-3)+\frac{8}{(x+2)} $$ Therefore your solving: $$ (x-3)+\frac{8}{x+2}=4 \\ x^2-5x-6+8=4x+8\\ x^2-5x-6=0$$ Can you solve this?

Jay
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$\dfrac{x^3-4x^2+5x-6}{x^2-x-6}=\dfrac{x^3-4x^2+5x-6}{(x-3)(x+2)}$

Now we are going to use Factor Theorem, to determine one of the factors of the numerator, which I call call $h(x)$.

(Note: Just in case you're not familiar with the Factor Theorem, it says that a polynomial $p(x)$ has a factor $(x-c) \iff p(c)=0$ )

We note that $h(3)=0$ and hence $x-3$ is a factor. Then we apply polynomial long division to obtain the other factors. You should get $x^2-x+2$ as your quotient.

$\dfrac{x^3-4x^2+5x-6}{(x-3)(x+2)}=\dfrac{(x-3)(x^2-x+2)}{(x-3)(x+2)}$ .

Then the $(x-3)$'s will cancel leaving

$\dfrac{(x^2-x+2)}{(x+2)}$

Going back to the original question,

$\dfrac{(x^2-x+2)}{(x+2)}=4 \\ \implies x^2-x+2=4x+8\\ \implies x^2-5x-6=0\\ \implies (x-6)(x+1)=0\\ \implies x=6\text{ or } x=-1$

Sujaan Kunalan
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