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I’m doing partial fractions and need to factorize the denominator. They are quadratic. However there are some that aren’t so easy to factorize and my first choice was to use the quadratic equation to find the roots however comparing my answer with the correct one the signs are different. Is the quadratic formula only to be used when the equation is equal to zero? The answer used another method of factorizing that didn’t involve equating anything to zero and I can’t find anything about it online. Where did I go wrong? My denominator is:

$-3z^2 -4z-1$

the correct answer is: $-(3z+1)(z+1)$

while if I do this using the quadratic formula I get: $(3z+1)(z+1)$

however if I factorize the negative sign then use the quadratic formula I get the correct answer which is confusing to me.

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  • $\begingroup$ It's not very clear what you mean, maybe you could give a practical example and write some of the terms that you want to factor? $\endgroup$ – Matti P. Oct 29 '18 at 12:20
  • $\begingroup$ I have edited the question. $\endgroup$ – Ian Oct 29 '18 at 12:27
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Suppose I have a mystery quadratic $p(x) = -2x^2 + x + 1$, and want to factorise it using the quadratic formula. I can find the roots easily enough:

$$ x = \frac{-1 \pm \sqrt{1 + 8}}{-4}$$ and so the roots are $-\frac{1}{2}$ and $1$, and so I think that the polynomial should factorise as $p(x) = (x - 1)(x+\frac{1}{2})$. But this is wrong, since I forgot that $p(x)$ and $2p(x)$ and $\frac{-p(x)}{9}$ and so on all have the same roots! And so knowing the two roots only determines $p(x)$ up to scaling. But it's easy to find the right scaling factor, by just looking at the $-2x^2$ term. So I arrive at the actual answer, of $$ p(x) = -2x^2 + x + 1 = -2(x-1)(x + \frac{1}{2})$$

So after you find the roots, remember to multiply the whole thing by the right number to fix up the $x^2$ term.

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  • $\begingroup$ what is this phenomenon called? where can I read up more on it? $\endgroup$ – Ian Oct 29 '18 at 12:36
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    $\begingroup$ I don't know if it's called anything particular. You just have to notice that scaling a function changes the function but not the roots, since if $p(\alpha) = 0$, then $2 p(\alpha) = 0$ also. But $p$ and $2p$ are not the same function. So knowing the roots is not enough to determine the quadratic, you have to know something about how steep it is. $\endgroup$ – Joppy Oct 29 '18 at 12:38
  • $\begingroup$ does this only apply if the coefficient of the $x^2$ is non zero???? $\endgroup$ – Ian Oct 29 '18 at 12:39
  • $\begingroup$ If the coefficient of the $x^2$ is zero, then you have a linear function $ax + b$, not a quadratic. $\endgroup$ – Joppy Oct 29 '18 at 12:40
  • $\begingroup$ so as a rule of thumb, once the coefficient isn't 1 I should multiply by the coefficient after I use the quadratic formula? $\endgroup$ – Ian Oct 29 '18 at 12:41
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I'm guessing, what you have is the following fraction: $$ f(z) = \frac{N(z)}{D(z)}, $$ where $D(z) = -3z^2-4z-1$.

Basically, you may write $D(z) = -(3z^2+4z+1) = -D_2(z)$.

Finally, you can decompose the fraction $f$ as $f(z) = \frac{N(z)}{D(z)}$ or as $f(z) = \frac{-N(z)}{D_2(z)}$.

In other words, you'll get the same partial fraction in any case (you'll keep only the two factors in the denominators, and push the minus sign to the numerator)

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Factoring in $\mathbb Q[x]$ is unique up to associates. The units in $\mathbb Q[x]$ are all nonzero rational numbers. So $(x + \frac 12)$, $(2x+1), -(2x+1), (4x+2)$ and so on are all considered equivalent factors.

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