How to factorise algebraic quotients?

Question

I am trying to simplify an algebraic expression of the following:

$\frac{2f+g-h}{10f^2+3gf-fh-g^2+3gh-2h^2}$

the answer is apparently:

$\frac{1}{5f-g+2h}$

but I can't seem to get to that at all? Could someone give me an explanation as to how this is achieved?

Answer 1

A quadratic form (dimension three) factors (possibly with complex coefficents) if and only if its Hessian matrix has determinant zero. You have $$ 10 f^2 - g^2 - 2 h^2 + 3gh-hf+3fg $$ for $$ \left( \begin{array}{rrr} 20 & 3& -1\\ 3 & -2 & 3 \\ -1 & 3 & -4 \end{array} \right) $$

of determinant $0.$

One way to factor the form is to find a matrix congruence with a diagonal form, $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 3 }{ 20 } & 1 & 0 \\ - \frac{ 1 }{ 20 } & - \frac{ 9 }{ 7 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 20 & 0 & 0 \\ 0 & - \frac{ 49 }{ 20 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 20 } & - \frac{ 1 }{ 20 } \\ 0 & 1 & - \frac{ 9 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 20 & 3 & - 1 \\ 3 & - 2 & 3 \\ - 1 & 3 & - 4 \\ \end{array} \right) $$

This says that double the original form is $$ 20 \left( f+\frac{3g}{20} - \frac{h}{20}\right)^2 - \frac{49}{20} \left( g - \frac{9h}{7}\right)^2 $$ or $$ \frac{1}{20} \left( 20f+3g - h\right)^2 - \frac{1}{20} \left( 7g -9h \right)^2 $$ or $$ \frac{1}{20} (20f+3g-h+7g-9h)(20f+3g-h-7g+9h) $$ $$ \frac{1}{20} (20f+10g-10h)(20f-4g+8h) $$ $$ 2 (2f+g-h)(5f-g+2h) $$

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Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 20 & 3 & - 1 \\ 3 & - 2 & 3 \\ - 1 & 3 & - 4 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 20 & 3 & - 1 \\ 3 & - 2 & 3 \\ - 1 & 3 & - 4 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 20 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 20 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 20 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 20 & 0 & - 1 \\ 0 & - \frac{ 49 }{ 20 } & \frac{ 63 }{ 20 } \\ - 1 & \frac{ 63 }{ 20 } & - 4 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & \frac{ 1 }{ 20 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 20 } & \frac{ 1 }{ 20 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 20 } & - \frac{ 1 }{ 20 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 20 & 0 & 0 \\ 0 & - \frac{ 49 }{ 20 } & \frac{ 63 }{ 20 } \\ 0 & \frac{ 63 }{ 20 } & - \frac{ 81 }{ 20 } \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 9 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 20 } & - \frac{ 1 }{ 7 } \\ 0 & 1 & \frac{ 9 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 20 } & - \frac{ 1 }{ 20 } \\ 0 & 1 & - \frac{ 9 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 20 & 0 & 0 \\ 0 & - \frac{ 49 }{ 20 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 3 }{ 20 } & 1 & 0 \\ - \frac{ 1 }{ 7 } & \frac{ 9 }{ 7 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 20 & 3 & - 1 \\ 3 & - 2 & 3 \\ - 1 & 3 & - 4 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 20 } & - \frac{ 1 }{ 7 } \\ 0 & 1 & \frac{ 9 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 20 & 0 & 0 \\ 0 & - \frac{ 49 }{ 20 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 3 }{ 20 } & 1 & 0 \\ - \frac{ 1 }{ 20 } & - \frac{ 9 }{ 7 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 20 & 0 & 0 \\ 0 & - \frac{ 49 }{ 20 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 20 } & - \frac{ 1 }{ 20 } \\ 0 & 1 & - \frac{ 9 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 20 & 3 & - 1 \\ 3 & - 2 & 3 \\ - 1 & 3 & - 4 \\ \end{array} \right) $$

Answer 2

A priori, it is not clear whether that expression can be simplified. At first glance, you have two options, because the numerator is a polynomial of homogeneous degree $1$: either the numerator divides the denominator, or it doesn't.

Suppose the first option is true, that numerator divides denominator. Then because the denominator is a homogeneous polynomial of degree $2$, there exists a homogeneous polynomial $p(f,g,h) = a \cdot f + b \cdot g + c \cdot h$ such that $$\text{numerator} \times p(f,g,h) = \text{denominator}.$$ Expanding the above equation and equating coefficients on both sides gives a system of equations in $a,b,c$ which may have a solution (in which case the fraction can indeed be simplified), or may not have a solution (in which case the fraction is in simplest form).

Answer 3

Since you know it does simplify to that you know that the numerator divides the denominator. You could use multivariable polynomial long division to find the result.

I should also mention that your problem looks kind of reasonable that it would simplify since $5f$ times the first term gives the first term of the product and $2h$ times the last works also. Then it’s just a question of finding if there is a coefficient for $g$ that works. You could try $5f+ag+2h$ times your numerator and see if you could equate coefficients to find a.

Where did this come up by the way?