Can some one help me that how to reach this statement

Question

$$524384x^3−27204x^2y−40806xy^2−295763y^3=0$$

$$(8x−7y)(65568x^2+53954xy+42109y^2)=0$$ i.e, $$8x−7y=0$$

I just want to know how we reach from 1st line to 2nd line. I mean how to factorize it? How $$8x-7y$$ got separated out?

Answer 1

Notice when you multiply out the second line you get $8 \times 65548 = 524384$ (you have a typo in your second line where it says $65568x^2$) for the $x^3$ term and $-7 \times 42109 = -294763$ (you have a typo in your first line where it says -$295763y^3$) for the $y^3$ term. What was likely done was those coefficients were factored (note that $524384 = 2^5 \times 7 \times 2341$ (so $8$ is a factor due to the $2^5 = 32$ factor) and $294763 = 7 \times 17 \times 2477$), with the factors checked to see if any of them work with the linear term, with the middle coefficient in the second factor, i.e., $53954$, being determined & then checked to see if you got one consistent value for it when you compare the result against the coefficients of the middle $2$ terms in the first line, i.e., $-27204$ and $-40806$. In this case, $8x - 7y$ is what works.

In particular, you have (I used negative for the $b$ term due to the $y^3$ term being negative)

$$(ax - by)(cx^2 + dxy + ey^2) = (ac)x^3 + (ad - bc)x^2y + (ae - bd)xy^2 + (-be)y^3 \tag{1}\label{eq1A}$$

Matching coefficients gives

$$ac = 524384 \implies c = \frac{524384}{a} \tag{2}\label{eq2A}$$

$$ad - bc = -27204 \implies d = \frac{-27204 + bc}{a} \tag{3}\label{eq3A}$$

$$ae - bd = -40806 \implies d = \frac{40806 + ae}{b} \tag{4}\label{eq4A}$$

$$-be = -294763 \implies e = \frac{294763}{b} \tag{5}\label{eq5A}$$

Note when you choose $a$ and $b$ that you get $c$ from \eqref{eq2A} and $e$ from \eqref{eq5A}. However, you must then get the same $d$ value in \eqref{eq3A} and \eqref{eq4A}. There are many situations where no combination of integral $a$ and $b$ will work. However, in this case, you have that $a = 8$ and $b = 7$ do work.

FYI, this process is similar to what is stated in the Rational root theorem.

Answer 2

Another method is to set $y=tx$ to get the polynomial $f(t)=\sum\limits_{i=0}^3 a_it^i$

$$f(t)=524384-27204t-40806t^2-294763t^3$$

Possible rational roots of $f$ are to be searched in $$r\in\left\{\pm\dfrac{\operatorname{divisors}(a_0)}{\operatorname{divisors}(a_3)}\right\}$$

https://www.chilimath.com/lessons/intermediate-algebra/rational-roots-test/

In this case it is not a very interesting method since $\begin{cases}a_0=(2)^5(7)(2341) &\text{has 24 divisors}\\ a_3=-(7)(11)(2477)&\text{has 8 divisors}\end{cases}$

and there are $192$ possible $r$ to test ($\times 2$ for the sign minus duplicates). But for equations with coefficients that have less factors, it may be a suitable method. However in this particular case John's method is faster.

Anyway, we find that only $f(\frac 87)=0$

So $y=\frac 87 x\iff 7y=8x$ and you can factorize by $(8-7t)$ or equivalently by $(8x-7y)$.

The factorization is done incrementally, first divide the leading coefficient by $8x$ to get the first term. Multiply this term by $-7y$ and calculate the remaining. Then go on by dividing the remaining by $8x$ again and so forth until you get the complete factorization.

$\begin{array}{ll} 524384x^3−27204x^2y−40806xy^2−295763y^3\\\\ = (8x-7y)\times(65548x^2+\cdots) & \text{gives }-458836x^2y & \text{miss}\quad 431632x^2y\\ = (8x-7y)\times(65548x^2+53954xy+\cdots) & \text{gives }-377678xy^2 & \text{miss}\quad 336872xy^2\\ = (8x-7y)\times(65548x^2+53954xy+42109y^2) & \text{gives }-294763y^3 & \text{miss}\quad 0y^3 \end{array}$