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$8 + 15xy − 12x − 10y $

How would I go about factorizing this? I'm not sure if it is even possible.

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    $(3x-2)(5y-4)$. – 2017-01-09

6 Answers 6

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There is also the possibility of going brute force on it.

$(ax+b)(cy+d)=acxy+adx+bcy+bd$.

Identifying the coefficients to $8+15xy−12x−10y\;$ we get $\begin{cases} bd=8\\ ac=15\\ ad=-12\\ bc=-10\\ \end{cases}$

Notice that $(aqx+bq)(\frac{c}{q}y+\frac{d}{q})=(ax+b)(cy+d)$ so we can fix any of $a,b,c,d$ to a desired value by multiplying by a convenient rational $q$.

Let's have for instance $a=3$ which seems attractive. The system immediately solves to $\begin{cases} a=3\\ b=-2\\ c=5\\ d=-4\\ \end{cases}$

And $(3x-2)(5y-4)$ is the desired factorization.

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If the expression factors at all, its factorization must be of the form $15(x-b)(y-c)$.

Setting $y=0$ in the original equation leaves $8-12x=0$ so the root in $x$ must be $b=\frac{2}{3}\,$. Similarly, setting $x=0$ gives $c=\frac{4}{5}\,$.

Then, the tentative factorization is $15\left(x-\frac{2}{3}\right)\left(y-\frac{4}{5}\right) = (3x-2)(5y-4)\,$.

Since this was derived on the assumption that such a factorization does in fact exist, the result must be verified, and it is indeed easily verified that this is the correct factorization.

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    Could you give an example where we work on the assumption that such a factorization exists, find one, and then check and it is wrong? Or explain how that is possible? – 2017-01-09
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    @AndroidFish Good question. That happens in most real cases, unlike carefully set-up practice problems. Take for example $1+xy-x+y\,$, then following the same steps on the assumption that a factorization exists would (tentatively) point to $(x-1)(y+1)\,$. But that's obviously not equal to the original expression, and that's because that expression does in fact *not* have a polynomial factorization. – 2017-01-09
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$$8 + 15xy − 12x − 10y$$

$$=15xy-10y+8-12x$$

$$=5y(3x-2)+8-12x$$

$$=5y(3x-2)+4(2-3x)$$

$$=5y(3x-2)-4(3x-2)$$

$$=(5y-4)(3x-2)$$

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Observe that the $15xy$ and $-10y$ terms both have $5y$ as a common factor, so $$ 8 + 15xy - 12x - 10y = (3x-2) 5y + 8 - 12 x $$ Observe that also the $8$ and $-12x$ terms have $-4$ as a common factor, so $$ (3x-2) 5y + 8 - 12 x = (3x-2)5y - (3x-2)4 $$ Then $$ (3x-2)5y - (3x-2)4 = (3x-2)(5y-4) $$

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    Thank you, I saw the common factors but I was really stuck on whether or not I could rearrange the equation. – 2017-01-09
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We can write $$P= (15xy-12x) +(8-10y) =(3x)(5y-4) +(-2)(5y-4) =(3x-2)(5y-4) $$ Hope it helps.

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Though you have already got many answers, but alternately it can be done multiplying and dividing the entire expression by $2$ to simplify it, i.e., $8+15xy-12x-10y = [15xy-10y-12x+8]\times2/2$

$=[\frac{15xy}{2}-5y-6x+4]\times2$

$=[5y(\frac{3x}{2}-1)-2(3x-2)]\times2$

$=[5y(3x-2)-4(3x-2)]$

$=(5y-4)(3x-2)$