algebra precalculus - How to factor $ax^{2}+bxy+cy^{2},,aneq 0$?

Question: Factor: $3x^{2}-5xy-12y^{2}$

Answer: $(x-3y)(3x+4y)$

What are the exact steps to finding this answer from the original question (factored form from standard form, respectively)?

Answer

To factor $3x^2 - 5xy - 12y^2$, we first split the linear term, then factor by grouping. To split the linear term, we must find two numbers with product $3 \cdot -12 = -36$ and sum $-5$. They are $-9$ and $4$. Hence,
$$\begin{align*} 3x^2 - 5xy - 12y^2 & = 3x^2 - 9xy + 4xy - 12y^2 && \text{split the linear term}\\ & = 3x(x - 3y) + 4y(x - 3y) && \text{factor by grouping}\\ & = (3x + 4y)(x - 3y) && \text{extract the common factor} \end{align*}$$
To check that the answer is correct, we multiply the factors. Observe that doing so amounts to performing the steps of the factorization in reverse order.

In your second example of $5x^2 - 14x + 8$, to split the linear term, we must find two numbers with product $5 \cdot 8 = 40$ and sum $-14$. They are $-10$ and $-4$. Hence,
$$\begin{align*} 5x^2 - 14x + 8 & = 5x^2 - 10x - 4x + 8 && \text{split the linear term}\\ & = 5x(x - 2) - 4(x - 2) && \text{factor by grouping}\\ & = (5x - 4)(x - 2) && \text{extract the common factor} \end{align*}$$

In general, if $ax^2 + bx + c$, $a \neq 0$, is a quadratic polynomial with rational coefficients, then it factors with respect to the rationals if there exist two rational numbers with product $ac$ and sum $b$. In particular, if $r$, $s$, $t$, and $u$ are rational numbers such that
$$ax^2 + bx + c = (rx + s)(tx + u) \tag{1}$$
then $a = rt$, $b = ru + st$, and $c = su$. If you perform the multiplication
$$\begin{align*} (rx + s)(tx + u) & = rx(tx + u) + s(tx + u)\\ & = rtx^2 + rux + stx + su\\ & = rtx^2 + (ru + st)x + su\\ & = ax^2 + bx + su \end{align*}$$

you will notice that we can obtain $a = rt$, $b = ru + st$, and $c = su$ by matching coefficients, as Dietrich Burde stated.

We can prove this assertion by treating equation 1 as an algebraic identity. Since it is an identity, equation 1 holds for each value of the variable. In particular, it holds for $x = 0$, $x = 1$, and $x = -1$. Setting $x = 0$ in equation 1 yields
$$c = su \tag{2}$$
Setting $x = 1$ in equation 1 yields
$$\begin{align*} a + b + c & = (r + s)(t + u)\\ & = r(t + u) + s(t + u)\\ & = rt + ru + st + su \tag{3} \end{align*}$$

Since $c = su$, we can cancel $c$ from the left hand side and $su$ from the right hand side of equation 3 to obtain
$$a + b = rt + ru + st \tag{4}$$
Setting $t = -1$ in equation 2 yields
$$\begin{align*} a - b + c & = (r - s)(t - u)\\ & = r(t - u) - s(t - u)\\ & = rt - ru - st + su \tag{5} \end{align*}$$
Since $c = su$, we can cancel $c$ from the LHS and $su$ from the RHS of equation 5 to obtain
$$a - b = rt - ru - st \tag{6}$$

Adding equations $4$ and $6$ yields
$$2a = 2rt \tag{7}$$
Dividing both sides of equation 7 by $2$ yields
$$a = rt \tag{8}$$
Since $a = rt$, we can cancel an $a$ from the LHS of equation 4 and $rt$ from the RHS of equation 4 to obtain
$$b = ru + st \tag{9}$$
Our derivation of equations 2, 8, and 9 proves the claim.