How to factor $ax^{2}+bxy+cy^{2},\,a\neq 0$?

Question

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 1

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.

Answer 2

HINT: dividing by $xy\ne 0$ we get $$3\frac{x}{y}-5-12\frac{y}{x}=0$$ and now set $$t=\frac{x}{y}$$ this is equivalent to $$3t^2-5t-12=0$$

Answer 3

I propose a different (I think more direct for early students) way to factor it.

You can treat one of the variables as if it was another number, for example $y$. You can just then:

$$3x^2-5xy-12y^2=0,$$ and using the general solution: $$x = \frac{5y\pm y\sqrt{25+144}}{6}=y\frac{5\pm 13}{6}$$

The solutions are $3y$ and $-4y/3$, and so the factoring is: $$3(x-3y)(x+\frac{4}{3}y)=0$$

Answer 4

If $y=0$, then $(x,y)=(0,0)$ is the only solution. Otherwise:

$$ax^2+bxy+cy^2=y^2\left(a\left(\frac{x}{y}\right)^2+b\left(\frac{x}{y}\right)+c\right)$$

The paranthesised expression is a quadratic in $\frac{x}{y}$, which you should know how to factor.

Answer 5

Solve the equation for $x$, with the usual formula.

$$x=\frac{-by\pm\sqrt{(by)^2-4a(cy^2)}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}y.$$

Then factor using these roots

$$ax^2+bxy+cy^2=a\left(x-\frac{-b+\sqrt{b^2-4ac}}{2a}y\right)\left(x-\frac{-b-\sqrt{b^2-4ac}}{2a}y\right).$$

With the given example,

$$3x^2-5xy+12y^2=3(x-3)(x+\frac43y)=(x-3)(3x+4).$$

Answer 6

You can look at $f(x,y)=ax^2+bxy+cy^2$ as the homogeneous form of $g(x)=ax^2+bx+c$

If $g(x)=(px+q)(rx+s)$ then $f(x,y)=(px+qy)(rx+sy)$

This works because, from the factorisation of $g(x)$ it is clear that $pr=a, ps+qr=b, qs=c$.

It is normally straightforward, as in other answers, to prove that this works, but if all you want is a solution, then just take the $y$s out at the beginning, and put them back at the end so that each term or expression has the same order.

Answer 7

Here is a an elementary way to find the factorization, of, say $3x^2-5xy-12y^2$. Just make the Ansatz $$ 3x^2-5xy-12y^2=(ax+by)(cx+dy)=acx^2 + (ad + bc)xy + bdy^2. $$ Comparing the coefficients we immeadiately see that $ac=3, ad+bc=-5, bd=-12$. A solution is $(a,b,c,d)=(1,-3,3,4)$.