Quadratics in Two Variables

What this move does

The cross method handles $a x^{2} + b x + c$ . Now there is one more letter:

x^{2} + 3 x y + 2 y^{2} + 2 x + 5 y - 3

If it factors, it must be a product of two linear expressions, each with three terms: a multiple of $x$ , a multiple of $y$ and a constant. So the cross needs three columns:

Columns 1 and 2: treat the quadratic part $x^{2} + 3 x y + 2 y^{2}$ as an ordinary cross-method problem and factor it as $(x + 2 y) (x + y)$ .
Column 3: split the constant term into two numbers, one to go at the end of each bracket.
Check two more crosses: crossing columns 2 and 3 must give the coefficient of $y$ , and crossing columns 1 and 3 must give the coefficient of $x$ .

Predict first

Multiply out $(x + 2 y + r_{1}) (x + y + r_{2})$ . What is the coefficient of the $x$ term?

r_{1} + r_{2}

r_{1} r_{2}

2 r_{1} + r_{2}

r_{1} + 2 r_{2}

Choose an answer first; then you can see the result.

Work through one

Example: factor x² + 3xy + 2y² + 2x + 5y − 3

Step 1 / 5

The first two columns
Look only at $x^{2} + 3 x y + 2 y^{2}$ . With $1 \times 1$ and $2 \times 1$ , the cross sum is $1 \cdot 1 + 1 \cdot 2 = 3$ , which matches the coefficient of $x y$ .

$x^{2} + 3 x y + 2 y^{2} = (x + 2 y) (x + y)$
Split the constant: try one pair
Swap the signs
Do not forget the third cross
Read the answer across

Another angle: make one letter the main variable

One problem, several solutions

Factor $x^{2} + 3 x y + 2 y^{2} + 2 x + 5 y - 3$ .

The double cross method

First two columns: $(x + 2 y) (x + y)$ . Third column: $- 1$ and $3$ .

Check the $y$ term: $2 \cdot 3 + 1 \cdot (- 1) = 5$ ; check the $x$ term: $3 - 1 = 2$ .

Result: $(x + 2 y - 1) (x + y + 3)$ .

CostFast, but all three crosses must be checked, and it is easy to miss one.

Treating one letter as the main variable

Treat $x$ as the only letter and $y$ as “a number”, and write it in descending powers of $x$ :

x^{2} + (3 y + 2) x + (2 y^{2} + 5 y - 3)

The “constant term” is $2 y^{2} + 5 y - 3 = (2 y - 1) (y + 3)$ . The sum of these two factors is $3 y + 2$ , exactly the coefficient of $x$ .

So the original expression $= (x + (2 y - 1)) (x + (y + 3))$ .

CostTwo more lines of writing, but it only needs the ordinary cross method.

The picture: a curve that breaks into two lines

Set the expression equal to $0$ and you get a curve in the plane. If the left side can be written as $(\dots) (\dots)$ , then “the product is $0$ ” means “one of the factors is $0$ ”, and each linear factor is a straight line.

Predict first

Keep the other coefficients fixed and change only the constant: $x^{2} + 3 x y + 2 y^{2} + 2 x + 5 y + f$ . For how many values of $f$ does it factor into two linear factors?

Only one,

f = - 3

Exactly twoInfinitely many

Choose an answer first; then you can see the result.

Practice

Exercise 1Basic

Factor $x^{2} + 5 x y + 6 y^{2}$ .

Exercise 2Basic

Factor $2 x^{2} - x y - 3 y^{2}$ .

Exercise 3Advanced

Factor $x^{2} + x y - 2 y^{2} + 2 x + 7 y - 3$ .

Exercise 4Challenge

Given that $x^{2} - y^{2} + 3 x + y + k$ factors into a product of two linear expressions, find $k$ .

What this move does

Work through one

Example: factor x² + 3xy + 2y² + 2x + 5y − 3

Another angle: make one letter the main variable

The double cross method

Treating one letter as the main variable

The picture: a curve that breaks into two lines

Practice

Mistake Museum · Exhibits from this stop

Checking only one cross

Factoring only the quadratic part