Factor x^2+3xy-18y^2

Problem

x2+3*x*y−18*y2

Solution

1. Identify the expression as a quadratic trinomial in the form A*x2+B*x*y+C*y2 where A=1 B=3 and C=−18

2. Find two numbers that multiply to the constant term C=−18 and add to the middle coefficient B=3

3. Determine the pair of numbers, which are 6 and −3 since 6⋅(−3)=−18 and 6 + (-3) = 3$.

4. Write the factored form by placing these numbers as coefficients of y within the binomial factors (x+p*y)*(x+q*y)

Final Answer

x2+3*x*y−18*y2=(x+6*y)*(x−3*y)

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