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Begin by considering only one variable of the expression.
(x-y)(x-5y)
We want to factor the given expression.
x^2-6xy+5y^2
To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. The given expression has two variables. Factoring it is similar to factoring a trinomial that has only one variable, so we will consider x^2-6x+5 first. We will start by identifying the values of a, b, and c.
| Factor Pair | Product | Sum |
|---|---|---|
| 1 and 5 | ^(1* 5) 5 | 1+5 6 |
| -1 and -5 | ^(-1* (-5)) 5 | ^(-1+(-5)) -6 |
The integers whose product is 5 and whose sum is -6 are -1 and -5. x^2-6x+5 ⇔ (x -1)(x-5) The original expression has two variables. We want the linear term to be the sum and the constant to be the product of the factors. The factors of one variable equation are - 1 and - 5. Notice that if we multiply each of them by y, these conditions will be satisfied for the two-variable equation. - y+(- 5y)=- 6y - y * (- 5y)=5y^2 This means that the factors of the original equation are - y and - 5y. x^2-6xy+5y^2 ⇔ (x-y)(x-5y)
Distribute (x-y)
Distribute x
Distribute - 5y
Subtract term
After applying the Distributive Property and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!