McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Solving x^2+bx+c=0
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Exercise 35 Page 49

Begin by considering only one variable of the expression.

(x-y)(x-5y)

Practice makes perfect

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. x^2-6x+5 ⇔ 1x^2+( -6)x+ 5 To factor a quadratic expression with leading coefficient a= 1, we need to find two factors of c= 5 whose sum is b= -6. Since 5 is a positive number, we will only consider factors with the same sign — both positive or both negative — so that their product is positive.

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)

Checking Our Answer

Check your answer ✓
We can check our answer by applying the Distributive Property and comparing the result with the given expression.

(x-y)(x-5y)
x(x-y)-5y(x-y)
x^2 - xy - 5y(x-y)
x^2 - xy - 5xy + 5y^2
x^2 - 6xy + 5y^2

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!