Factoring trinomials with a=1
x^2-18x-40
factors of 40 which differ by 18: 2 and 20
so,
(x-20)(x+2)
To factor a trinomial with a coefficient of 1 for the quadratic term (x^2 in this case), such as x^2-18x-40, we need to find two numbers whose product is equal to the product of the constant term (-40 in this case) and whose sum is equal to the coefficient of the linear term (-18 in this case).
In this example, the product of the constant term (-40) is -40 and the coefficient of the linear term (-18) is also -18.
We need to find two numbers whose product is -40 and whose sum is -18.
Let's list the possible pairs of numbers whose product is -40:
1, -40
-1, 40
2, -20
-2, 20
4, -10
-4, 10
5, -8
-5, 8
Now, let's check the sum of each pair of numbers:
1 + (-40) = -39
-1 + 40 = 39
2 + (-20) = -18 (This is a possible pair!)
-2 + 20 = 18
4 + (-10) = -6
-4 + 10 = 6
5 + (-8) = -3
-5 + 8 = 3
The pair of numbers that adds up to -18 is 2 and -20.
Now, we can write the original trinomial as the product of two binomials using these numbers:
x^2-18x-40 = (x + 2)(x - 20)
So, the factored form of the trinomial x^2-18x-40 is (x + 2)(x - 20).