The architect must factor several trinomials the are of the form x2- mx + n, where m & n are whole numbers greater than zero. She wonders if any of these trinomials factor as (x+a)(x+b), where a > 0 and b < 0. Is this possible? Why or why not?
I do not understand how to figure this out, at first I was thinking no this isn't possible because m and n are greater than zero and a>0 is less than? But then it made no sense to me... :/
(x+a)(x+b) = x^2 + (a+b)x + ab
If a>0 and b<0, then ab<0 so there's no way that n can be positive if a and b have different signs.
To determine if a trinomial of the form x2 - mx + n can be factored as (x+a)(x+b), where a > 0 and b < 0, we need to consider the relationship between the coefficients and the factors.
Let's break down the factors (x+a) and (x+b). When we expand the product of these two factors, we get:
(x+a)(x+b) = x^2 + bx + ax + ab
Simplifying further:
= x^2 + (a+b)x + ab
Now, we can compare this expression to the trinomial x2 - mx + n:
x^2 - mx + n
To check if they are equivalent, we compare the corresponding terms. In other words, we compare the coefficients of the x^2 term, x term, and constant term.
For x^2, we have:
1 (coefficient of x^2 in x^2 - mx + n) = 1 (coefficient of x^2 in x^2 + bx + ax + ab)
Since the coefficients are the same, there is no issue with the x^2 term.
For the x term, we have:
-m (coefficient of x in x^2 - mx + n) = a+b (coefficient of x in x^2 + bx + ax + ab)
This tells us that:
-m = a + b
Finally, for the constant term, we have:
n (constant term in x^2 - mx + n) = ab (constant term in x^2 + bx + ax + ab)
This implies:
n = ab
Now, if we go back to the original condition given, which is a > 0 and b < 0, we see that a is a positive value and b is a negative value. Therefore, the product ab will be negative.
Considering the relationship between a = -b and n = ab, if a > 0 and b < 0, then we would have a product of a positive and negative number, resulting in a negative value for n. However, the condition states that both m and n are whole numbers greater than zero. Since n cannot be negative, it means the trinomial cannot be factored as (x+a)(x+b) under these conditions.
So, in conclusion, it is not possible to factor a trinomial of the form x2 - mx + n as (x+a)(x+b) when a > 0 and b < 0, given that m and n are whole numbers greater than zero.