Step-by-step explanation:
Since both terms are perfect squares, factor using the difference of squares formula, a 2 − b 2 = ( a + b ) ( a − b ) where a = 7 and b = a + x . − ( 7 + a + x ) ( a + x − 7 )
Answer:
D. ![\frac{x-1}{x^2+3x+2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bx-1%7D%7Bx%5E2%2B3x%2B2%7D%7D)
Step-by-step explanation:
The given expression is;
![\frac{x}{x^2+3x+2}-\frac{1}{(x+2)(x+1)}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bx%5E2%2B3x%2B2%7D-%5Cfrac%7B1%7D%7B%28x%2B2%29%28x%2B1%29%7D)
This is the same as;
![\frac{x}{(x+2)(x+1)}-\frac{1}{(x+2)(x+1)}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B%28x%2B2%29%28x%2B1%29%7D-%5Cfrac%7B1%7D%7B%28x%2B2%29%28x%2B1%29%7D)
The denominators are the same.
Subtract the numerators and write one denominator;
![\frac{x-1}{(x+2)(x+1)}}](https://tex.z-dn.net/?f=%5Cfrac%7Bx-1%7D%7B%28x%2B2%29%28x%2B1%29%7D%7D)
Rewrite to get;
![\frac{x-1}{x^2+3x+2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bx-1%7D%7Bx%5E2%2B3x%2B2%7D%7D)
Hello from MrBillDoesMath!
Answer:
x = -3, 4
Discussion:
The zeroes of f(x) are the x values where
2x + 6 = 0 => x = -3
and
x - 4 = 0 => x = 4
Thank you,
MrB