It is so easy to write a fraction as 4/5 for "four over five"
so I will interpret your gibberish as
A(x-5) + B(x+1) = (-3x^2 + 12x + 22)/(x-5)
confirm if that is what you meant before I start my answer.
Let $A$ and $B$ be real numbers such that
$\frac{A}{x-5}+B(x+1)=\frac{-3x^2+12x+22}{x-5}$. What is $A+B$?
3 answers
Input it into a latex converter, and u will get your thing
It appears you meant
A/(x-5) + B(x+1) = (-3x^2 + 12x + 22)/(x-5)
If so, then
A + B(x+1)(x-5) = -3x^2 + 12x + 22
Now just expand the left side, and make sure coefficients of like exponents are the same. That gives you
Bx^2 - 4Bx - 5B+A = -3x^2 + 12x + 22
B = -3
-4B = 12 (good thing that's consistent!)
-5B+A = 22
Now just find A, and then A+B
This is clearly not a TeX- friendly forum. Henceforth, just use plain text.
A/(x-5) + B(x+1) = (-3x^2 + 12x + 22)/(x-5)
If so, then
A + B(x+1)(x-5) = -3x^2 + 12x + 22
Now just expand the left side, and make sure coefficients of like exponents are the same. That gives you
Bx^2 - 4Bx - 5B+A = -3x^2 + 12x + 22
B = -3
-4B = 12 (good thing that's consistent!)
-5B+A = 22
Now just find A, and then A+B
This is clearly not a TeX- friendly forum. Henceforth, just use plain text.
1 answer