Problem:
(x+2)/2x^2+5x+2 + (x-3)/2x^2-5x-3
What I Got:
0
at x=0, one gets 2/2 + -3/-3 which is not zero. You messed up a sign, I suspect.
you are right girl
To solve the problem, we need to simplify the expression:
(𝑥+2)/(2𝑥^2+5𝑥+2) + (𝑥−3)/(2𝑥^2−5𝑥−3)
To simplify this expression, we need to find a common denominator for both fractions. The common denominator will be the product of the two denominators:
Common denominator = (2𝑥^2+5𝑥+2)(2𝑥^2−5𝑥−3)
Then we can rewrite each fraction with the common denominator:
(𝑥+2)(2𝑥^2−5𝑥−3)/(2𝑥^2+5𝑥+2)(2𝑥^2−5𝑥−3) + (𝑥−3)(2𝑥^2+5𝑥+2)/(2𝑥^2+5𝑥+2)(2𝑥^2−5𝑥−3)
Now, we can combine the numerators over the common denominator:
(𝑥(2𝑥^2−5𝑥−3)+2(2𝑥^2−5𝑥−3))/(2𝑥^2+5𝑥+2)(2𝑥^2−5𝑥−3) + (𝑥(2𝑥^2+5𝑥+2)−3(2𝑥^2+5𝑥+2))/(2𝑥^2+5𝑥+2)(2𝑥^2−5𝑥−3)
Next, we can distribute and simplify the terms in the numerators:
(2𝑥^3−5𝑥^2−3𝑥+4𝑥^2−10𝑥−6)/(2𝑥^2+5𝑥+2)(2𝑥^2−5𝑥−3) + (2𝑥^3+5𝑥^2+2𝑥−3𝑥−6𝑥−6)/(2𝑥^2+5𝑥+2)(2𝑥^2−5𝑥−3)
Now, we can combine like terms in the numerators:
(2𝑥^3−𝑥^2−19𝑥−6)/(2𝑥^2+5𝑥+2)(2𝑥^2−5𝑥−3)
At this point, we cannot simplify the expression any further because the numerator and denominator do not share any common factors.
Therefore, the simplified expression is:
(2𝑥^3−𝑥^2−19𝑥−6)/(2𝑥^2+5𝑥+2)(2𝑥^2−5𝑥−3)
So, at x=0, the expression is:
(2(0)^3−(0)^2−19(0)−6)/(2(0)^2+5(0)+2)(2(0)^2−5(0)−3)
Simplifying this expression gives us:
(-6)/(0+0+2)(0−0−3)
The denominator is (0+0+2)(0−0−3) = 0, which means the expression is undefined at x=0. Therefore, the expression is not zero at x=0.