Sorry for that
Some of these got already clear now, only few left.
(x-2)(x+3)=x2-3x+2x+6? Im not sure if signs are right.
(x-1)(x+1)=x2-1, but how do I get x2+1 if (x+1)(x+1) is (x+1)2?
The first one is simply a case of multiplying it out. Some instructors/books teach it with the acronym FOIL (First Outer Inner Last). That works, but I personally don't like it because it only works for two terms. You'd need a whole new acronym to multiply (a+b+c)(d+e+f). Instead, think of it as an iterative process to get each term of the answer:
Begin with the first term of the first expression. Multiply it by the first term of the second expression.
Take the first term of the first expression and multiply it by the second term of the second expression. Repeat until there are no more terms in the second expression.
Now take the second term of the first expression and multiply it by the first term of the second expression. Again, iterate through the terms in the second expression until you run out.
For example (2+3+4)(5+6+7) = (2*5) + (2*6) + (2*7) + (3*5) + (3*6) + (3*7) + (4*5) + (4*6) + (4*7) = 162. Compare that with (2+3+4)(5+6+7) = (9)(18) = 162. It checks. Now you try.
As for the second part, you can't ever get x
2+1 by multiplying only real numbers. You can see why this is the case by using the quadratic formula, if you know it:
x
2 + 1 = x
2 + 0x + 1.
\(\displaystyle x=\frac{-0\pm \sqrt{0^2-4\left(1\right)\left(1\right)}}{2}=\frac{0\pm \sqrt{0-4}}{2}=\frac{0\pm \sqrt{-4}}{2}\)
You can't take the square root of negative 4 with only real numbers, so that means you can't factor x
2+1 using only real numbers.
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Now, I'm assuming you're still struggling with the two exercises at the bottom of your original post:
The reason im asking those is because im pretty sure I made somewhere a mistake with these.
(x+3)/(x-1)-(x-1)/(x+3)-((x+1)(x+1))/((x+3)(x-1)) | its a f(x)/g(x)=0
(x2+9-x2+1-x2-1)/(x-1)(x+3)=(-x2-9)/(x-1)(x+3) | I got answer x=3 and x=-3, but im supposed to get x=-1 and x=7. What did I do wrong ? Thank you !
In the first exercise, I don't understand what you mean by "its a f(x)/g(x) = 0" Were the instructions something like "Simplify the expression given?" If yes, you have three fractions being subtracted, so what did you get when you found a common denominator?
In the second exercise, were you given both sides of the equation and told to solve for
x? Or is the expression after the equals sign part of your work? Additionally, are both of the multiplied terms meant to be in the denominator? In other words, does either of these represent the given problem?
\(\displaystyle \frac{x^2+9-x^2+1-x^2-1}{\left(x-1\right)\left(x+3\right)}\) or \(\displaystyle \frac{x^2+9-x^2+1-x^2-1}{\left(x-1\right)}\left(x+3\right)\)
