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Have you factored each expression @Nicole?
Like all the numerators and denominators?
Yes
Ill do it right now
Okay, but you should post your work here.
Can u show me the first one and I do the other ones?
Pretty sure you already know how to factor \(5x + 10\).
Ohhhh wait I got confused for a sec. So it would be 5x+10: 5(x+3) x^2+4x-5= (x-1)(x+5) and x^2+10x+25= (x+5)^2 x^2+x-2= (x-1)(x+2)
And explain how you got those answers.
I just factored all of them out
I did it on paper and just typed the answer here.
Explain how you factored it though.
If you can't explain it then you don't really know how to do it.
for the first numerator: the GCF is 5 so I did \[5\left(\begin{matrix}5x \\ 5\end{matrix}\right)+\left(\begin{matrix}10 \\ 5\end{matrix}\right)\] and comes out as 5(x+2)
correct?
Yeah, that one is easy. I was referring to the quadratic trinomial expressions
So for x^2+4x-5 I thought what 2 numbers add up to 4 and multiply to -5 and got -1 and 5 then I just rewrote the expression to (x−1)(x+5)
Very good
For x^2+10x+25 I rewrite the form in a^2+2ab+b^2 where a was x and b was 5 then I used the square of sum and got (x+5)^2
rewrote*
Okay, so you factored all your expressions and now you have : \(\dfrac{5(x + 2)}{(x - 1)(x + 5)} \cdot\ \dfrac{(x + 5)^2}{(x +2)(x - 1)}\) What factors of one can we cancel at this point?
For x^2+x-2 I thought what 2 numbers add up to 1 and multiply to -2 then I rewrote the expression and got (x−1)(x+2)
Um (x+2)
And what else?
(x+5)
Very good: \(\dfrac{5(\cancel{x + 2})}{(x - 1)(\cancel{x + 5})} \cdot\ \dfrac{(x + 5)^{\cancel{2}}}{(\cancel{x +2})(x - 1)}\)
So you're left with just \(\dfrac{5}{x - 1} \cdot \dfrac{x + 5}{x - 1}\)
so A is one of them
is their another?
Well, what happens if you multiply them together?
Do you know how to multiply fractions?
It would be like D
D is what it would look like if you were to multiply them together
You should use the property \(\dfrac{a}{b} \cdot\ \dfrac{c}{d} = \dfrac{ac}{bd}\) Correct.
Great job
Post as a separate question always.
Okay
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