Simplify. (4x - 12)/(x^2 - 9) * (x+3)/(5x^2-25x) / (28)/(x^2-25)
I need to factor out like terms
also These are fractions.
Okay, so @Kaihalojr, what does \(4x - 12\) factor to?
@Hero it can factor to x - 3 right?
Yes, but what did you factor out in order to get \(x - 3\) as one of the factors?
4 because that's a number that they are both divisible by.
Okay, so wouldn't that make both \(4\) and \(x - 3\) factors of \(4(x - 3)\)?
Oops, wrote that wrong. Guess that's a freebee for you.
So yeah \(4x - 12\) factors to \(4(x - 3)\) since \(4(x - 3) = 4x - 12\)
so for the bottom half of the fraction isn't x² - 9 a difference of squares?
Correct and how would you write that in factored form?
x - 3
so this whole fraction is basically x - 3/x - 3
Is it?
wait no wouldn't it actually be 4(x-3)/(x-3)²?
Yes, but doesn't \(\dfrac{x - 3}{x - 3} = 1\) And knowing that couldn't we just re-write \((4(x - 3))/(x - 3)^2\) as \(\dfrac{4}{x - 3} \cdot\ \dfrac{x - 3}{x - 3}\) and then as \(\dfrac{4}{x - 3} \cdot\ 1\)
I think so yes
Okay, so that should tell you what the first fraction should be
Any ideas on how to factor the second fraction?
I think the upper half doesn't need to be factored right? just the bottom?
Are you sure about that?
no that's why I'm asking you
Okay, so basically the denominators need to be factored out in this case.
Let's write it out
I'm thinking we divide 25 and 5 by 5 so it's 1 and 5 then we divide x^2 and x by x so it's x and 1 correct?
\(\dfrac{\dfrac{x + 3}{5x^2 -25x}}{\dfrac{28}{x^2 - 25}}\)
What can we factor from the denominator of the top fraction?
oh we can divide the whole thing by x right so it becomes 3/ 5x - 25?
What can we FACTOR from the denominator of the top fraction?
5 right because 5 and 25 can each be factored by 5 or x for x^2 and x
You can factor \(5x\) not \(5\) or \(x\).
oh yeah that makes sense so then it's x - 5 right?
Yes, and you know the denominator of the bottom fraction is difference of squares.
By the way, the first fraction is \(\dfrac{4}{x + 3}\)
which multiplied by (x+3)/(x-5) is now 4(x-3)/(x+3)(x-5) correct?
How do you figure that?
because after we factor the first two fractions that's the only idea that seems fitting plus with fraction number 3 the numerator is x^2 - 25 which if squared is x - 5 so both (x - 5)s can cancel out.
I say it's the only thing that fits because we must multiply
You should write out your steps one by one here. Using the draw tool
\(\dfrac{\dfrac{x + 3}{5x^2 -25x}}{\dfrac{28}{x^2 - 25}} = \) Write what the above equals after factoring the denominators
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