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Factorising numerator, need explanation
- Thread starter Henrik
- Start date
D
Deleted member 4993
Guest
Here's an example from my book
it starts off from this
View attachment 2500
and jumps immediately to this
View attachment 2501
Question is: what happened to x^5/2
x 5/2 = 1/2 * x 1/2 * [ .....2x2]
and where did 2x^3 where is it?? come from?
This is an example of the quotient rule if it is relevant.
.
Hello, Henrik!
Look at the numerator.
\(\displaystyle \text{It has two terms: }\;\underbrace{\frac{3}{2}x^{\frac{1}{2}}(x^2-1)^{\frac{1}{2}}} - \underbrace{x^{\frac{5}{2}}(x^2-1)^{\text{-}\frac{1}{2}}}\)
\(\displaystyle \text{Factor out }\,\frac{1}{2}x^{\frac{1}{2}}(x^2-1)^{\text{-}\frac{1}{2}}\,\text{ from both terms}\)
. . and we have: .\(\displaystyle \frac{1}{2}x^{\frac{1}{2}}(x^2-1)^{\text{-}\frac{1}{2}}\bigg[3(x^2-1) - 2x^2\bigg] \)
If you don't believe it, multiply it out.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I had found this type of factoring to be confusing and difficult to teach.
So I baby-talked my way through it, so I could explain it to my students.
Example: .Factor \(\displaystyle x^5 + x^2\)
We see that both terms have a factor of \(\displaystyle x^2.\)
How do we know that it is \(\displaystyle x^{\color{red}2}\) ?
Because \(\displaystyle x^2\) has the least exponent.
Exactly how do we factor out \(\displaystyle x^2\) ?
We divide it out.
Here it is in baby steps:
. . \(\displaystyle x^5 + x^2 \;=\;\frac{x^2}{x^2}(x^5 + x^2) \;=\;x^2\left(\frac{x^5}{x^2} + \frac{x^2}{x^2}\right) \;=\;x^2\left(x^{5-2} + x^{2-2}\right) \;=\;x^2(x^3 + 1)\)
Now consider: .\(\displaystyle x^3 + x^{-2}\)
\(\displaystyle x^{-2}\) has the least exponent; we will factor it out.
. . \(\displaystyle x^3 + x^{-2} \;=\;x^{-2}\left(\frac{x^3}{x^{-2}} + \frac{x^{-2}}{x^{-2}}\right) \;=\; x^{-2}\left(x^{3-(-2)} + x^{-2-(-2)}\right) \;=\;x^{-2}\left(x^5 + 1\right) \)
After a little practice, you can omit the intermediate steps.
Here's an example from my book.
It starts off from this: .\(\displaystyle \dfrac{dy}{dx} \;=\;\dfrac{\frac{3}{2}x^{\frac{1}{2}}(x^2-1)^{\frac{1}{2}} - x^{\frac{5}{2}}(x^2-1)^{\text{-}\frac{1}{2}}}{x^2-1}\)
and jumps immediately to this: .\(\displaystyle \dfrac{dy}{dx} \;=\;\dfrac{\frac{1}{2}x^{\text{-}\frac{1}{2}}(x^2-1)^{-\frac{1}{2}}\big[3(x^2-1) - 2x^2\big]}{x^2-1} \)
Question: what happened to x^5/2 and where did 2x^2 come from?
Look at the numerator.
\(\displaystyle \text{It has two terms: }\;\underbrace{\frac{3}{2}x^{\frac{1}{2}}(x^2-1)^{\frac{1}{2}}} - \underbrace{x^{\frac{5}{2}}(x^2-1)^{\text{-}\frac{1}{2}}}\)
\(\displaystyle \text{Factor out }\,\frac{1}{2}x^{\frac{1}{2}}(x^2-1)^{\text{-}\frac{1}{2}}\,\text{ from both terms}\)
. . and we have: .\(\displaystyle \frac{1}{2}x^{\frac{1}{2}}(x^2-1)^{\text{-}\frac{1}{2}}\bigg[3(x^2-1) - 2x^2\bigg] \)
If you don't believe it, multiply it out.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I had found this type of factoring to be confusing and difficult to teach.
So I baby-talked my way through it, so I could explain it to my students.
Example: .Factor \(\displaystyle x^5 + x^2\)
We see that both terms have a factor of \(\displaystyle x^2.\)
How do we know that it is \(\displaystyle x^{\color{red}2}\) ?
Because \(\displaystyle x^2\) has the least exponent.
Exactly how do we factor out \(\displaystyle x^2\) ?
We divide it out.
Here it is in baby steps:
. . \(\displaystyle x^5 + x^2 \;=\;\frac{x^2}{x^2}(x^5 + x^2) \;=\;x^2\left(\frac{x^5}{x^2} + \frac{x^2}{x^2}\right) \;=\;x^2\left(x^{5-2} + x^{2-2}\right) \;=\;x^2(x^3 + 1)\)
Now consider: .\(\displaystyle x^3 + x^{-2}\)
\(\displaystyle x^{-2}\) has the least exponent; we will factor it out.
. . \(\displaystyle x^3 + x^{-2} \;=\;x^{-2}\left(\frac{x^3}{x^{-2}} + \frac{x^{-2}}{x^{-2}}\right) \;=\; x^{-2}\left(x^{3-(-2)} + x^{-2-(-2)}\right) \;=\;x^{-2}\left(x^5 + 1\right) \)
After a little practice, you can omit the intermediate steps.