Algebra simplification

Andrew Rubin

New member
Joined
Jun 24, 2019
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22
Hi everyone

I'm working on an optimalization problem, although I'm stuck on an algebra simplication issue.

First, I obtain the derivative of
[MATH]f\left(x\right)\:=\:x^3-3x^2+3x[/MATH]as,
[MATH]f'\left(x\right)\:=\:3x^2-6x+3[/MATH]
Then my textbook says that this can be further simplified to,
[MATH]f'\left(x\right)\:=\:3\left(x-1\right)^2[/MATH]
I can't crack how to arrive at this simplication. I've tried to first cancel out the common factor 3 (don't know if this is correct terminology for my use here) from
[MATH]f'\left(x\right)\:=\:3x^2-6x+3[/MATH]to arrive at,
[MATH]f'\left(x\right)\:=\:3\left(x^2-6x\right)[/MATH]
Then I progress using the law of distribution of multiplication, [MATH]a(b±c)≡ab±ac[/MATH], and arrive at
[MATH]=\:3\cdot x^2-3\cdot 6x[/MATH]
But I'm not sure if this is the right path nor how to move forward. I would greatly appreciate input on this.
 
Hi everyone

I'm working on an optimalization problem, although I'm stuck on an algebra simplication issue.

First, I obtain the derivative of
[MATH]f\left(x\right)\:=\:x^3-3x^2+3x[/MATH]as,
[MATH]f'\left(x\right)\:=\:3x^2-6x+3[/MATH]
Then my textbook says that this can be further simplified to,
[MATH]f'\left(x\right)\:=\:3\left(x-1\right)^2[/MATH]I can't crack how to arrive at this simplication. I've tried to first cancel out the common factor 3 (don't know if this is correct terminology for my use here) from
[MATH]f'\left(x\right)\:=\:3x^2-6x+3[/MATH]to arrive at,
[MATH]f'\left(x\right)\:=\:3\left(x^2-6x\right)[/MATH]
Then I progress using the law of distribution of multiplication, [MATH]a(b±c)≡ab±ac[/MATH], and arrive at
[MATH]=\:3\cdot x^2-3\cdot 6x[/MATH]
But I'm not sure if this is the right path nor how to move forward. I would greatly appreciate input on this.
How would you expand (x-1)2
 
How would you expand (x-1)2
Thanks for your reply.

I guess by the perfect square formula: [MATH]\left(a-b\right)^2=a^2-2ab+b^2[/MATH],

[MATH]=x^2-2x\cdot \:1+1^2[/MATH][MATH]=x^2-2x+1[/MATH]
But I can't see how I arrive at the above simplication of the derivation provided by my textbook using this?
 
Starting from
...but I can't see how I arrive at the above simplication of the derivation provided by my textbook using this?

Starting from what you already got to:

\(f^{\prime}(x) = 3x^2-6x+3\)

You've correctly identified the next step of "[factor] out the common factor 3," although you've executed it incorrectly. To factor out a 3 from the expression, that just means dividing and multiplying the expression by 3, possibly rearranging or rewriting some terms in the interim. But because multiplication and division are distributive over addition, that's the same as individually dividing each term by 3 but multiplying the entire expression by 3. Let's do that now:

\(\displaystyle f^{\prime}(x) = 3 \left(\frac{3x^2}{3} - \frac{6x}{3} + \frac{3}{3}\right)\)

\(f^{\prime}(x) = 3(x^2 - 2x + {\color{red}\textbf{1}} ) \)

This is where you went wrong, because the 3 term disappeared completely, rather than get divided by 3. This, then, exactly matches the expression you found was equivalent to \((x - 1)^2\). Essentially, there's two important things here. One is to be careful not to lose terms, but more importantly, practice practice practice. After you work enough of these types of problems, you too will begin to just "see" the answer right away.
 
My answer is fundamentally the same as #4, but from a different direction.

When you have completed a factoring, multiply it out to make sure that you have not made an error. Check your work as you go along: it is much easier than looking for an error in dozens of lines of math.

[MATH]3(x^2 - 6x) = 3x^2 - 3(6x) = 3x^2 - 18x \ne 3x^2 - 6x + 3.[/MATH]
 
Starting from

Starting from what you already got to:

\(f^{\prime}(x) = 3x^2-6x+3\)

You've correctly identified the next step of "[factor] out the common factor 3," although you've executed it incorrectly. To factor out a 3 from the expression, that just means dividing and multiplying the expression by 3, possibly rearranging or rewriting some terms in the interim. But because multiplication and division are distributive over addition, that's the same as individually dividing each term by 3 but multiplying the entire expression by 3. Let's do that now:

\(\displaystyle f^{\prime}(x) = 3 \left(\frac{3x^2}{3} - \frac{6x}{3} + \frac{3}{3}\right)\)

\(f^{\prime}(x) = 3(x^2 - 2x + {\color{red}\textbf{1}} ) \)

This is where you went wrong, because the 3 term disappeared completely, rather than get divided by 3. This, then, exactly matches the expression you found was equivalent to \((x - 1)^2\). Essentially, there's two important things here. One is to be careful not to lose terms, but more importantly, practice practice practice. After you work enough of these types of problems, you too will begin to just "see" the answer right away.
Thank you - this really made it clear to me and I will definitely practice more to get the hang of it!
 
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