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I have:

$(z - 1)(1 + z + z^2 + z^3)$

As, I have tried my own methods and enlisted the help on online software, but as well as them not all arriving at the same solution, I can't follow their reasoning.

I tried to gather all the like terms:

$(z - 1)(z^6+1)$

And then thought it may offer a route to a difference or two cubes if I shifted powers across the brackets, but that's $z^7$ in total, isn't it? I can't just do $z^7-1$, can I?

If I try to expand, I see:

$(z - 1)(z^6 + 1) = z^7+z-z^6-1$

Well...

$z^7-z^6=z$

Add the other z leaves:

$z^2-1$?

I expect this is all wrong.

Will you help please?

On a side note, I have difficulty with understanding how I'm expected to tag this post accurately when I can't used "expand", "multiply", or use any of the subjects it covers as tags?

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    $\begingroup$ $z+z^2+z^3\ne z^6$ and $1+z+z^2+z^3\ne z^7$. You can't "add powers" like that. $\endgroup$ – David Mitra Oct 27 '15 at 13:41
  • $\begingroup$ True, but you can add powers like $z^2+z^3=z^5$ and there is a z in the left bracket, which makes $z^6$, doesn't it? Can I then do $z^3*z^3$ and solve the difference of two cubes? I must add, it would be very much more helpful if you'd say what does work, rather than what doesn't! Thanks. :-) $\endgroup$ – Georgina Davenport Oct 27 '15 at 14:15
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You seem to have some very unfortunate ideas about algebra! As David Mitra said, When you are adding powers of z you do not add the powers themselves. That is a property of multiplication: $z^n\cdot z^m= z^{n+ m}$.

To multiply $(z- 1)(1+ z+ z^2+ z^3)$ use the "distributive law" $a(b+ c)= ab+ ac$ and $(b+ c)a= ab+ ac$.

Think of $z- 1$ as $(b+ c)$ with $b= z$ and $c= -1$ and a as $(1+ z+ z^2+ z^3)$. $(z- 1)(1+ z+ z^2+ z^3)= z(1+ z+ z^2+ z^3)- 1(1+ z+ z^2+ z^3)$.

Now, for each of those, use the distributive law again: $z(1+ z+ z^2+ z^3)= z(1)+ z(z)+ z(z^2)+ z(z^4)$.

NOW use the rule for adding exponents (with $1= z^0$ and $z= z^1$): $z(1)+ z(z)+ z(z^2)+ z(z^3)= z+ z^2+ z^3+ z^4$.

And, of course, $-1(1+ z+ z^2+ z^3)= -1- z- z^2- z^3$.

Putting those together, $(z- 1)(1+ z+ z^2+ z^3)= z+ z^2+ z^3+ z^4- 1- z- z^2- z^3= -1+ (z- z)+ (z^2- z^2)+ (z^3- z^3)+ z^4= z^4- 1$.

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  • $\begingroup$ You are quite right user247327, my ideas about it are extremely unfortunate, thank you very much for your help accumulating better ideas about it. $\endgroup$ – Georgina Davenport Oct 27 '15 at 14:24
  • $\begingroup$ Unfortunately I can't +1 responses yet, but this answer is excellent. Thank you very much for your help! Most appreciated. $\endgroup$ – Georgina Davenport Oct 27 '15 at 15:00
  • $\begingroup$ @GeorginaDavenport You can upvote an answer once you receive at least 15 reputation points. See math.stackexchange.com/help/privileges. You also have the option of awarding a best answer to a response that does a particularly good job of answering your question. See math.stackexchange.com/help/someone-answers. $\endgroup$ – N. F. Taussig Nov 10 '15 at 12:50
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I like to do factorizations like this by writing $$ z^7-1=(z-1)(\cdots) $$ and try to figure out what should replace the dots. First, we'd like to have a factor of $z^7$ in the end result, so the $z$ on the right side should be multiplied by $z^6$ to get this. Then, we write $$ z^7-1=(z-1)(z^6+\cdots). $$

At this point, by the distributive law, the right side will multiply out to $z^7-z^6+\cdots$. Since we don't want the $z^6$ in the final product, it must cancel with something. We can cancel the $-z^6$ by multiplying $z$ by $z^5$ to get $$ z^7-1=(z-1)(z^6+z^5+\cdots) $$ because after distributing, the product is $z^7-z^6+z^6-z^5+\cdots$ and the $z^6$'s cancel. Now, we're left with $z^7-z^5+\cdots$, so the $-z^5$ must cancel with another term, since $z^5=z\cdot z^4$, we have $$ z^7-1=(z-1)(z^6+z^5+z^4+\cdots). $$

At this point, the product is $z^7-z^4+\cdots$. In order to cancel the $-z^4$, we introduce a $z^3$, which results in an extra $-z^3$. This extra term is cancelled with a $z^2$, which results in an extra $-z^2$ in the product. This $-z^2$ is cancelled with a $z$, but that gives an extra $-z$ in the product. The extra $-z$ is cancelled with a $-1$. Therefore, we have $$ z^7=(z-1)(z^6+z^5+z^4+z^3+z^2+z^1+1+\cdots). $$ But, after multiplying everything out, we find that the left side already matches the right side and there is nothing more in the dots! Therefore, $$ z^7=(z-1)(z^6+z^5+z^4+z^3+z^2+z^1+1). $$

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First off, user247327 gives an excellent answer. I don't have the reputation to comment on his comment, so here is an explanation of a critical fact you are missing. Q: why doesn't $$(1+z+z^2+z^3)=(z^6+1)?$$

Answer: When we write $z^3$, we mean $z*z*z$. Likewise $z^2=z*z$. Therefore: $$z^3*z^2=(z*z*z)*(z*z)$$ Well, how to we write this compactly? Simple! count the number of times we multiply z. Thus we get: $$z^3*z^2=(z*z*z)*(z*z)=z^5$$ This is why it is true in general that: $$z^a*z^b=z^{a+b}$$ If you ever forget this, just google "exponent rules", and you can see all the neat tricks. HOWEVER, there is no identity I can think of for simplifying $z^a+z^b$.

Edit: PS, in your comment you state $z^3+z^2=z^5$. Substitute z=2 to see that this is wrong. Note $2^5=32$, but: $$2^3+2^2=8+4=12 \neq 32$$ This one counterexample is enough to show that $z^a+z^b \neq z^{a+b}$

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First remove the left brackets and distribute the left terms on the right brackets:

$$(z - 1)(1 + z + z^2 + z^3)=z(1 + z + z^2 + z^3)-(1 + z + z^2 + z^3).$$

Then remove the remaining brackets and distribute what needs to be:

$$z(1 + z + z^2 + z^3)-(1 + z + z^2 + z^3)=z+z^2+z^3+z^4-1-z-z^2-z^3.$$

Finally, simplify:

$$z+z^2+z^3+z^4-1-z-z^2-z^3=z^4-1.$$

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