How can we factor $ q^4 + q^2 + 1$
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Alright, I met this example solved in a book, but I want to know how did they come up with the answer:
$$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$
Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.
elementary-number-theory factoring
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add a comment |
$begingroup$
Alright, I met this example solved in a book, but I want to know how did they come up with the answer:
$$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$
Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.
elementary-number-theory factoring
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$begingroup$
here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
$endgroup$
– user25406
Nov 8 '16 at 20:20
add a comment |
$begingroup$
Alright, I met this example solved in a book, but I want to know how did they come up with the answer:
$$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$
Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.
elementary-number-theory factoring
$endgroup$
Alright, I met this example solved in a book, but I want to know how did they come up with the answer:
$$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$
Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.
elementary-number-theory factoring
elementary-number-theory factoring
edited Jan 1 at 15:59
Bill Dubuque
211k29194648
211k29194648
asked Nov 8 '16 at 16:01
G. NickG. Nick
82
82
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here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
$endgroup$
– user25406
Nov 8 '16 at 20:20
add a comment |
$begingroup$
here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
$endgroup$
– user25406
Nov 8 '16 at 20:20
$begingroup$
here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
$endgroup$
– user25406
Nov 8 '16 at 20:20
$begingroup$
here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
$endgroup$
– user25406
Nov 8 '16 at 20:20
add a comment |
2 Answers
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votes
$begingroup$
Hint $ $ completing the square leads to a difference of squares, e.g.
$$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
end{eqnarray}$$ Here is another common example
$$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
&,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$
$endgroup$
add a comment |
$begingroup$
$$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
$$(1+q^2)^2=1+q^4+2q^2$$
Hence:
$$1+q^2+q^4=(1+q^2)^2-q^2$$
Which is the difference of two squares, something we know we can factor:
$$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$
Some often used rules for factoring are:
$$(a+b)^2=a^2+b^2+2ab$$
$$(a-b)(a+b)=a^2-b^2$$
$$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Hint $ $ completing the square leads to a difference of squares, e.g.
$$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
end{eqnarray}$$ Here is another common example
$$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
&,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$
$endgroup$
add a comment |
$begingroup$
Hint $ $ completing the square leads to a difference of squares, e.g.
$$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
end{eqnarray}$$ Here is another common example
$$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
&,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$
$endgroup$
add a comment |
$begingroup$
Hint $ $ completing the square leads to a difference of squares, e.g.
$$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
end{eqnarray}$$ Here is another common example
$$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
&,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$
$endgroup$
Hint $ $ completing the square leads to a difference of squares, e.g.
$$begin{eqnarray}overbrace{q^4+1}^{rm incomplete large Box!!}!!! + q^2!&,=,&!! overbrace{(q^2+1)^2}^{rm complete the LargeBox !!!}!!!- color{#c00}{q}^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (q^2+1, -, color{#c00}{q}),(q^2+1+,color{#c00}{q})\
end{eqnarray}$$ Here is another common example
$$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
&,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$
edited Jan 1 at 16:08
answered Nov 8 '16 at 16:05
Bill DubuqueBill Dubuque
211k29194648
211k29194648
add a comment |
add a comment |
$begingroup$
$$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
$$(1+q^2)^2=1+q^4+2q^2$$
Hence:
$$1+q^2+q^4=(1+q^2)^2-q^2$$
Which is the difference of two squares, something we know we can factor:
$$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$
Some often used rules for factoring are:
$$(a+b)^2=a^2+b^2+2ab$$
$$(a-b)(a+b)=a^2-b^2$$
$$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$
$endgroup$
add a comment |
$begingroup$
$$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
$$(1+q^2)^2=1+q^4+2q^2$$
Hence:
$$1+q^2+q^4=(1+q^2)^2-q^2$$
Which is the difference of two squares, something we know we can factor:
$$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$
Some often used rules for factoring are:
$$(a+b)^2=a^2+b^2+2ab$$
$$(a-b)(a+b)=a^2-b^2$$
$$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$
$endgroup$
add a comment |
$begingroup$
$$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
$$(1+q^2)^2=1+q^4+2q^2$$
Hence:
$$1+q^2+q^4=(1+q^2)^2-q^2$$
Which is the difference of two squares, something we know we can factor:
$$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$
Some often used rules for factoring are:
$$(a+b)^2=a^2+b^2+2ab$$
$$(a-b)(a+b)=a^2-b^2$$
$$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$
$endgroup$
$$1+q^2+q^4=(1+q+q^2)(1+q^2-q)$$
Notice how $1^2=1$ and $(q^2)^2=q^4$, so $(1+q^2)^2$ will get us a few terms already:
$$(1+q^2)^2=1+q^4+2q^2$$
Hence:
$$1+q^2+q^4=(1+q^2)^2-q^2$$
Which is the difference of two squares, something we know we can factor:
$$1+q^2+q^4=(1+q^2)^2-q^2=((1+q^2)+q)((1+q^2)-q)=(1+q^2+q)(1+q^2-q)$$
Some often used rules for factoring are:
$$(a+b)^2=a^2+b^2+2ab$$
$$(a-b)(a+b)=a^2-b^2$$
$$(a-1)(1+a+a^2+...+a^n)=a^{n+1}-1$$
answered Nov 8 '16 at 16:05
MastremMastrem
3,79711230
3,79711230
add a comment |
add a comment |
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$begingroup$
here's something you may want to check and enjoy. en.wikipedia.org/wiki/Factorization
$endgroup$
– user25406
Nov 8 '16 at 20:20