how to factor $f(x)=x^4-7x^2+1$ [closed]
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How do you factor this equation? I have tried to use factoring method where the expanded b term factors add to B and multiply to get C.
polynomials quadratics factoring
edited Aug 31 at 0:21
dxiv
55.9k64798
asked Aug 30 at 23:56
Brian Betts
261
closed as off-topic by Jam, Leucippus, Theoretical Economist, Xander Henderson, José Carlos Santos Aug 31 at 3:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJam, Leucippus, Theoretical Economist, Xander Henderson, Jos̩ Carlos Santos
 |Â
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up vote
3
down vote
favorite
How do you factor this equation? I have tried to use factoring method where the expanded b term factors add to B and multiply to get C.
polynomials quadratics factoring
edited Aug 31 at 0:21
dxiv
55.9k64798
asked Aug 30 at 23:56
Brian Betts
261
closed as off-topic by Jam, Leucippus, Theoretical Economist, Xander Henderson, José Carlos Santos Aug 31 at 3:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJam, Leucippus, Theoretical Economist, Xander Henderson, Jos̩ Carlos Santos
4
Hint: rewrite it as $(x^4 + 2x^2 + 1) - 9x^2$.
– Dave Radcliffe
Aug 30 at 23:59
2
Hint: Let's name a new variable $u$ and let $u=x^2$. We have now $u^2-7u+1$ instead of your original expression, which should now be more easily handled. Once you've factored in terms of $u$, you may then take put $u$ back in terms of $x$ and continue from there.
– JMoravitz
Aug 30 at 23:59
2
I think it would be reasonable for down-voters to explain their actions. The OP has mentioned his efforts.
– Ahmad Bazzi
Aug 31 at 0:09
@Brian Please research how to solve quadratics and show us what you've tried. Any major technique should be useful and I'm sure you can find some resources online.
– Jam
Aug 31 at 0:11
@DaveRadcliffe's hint also follows if you first note that the polynomial is palindromic, then write it in terms of $,x + 1/x,$ as $,f(x)=x^2left(x^2+dfrac1x^2colorred+2-2-7right) = x^2left(left(x+dfrac1xright)^2-9right) = ldots$.
– dxiv
Aug 31 at 0:18
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How do you factor this equation? I have tried to use factoring method where the expanded b term factors add to B and multiply to get C.
polynomials quadratics factoring
edited Aug 31 at 0:21
dxiv
55.9k64798
asked Aug 30 at 23:56
Brian Betts
261
How do you factor this equation? I have tried to use factoring method where the expanded b term factors add to B and multiply to get C.
polynomials quadratics factoring
polynomials quadratics factoring
edited Aug 31 at 0:21
dxiv
55.9k64798
asked Aug 30 at 23:56
Brian Betts
261
edited Aug 31 at 0:21
dxiv
55.9k64798
asked Aug 30 at 23:56
Brian Betts
261
edited Aug 31 at 0:21
dxiv
55.9k64798
edited Aug 31 at 0:21
dxiv
55.9k64798
edited Aug 31 at 0:21
dxiv
55.9k64798
55.9k64798
asked Aug 30 at 23:56
Brian Betts
261
asked Aug 30 at 23:56
Brian Betts
261
asked Aug 30 at 23:56
Brian Betts
261
261
closed as off-topic by Jam, Leucippus, Theoretical Economist, Xander Henderson, José Carlos Santos Aug 31 at 3:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJam, Leucippus, Theoretical Economist, Xander Henderson, Jos̩ Carlos Santos
closed as off-topic by Jam, Leucippus, Theoretical Economist, Xander Henderson, José Carlos Santos Aug 31 at 3:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJam, Leucippus, Theoretical Economist, Xander Henderson, Jos̩ Carlos Santos
4
Hint: rewrite it as $(x^4 + 2x^2 + 1) - 9x^2$.
– Dave Radcliffe
Aug 30 at 23:59
2
Hint: Let's name a new variable $u$ and let $u=x^2$. We have now $u^2-7u+1$ instead of your original expression, which should now be more easily handled. Once you've factored in terms of $u$, you may then take put $u$ back in terms of $x$ and continue from there.
– JMoravitz
Aug 30 at 23:59
2
I think it would be reasonable for down-voters to explain their actions. The OP has mentioned his efforts.
– Ahmad Bazzi
Aug 31 at 0:09
@Brian Please research how to solve quadratics and show us what you've tried. Any major technique should be useful and I'm sure you can find some resources online.
– Jam
Aug 31 at 0:11
@DaveRadcliffe's hint also follows if you first note that the polynomial is palindromic, then write it in terms of $,x + 1/x,$ as $,f(x)=x^2left(x^2+dfrac1x^2colorred+2-2-7right) = x^2left(left(x+dfrac1xright)^2-9right) = ldots$.
– dxiv
Aug 31 at 0:18
 |Â
show 2 more comments
4
Hint: rewrite it as $(x^4 + 2x^2 + 1) - 9x^2$.
– Dave Radcliffe
Aug 30 at 23:59
2
Hint: Let's name a new variable $u$ and let $u=x^2$. We have now $u^2-7u+1$ instead of your original expression, which should now be more easily handled. Once you've factored in terms of $u$, you may then take put $u$ back in terms of $x$ and continue from there.
– JMoravitz
Aug 30 at 23:59
2
I think it would be reasonable for down-voters to explain their actions. The OP has mentioned his efforts.
– Ahmad Bazzi
Aug 31 at 0:09
@Brian Please research how to solve quadratics and show us what you've tried. Any major technique should be useful and I'm sure you can find some resources online.
– Jam
Aug 31 at 0:11
@DaveRadcliffe's hint also follows if you first note that the polynomial is palindromic, then write it in terms of $,x + 1/x,$ as $,f(x)=x^2left(x^2+dfrac1x^2colorred+2-2-7right) = x^2left(left(x+dfrac1xright)^2-9right) = ldots$.
– dxiv
Aug 31 at 0:18
4
4
Hint: rewrite it as $(x^4 + 2x^2 + 1) - 9x^2$.
– Dave Radcliffe
Aug 30 at 23:59
Hint: rewrite it as $(x^4 + 2x^2 + 1) - 9x^2$.
– Dave Radcliffe
Aug 30 at 23:59
2
2
Hint: Let's name a new variable $u$ and let $u=x^2$. We have now $u^2-7u+1$ instead of your original expression, which should now be more easily handled. Once you've factored in terms of $u$, you may then take put $u$ back in terms of $x$ and continue from there.
– JMoravitz
Aug 30 at 23:59
Hint: Let's name a new variable $u$ and let $u=x^2$. We have now $u^2-7u+1$ instead of your original expression, which should now be more easily handled. Once you've factored in terms of $u$, you may then take put $u$ back in terms of $x$ and continue from there.
– JMoravitz
Aug 30 at 23:59
2
2
I think it would be reasonable for down-voters to explain their actions. The OP has mentioned his efforts.
– Ahmad Bazzi
Aug 31 at 0:09
I think it would be reasonable for down-voters to explain their actions. The OP has mentioned his efforts.
– Ahmad Bazzi
Aug 31 at 0:09
@Brian Please research how to solve quadratics and show us what you've tried. Any major technique should be useful and I'm sure you can find some resources online.
– Jam
Aug 31 at 0:11
@Brian Please research how to solve quadratics and show us what you've tried. Any major technique should be useful and I'm sure you can find some resources online.
– Jam
Aug 31 at 0:11
@DaveRadcliffe's hint also follows if you first note that the polynomial is palindromic, then write it in terms of $,x + 1/x,$ as $,f(x)=x^2left(x^2+dfrac1x^2colorred+2-2-7right) = x^2left(left(x+dfrac1xright)^2-9right) = ldots$.
– dxiv
Aug 31 at 0:18
@DaveRadcliffe's hint also follows if you first note that the polynomial is palindromic, then write it in terms of $,x + 1/x,$ as $,f(x)=x^2left(x^2+dfrac1x^2colorred+2-2-7right) = x^2left(left(x+dfrac1xright)^2-9right) = ldots$.
– dxiv
Aug 31 at 0:18
 |Â
show 2 more comments
1 Answer
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It's easy to see that $x^4-7x^2+1$ has no rational roots. So if it factors (over the integers), it can only do so as the product of two quadratics. Since the lead and constant coefficients are both $1$ and there are no terms of odd degree, the factorization must be of the form $(x^2+bx+sigma)(x^2-bx+sigma)$ with $sigma=pm1$. We have
$$(x^2+bx+sigma)(x^2-bx+sigma)=x^4+(2sigma-b^2)x^2+1$$
so we need $2sigma-b^2=-7$, which rewrites as $b^2=7+2sigma$. Letting $sigma=1$ gives $b=pm3$, so we have the factorization
$$x^4-7x^2+1=(x^2+3x+1)(x^2-3x+1)$$
Note, the other option, $sigma=-1$, gives another valid factorization, just not over the integers:
$$x^4-7x^2+1=(x^2+sqrt5x-1)(x^2-sqrt5x-1)$$
answered Aug 31 at 2:40
Barry Cipra
56.9k652120
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
It's easy to see that $x^4-7x^2+1$ has no rational roots. So if it factors (over the integers), it can only do so as the product of two quadratics. Since the lead and constant coefficients are both $1$ and there are no terms of odd degree, the factorization must be of the form $(x^2+bx+sigma)(x^2-bx+sigma)$ with $sigma=pm1$. We have
$$(x^2+bx+sigma)(x^2-bx+sigma)=x^4+(2sigma-b^2)x^2+1$$
so we need $2sigma-b^2=-7$, which rewrites as $b^2=7+2sigma$. Letting $sigma=1$ gives $b=pm3$, so we have the factorization
$$x^4-7x^2+1=(x^2+3x+1)(x^2-3x+1)$$
Note, the other option, $sigma=-1$, gives another valid factorization, just not over the integers:
$$x^4-7x^2+1=(x^2+sqrt5x-1)(x^2-sqrt5x-1)$$
answered Aug 31 at 2:40
Barry Cipra
56.9k652120
add a comment |Â
up vote
3
down vote
It's easy to see that $x^4-7x^2+1$ has no rational roots. So if it factors (over the integers), it can only do so as the product of two quadratics. Since the lead and constant coefficients are both $1$ and there are no terms of odd degree, the factorization must be of the form $(x^2+bx+sigma)(x^2-bx+sigma)$ with $sigma=pm1$. We have
$$(x^2+bx+sigma)(x^2-bx+sigma)=x^4+(2sigma-b^2)x^2+1$$
so we need $2sigma-b^2=-7$, which rewrites as $b^2=7+2sigma$. Letting $sigma=1$ gives $b=pm3$, so we have the factorization
$$x^4-7x^2+1=(x^2+3x+1)(x^2-3x+1)$$
Note, the other option, $sigma=-1$, gives another valid factorization, just not over the integers:
$$x^4-7x^2+1=(x^2+sqrt5x-1)(x^2-sqrt5x-1)$$
answered Aug 31 at 2:40
Barry Cipra
56.9k652120
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It's easy to see that $x^4-7x^2+1$ has no rational roots. So if it factors (over the integers), it can only do so as the product of two quadratics. Since the lead and constant coefficients are both $1$ and there are no terms of odd degree, the factorization must be of the form $(x^2+bx+sigma)(x^2-bx+sigma)$ with $sigma=pm1$. We have
$$(x^2+bx+sigma)(x^2-bx+sigma)=x^4+(2sigma-b^2)x^2+1$$
so we need $2sigma-b^2=-7$, which rewrites as $b^2=7+2sigma$. Letting $sigma=1$ gives $b=pm3$, so we have the factorization
$$x^4-7x^2+1=(x^2+3x+1)(x^2-3x+1)$$
Note, the other option, $sigma=-1$, gives another valid factorization, just not over the integers:
$$x^4-7x^2+1=(x^2+sqrt5x-1)(x^2-sqrt5x-1)$$
answered Aug 31 at 2:40
Barry Cipra
56.9k652120
It's easy to see that $x^4-7x^2+1$ has no rational roots. So if it factors (over the integers), it can only do so as the product of two quadratics. Since the lead and constant coefficients are both $1$ and there are no terms of odd degree, the factorization must be of the form $(x^2+bx+sigma)(x^2-bx+sigma)$ with $sigma=pm1$. We have
$$(x^2+bx+sigma)(x^2-bx+sigma)=x^4+(2sigma-b^2)x^2+1$$
so we need $2sigma-b^2=-7$, which rewrites as $b^2=7+2sigma$. Letting $sigma=1$ gives $b=pm3$, so we have the factorization
$$x^4-7x^2+1=(x^2+3x+1)(x^2-3x+1)$$
Note, the other option, $sigma=-1$, gives another valid factorization, just not over the integers:
$$x^4-7x^2+1=(x^2+sqrt5x-1)(x^2-sqrt5x-1)$$
answered Aug 31 at 2:40
Barry Cipra
56.9k652120
answered Aug 31 at 2:40
Barry Cipra
56.9k652120
answered Aug 31 at 2:40
Barry Cipra
56.9k652120
answered Aug 31 at 2:40
Barry Cipra
56.9k652120
56.9k652120
add a comment |Â
add a comment |Â
4
Hint: rewrite it as $(x^4 + 2x^2 + 1) - 9x^2$.
– Dave Radcliffe
Aug 30 at 23:59
2
Hint: Let's name a new variable $u$ and let $u=x^2$. We have now $u^2-7u+1$ instead of your original expression, which should now be more easily handled. Once you've factored in terms of $u$, you may then take put $u$ back in terms of $x$ and continue from there.
– JMoravitz
Aug 30 at 23:59
2
I think it would be reasonable for down-voters to explain their actions. The OP has mentioned his efforts.
– Ahmad Bazzi
Aug 31 at 0:09
@Brian Please research how to solve quadratics and show us what you've tried. Any major technique should be useful and I'm sure you can find some resources online.
– Jam
Aug 31 at 0:11
@DaveRadcliffe's hint also follows if you first note that the polynomial is palindromic, then write it in terms of $,x + 1/x,$ as $,f(x)=x^2left(x^2+dfrac1x^2colorred+2-2-7right) = x^2left(left(x+dfrac1xright)^2-9right) = ldots$.
– dxiv
Aug 31 at 0:18