How do you factor: $2x^4+x^3-7x^2-x+5$ [closed]
$begingroup$
I don’t know how factor this, I’ve never factored something with a degree higher than 3.
factoring
edited Jan 11 at 0:30
David G. Stork
10.7k31332
asked Jan 11 at 0:16
annaîs mendoxaannaîs mendoxa
1
$endgroup$
closed as off-topic by amWhy, anomaly, Shailesh, José Carlos Santos, KReiser Jan 11 at 9:05
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, anomaly, Shailesh, José Carlos Santos, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I don’t know how factor this, I’ve never factored something with a degree higher than 3.
factoring
edited Jan 11 at 0:30
David G. Stork
10.7k31332
asked Jan 11 at 0:16
annaîs mendoxaannaîs mendoxa
1
$endgroup$
closed as off-topic by amWhy, anomaly, Shailesh, José Carlos Santos, KReiser Jan 11 at 9:05
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, anomaly, Shailesh, José Carlos Santos, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
$endgroup$
– Doug M
Jan 11 at 0:19
$begingroup$
Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
$endgroup$
– Lalaloopsy
Jan 11 at 0:27
$begingroup$
Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
$endgroup$
– fleablood
Jan 11 at 0:27
1
$begingroup$
$(x-1) (x+1) left(2 x^2+x-5right)$
$endgroup$
– David G. Stork
Jan 11 at 0:30
add a comment |
$begingroup$
I don’t know how factor this, I’ve never factored something with a degree higher than 3.
factoring
edited Jan 11 at 0:30
David G. Stork
10.7k31332
asked Jan 11 at 0:16
annaîs mendoxaannaîs mendoxa
1
$endgroup$
I don’t know how factor this, I’ve never factored something with a degree higher than 3.
factoring
factoring
edited Jan 11 at 0:30
David G. Stork
10.7k31332
asked Jan 11 at 0:16
annaîs mendoxaannaîs mendoxa
1
edited Jan 11 at 0:30
David G. Stork
10.7k31332
asked Jan 11 at 0:16
annaîs mendoxaannaîs mendoxa
1
edited Jan 11 at 0:30
David G. Stork
10.7k31332
edited Jan 11 at 0:30
David G. Stork
10.7k31332
edited Jan 11 at 0:30
David G. Stork
10.7k31332
10.7k31332
asked Jan 11 at 0:16
annaîs mendoxaannaîs mendoxa
1
asked Jan 11 at 0:16
annaîs mendoxaannaîs mendoxa
1
asked Jan 11 at 0:16
annaîs mendoxaannaîs mendoxa
1
1
closed as off-topic by amWhy, anomaly, Shailesh, José Carlos Santos, KReiser Jan 11 at 9:05
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, anomaly, Shailesh, José Carlos Santos, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, anomaly, Shailesh, José Carlos Santos, KReiser Jan 11 at 9:05
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, anomaly, Shailesh, José Carlos Santos, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
$endgroup$
– Doug M
Jan 11 at 0:19
$begingroup$
Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
$endgroup$
– Lalaloopsy
Jan 11 at 0:27
$begingroup$
Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
$endgroup$
– fleablood
Jan 11 at 0:27
1
$begingroup$
$(x-1) (x+1) left(2 x^2+x-5right)$
$endgroup$
– David G. Stork
Jan 11 at 0:30
add a comment |
2
$begingroup$
I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
$endgroup$
– Doug M
Jan 11 at 0:19
$begingroup$
Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
$endgroup$
– Lalaloopsy
Jan 11 at 0:27
$begingroup$
Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
$endgroup$
– fleablood
Jan 11 at 0:27
1
$begingroup$
$(x-1) (x+1) left(2 x^2+x-5right)$
$endgroup$
– David G. Stork
Jan 11 at 0:30
2
2
$begingroup$
I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
$endgroup$
– Doug M
Jan 11 at 0:19
$begingroup$
I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
$endgroup$
– Doug M
Jan 11 at 0:19
$begingroup$
Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
$endgroup$
– Lalaloopsy
Jan 11 at 0:27
$begingroup$
Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
$endgroup$
– Lalaloopsy
Jan 11 at 0:27
$begingroup$
Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
$endgroup$
– fleablood
Jan 11 at 0:27
$begingroup$
Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
$endgroup$
– fleablood
Jan 11 at 0:27
1
1
$begingroup$
$(x-1) (x+1) left(2 x^2+x-5right)$
$endgroup$
– David G. Stork
Jan 11 at 0:30
$begingroup$
$(x-1) (x+1) left(2 x^2+x-5right)$
$endgroup$
– David G. Stork
Jan 11 at 0:30
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.
You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.
The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.
A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.
Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.
answered Jan 11 at 0:27
GnumbertesterGnumbertester
37518
$endgroup$
add a comment |
$begingroup$
Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$
answered Jan 11 at 0:32
Bill DubuqueBill Dubuque
209k29191638
$endgroup$
add a comment |
$begingroup$
Hint:
This polynoial is divisible by $x^2-1$:
begin{align}
2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
&=2x^2(x^2-1)+ x^2(x-5)-x+5 \
&= (x^2-1)(2x^2+x-5).
end{align}+
answered Jan 11 at 0:34
BernardBernard
119k740113
$endgroup$
$begingroup$
But you pulled the factor out of a hat.
$endgroup$
– Bill Dubuque
Jan 11 at 0:35
$begingroup$
The rational roots theorem helped for the possible quadratic factors.
$endgroup$
– Bernard
Jan 11 at 0:36
add a comment |
$begingroup$
Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:
Divide out:
$${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$
Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:
$${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$
So:
$$(x-1)(x+1)(2 x^2+x-5)$$
Use the quadratic equation:
$$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$
answered Jan 11 at 0:37
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.
You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.
The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.
A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.
Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.
answered Jan 11 at 0:27
GnumbertesterGnumbertester
37518
$endgroup$
add a comment |
$begingroup$
In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.
You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.
The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.
A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.
Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.
answered Jan 11 at 0:27
GnumbertesterGnumbertester
37518
$endgroup$
add a comment |
$begingroup$
In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.
You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.
The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.
A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.
Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.
answered Jan 11 at 0:27
GnumbertesterGnumbertester
37518
$endgroup$
In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.
You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.
The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.
A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.
Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.
answered Jan 11 at 0:27
GnumbertesterGnumbertester
37518
answered Jan 11 at 0:27
GnumbertesterGnumbertester
37518
answered Jan 11 at 0:27
GnumbertesterGnumbertester
37518
answered Jan 11 at 0:27
GnumbertesterGnumbertester
37518
37518
add a comment |
add a comment |
$begingroup$
Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$
answered Jan 11 at 0:32
Bill DubuqueBill Dubuque
209k29191638
$endgroup$
add a comment |
$begingroup$
Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$
answered Jan 11 at 0:32
Bill DubuqueBill Dubuque
209k29191638
$endgroup$
add a comment |
$begingroup$
Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$
answered Jan 11 at 0:32
Bill DubuqueBill Dubuque
209k29191638
$endgroup$
Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$
answered Jan 11 at 0:32
Bill DubuqueBill Dubuque
209k29191638
answered Jan 11 at 0:32
Bill DubuqueBill Dubuque
209k29191638
answered Jan 11 at 0:32
Bill DubuqueBill Dubuque
209k29191638
answered Jan 11 at 0:32
Bill DubuqueBill Dubuque
209k29191638
209k29191638
add a comment |
add a comment |
$begingroup$
Hint:
This polynoial is divisible by $x^2-1$:
begin{align}
2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
&=2x^2(x^2-1)+ x^2(x-5)-x+5 \
&= (x^2-1)(2x^2+x-5).
end{align}+
answered Jan 11 at 0:34
BernardBernard
119k740113
$endgroup$
$begingroup$
But you pulled the factor out of a hat.
$endgroup$
– Bill Dubuque
Jan 11 at 0:35
$begingroup$
The rational roots theorem helped for the possible quadratic factors.
$endgroup$
– Bernard
Jan 11 at 0:36
add a comment |
$begingroup$
Hint:
This polynoial is divisible by $x^2-1$:
begin{align}
2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
&=2x^2(x^2-1)+ x^2(x-5)-x+5 \
&= (x^2-1)(2x^2+x-5).
end{align}+
answered Jan 11 at 0:34
BernardBernard
119k740113
$endgroup$
$begingroup$
But you pulled the factor out of a hat.
$endgroup$
– Bill Dubuque
Jan 11 at 0:35
$begingroup$
The rational roots theorem helped for the possible quadratic factors.
$endgroup$
– Bernard
Jan 11 at 0:36
add a comment |
$begingroup$
Hint:
This polynoial is divisible by $x^2-1$:
begin{align}
2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
&=2x^2(x^2-1)+ x^2(x-5)-x+5 \
&= (x^2-1)(2x^2+x-5).
end{align}+
answered Jan 11 at 0:34
BernardBernard
119k740113
$endgroup$
Hint:
This polynoial is divisible by $x^2-1$:
begin{align}
2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
&=2x^2(x^2-1)+ x^2(x-5)-x+5 \
&= (x^2-1)(2x^2+x-5).
end{align}+
answered Jan 11 at 0:34
BernardBernard
119k740113
answered Jan 11 at 0:34
BernardBernard
119k740113
answered Jan 11 at 0:34
BernardBernard
119k740113
answered Jan 11 at 0:34
BernardBernard
119k740113
119k740113
$begingroup$
But you pulled the factor out of a hat.
$endgroup$
– Bill Dubuque
Jan 11 at 0:35
$begingroup$
The rational roots theorem helped for the possible quadratic factors.
$endgroup$
– Bernard
Jan 11 at 0:36
add a comment |
$begingroup$
But you pulled the factor out of a hat.
$endgroup$
– Bill Dubuque
Jan 11 at 0:35
$begingroup$
The rational roots theorem helped for the possible quadratic factors.
$endgroup$
– Bernard
Jan 11 at 0:36
$begingroup$
But you pulled the factor out of a hat.
$endgroup$
– Bill Dubuque
Jan 11 at 0:35
$begingroup$
But you pulled the factor out of a hat.
$endgroup$
– Bill Dubuque
Jan 11 at 0:35
$begingroup$
The rational roots theorem helped for the possible quadratic factors.
$endgroup$
– Bernard
Jan 11 at 0:36
$begingroup$
The rational roots theorem helped for the possible quadratic factors.
$endgroup$
– Bernard
Jan 11 at 0:36
add a comment |
$begingroup$
Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:
Divide out:
$${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$
Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:
$${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$
So:
$$(x-1)(x+1)(2 x^2+x-5)$$
Use the quadratic equation:
$$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$
answered Jan 11 at 0:37
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
add a comment |
$begingroup$
Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:
Divide out:
$${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$
Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:
$${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$
So:
$$(x-1)(x+1)(2 x^2+x-5)$$
Use the quadratic equation:
$$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$
answered Jan 11 at 0:37
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
add a comment |
$begingroup$
Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:
Divide out:
$${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$
Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:
$${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$
So:
$$(x-1)(x+1)(2 x^2+x-5)$$
Use the quadratic equation:
$$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$
answered Jan 11 at 0:37
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:
Divide out:
$${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$
Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:
$${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$
So:
$$(x-1)(x+1)(2 x^2+x-5)$$
Use the quadratic equation:
$$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$
answered Jan 11 at 0:37
David G. StorkDavid G. Stork
10.7k31332
answered Jan 11 at 0:37
David G. StorkDavid G. Stork
10.7k31332
answered Jan 11 at 0:37
David G. StorkDavid G. Stork
10.7k31332
answered Jan 11 at 0:37
David G. StorkDavid G. Stork
10.7k31332
10.7k31332
add a comment |
add a comment |
2
$begingroup$
I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
$endgroup$
– Doug M
Jan 11 at 0:19
$begingroup$
Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
$endgroup$
– Lalaloopsy
Jan 11 at 0:27
$begingroup$
Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
$endgroup$
– fleablood
Jan 11 at 0:27
1
$begingroup$
$(x-1) (x+1) left(2 x^2+x-5right)$
$endgroup$
– David G. Stork
Jan 11 at 0:30