How do you find the factor of a polynomial with the remainders of polynomials that are being divided? [on...
I don't understand how to do this problem. I need to determine constants a and b based of the remainders of two polynomials divided by f(x)
When a polynomial expression $f(x)$ is divided by $x^2-4$ the remainder is $ax+b$. Determine the constants $a$ and $b$, given that $x-2$ is a factor of $f(x)$ and also that when $f(x)$ is divided by $x+2$, the remainder is $8$.
polynomials
edited Jan 4 at 7:07
jmerry
2,521312
asked Jan 4 at 7:00
CosmoCrashCosmoCrash
13
New contributor
CosmoCrash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Nosrati, Paul Frost, Pierre-Guy Plamondon, mrtaurho, amWhy Jan 4 at 14:13
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." – Nosrati, Paul Frost, Pierre-Guy Plamondon, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I don't understand how to do this problem. I need to determine constants a and b based of the remainders of two polynomials divided by f(x)
When a polynomial expression $f(x)$ is divided by $x^2-4$ the remainder is $ax+b$. Determine the constants $a$ and $b$, given that $x-2$ is a factor of $f(x)$ and also that when $f(x)$ is divided by $x+2$, the remainder is $8$.
polynomials
edited Jan 4 at 7:07
jmerry
2,521312
asked Jan 4 at 7:00
CosmoCrashCosmoCrash
13
New contributor
CosmoCrash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Nosrati, Paul Frost, Pierre-Guy Plamondon, mrtaurho, amWhy Jan 4 at 14:13
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." – Nosrati, Paul Frost, Pierre-Guy Plamondon, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
See this answer for an easy way
– Bill Dubuque
Jan 4 at 19:32
add a comment |
I don't understand how to do this problem. I need to determine constants a and b based of the remainders of two polynomials divided by f(x)
When a polynomial expression $f(x)$ is divided by $x^2-4$ the remainder is $ax+b$. Determine the constants $a$ and $b$, given that $x-2$ is a factor of $f(x)$ and also that when $f(x)$ is divided by $x+2$, the remainder is $8$.
polynomials
edited Jan 4 at 7:07
jmerry
2,521312
asked Jan 4 at 7:00
CosmoCrashCosmoCrash
13
New contributor
CosmoCrash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I don't understand how to do this problem. I need to determine constants a and b based of the remainders of two polynomials divided by f(x)
When a polynomial expression $f(x)$ is divided by $x^2-4$ the remainder is $ax+b$. Determine the constants $a$ and $b$, given that $x-2$ is a factor of $f(x)$ and also that when $f(x)$ is divided by $x+2$, the remainder is $8$.
polynomials
polynomials
edited Jan 4 at 7:07
jmerry
2,521312
asked Jan 4 at 7:00
CosmoCrashCosmoCrash
13
New contributor
CosmoCrash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 7:07
jmerry
2,521312
asked Jan 4 at 7:00
CosmoCrashCosmoCrash
13
New contributor
CosmoCrash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 7:07
jmerry
2,521312
edited Jan 4 at 7:07
jmerry
2,521312
edited Jan 4 at 7:07
jmerry
2,521312
2,521312
asked Jan 4 at 7:00
CosmoCrashCosmoCrash
13
New contributor
CosmoCrash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 7:00
CosmoCrashCosmoCrash
13
asked Jan 4 at 7:00
CosmoCrashCosmoCrash
13
13
New contributor
CosmoCrash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
CosmoCrash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
CosmoCrash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Nosrati, Paul Frost, Pierre-Guy Plamondon, mrtaurho, amWhy Jan 4 at 14:13
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." – Nosrati, Paul Frost, Pierre-Guy Plamondon, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Nosrati, Paul Frost, Pierre-Guy Plamondon, mrtaurho, amWhy Jan 4 at 14:13
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." – Nosrati, Paul Frost, Pierre-Guy Plamondon, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
See this answer for an easy way
– Bill Dubuque
Jan 4 at 19:32
add a comment |
See this answer for an easy way
– Bill Dubuque
Jan 4 at 19:32
See this answer for an easy way
– Bill Dubuque
Jan 4 at 19:32
See this answer for an easy way
– Bill Dubuque
Jan 4 at 19:32
add a comment |
1 Answer
1
active
oldest
votes
The key to this problem is that $x-2$ and $x-4$ are the factors of $x^2-4$.
Suppose we split $f$ into two parts through that division: $f(x) = (x^2-4)g(x) + ax+b$. What does it look like if we divide that by $x-2$? By $x+2$?
answered Jan 4 at 7:10
jmerryjmerry
2,521312
I solved the problem and got a = 0 , b = 2 but it says that's wrong
– CosmoCrash
Jan 4 at 7:24
Indeed. That would be a remainder of $2$ when divided by $x-2$ and a remainder of $2$ when divided by $x+2$. We can't say more unless you say more about what you've tried to do.
– jmerry
Jan 4 at 8:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The key to this problem is that $x-2$ and $x-4$ are the factors of $x^2-4$.
Suppose we split $f$ into two parts through that division: $f(x) = (x^2-4)g(x) + ax+b$. What does it look like if we divide that by $x-2$? By $x+2$?
answered Jan 4 at 7:10
jmerryjmerry
2,521312
I solved the problem and got a = 0 , b = 2 but it says that's wrong
– CosmoCrash
Jan 4 at 7:24
Indeed. That would be a remainder of $2$ when divided by $x-2$ and a remainder of $2$ when divided by $x+2$. We can't say more unless you say more about what you've tried to do.
– jmerry
Jan 4 at 8:00
add a comment |
The key to this problem is that $x-2$ and $x-4$ are the factors of $x^2-4$.
Suppose we split $f$ into two parts through that division: $f(x) = (x^2-4)g(x) + ax+b$. What does it look like if we divide that by $x-2$? By $x+2$?
answered Jan 4 at 7:10
jmerryjmerry
2,521312
I solved the problem and got a = 0 , b = 2 but it says that's wrong
– CosmoCrash
Jan 4 at 7:24
Indeed. That would be a remainder of $2$ when divided by $x-2$ and a remainder of $2$ when divided by $x+2$. We can't say more unless you say more about what you've tried to do.
– jmerry
Jan 4 at 8:00
add a comment |
The key to this problem is that $x-2$ and $x-4$ are the factors of $x^2-4$.
Suppose we split $f$ into two parts through that division: $f(x) = (x^2-4)g(x) + ax+b$. What does it look like if we divide that by $x-2$? By $x+2$?
answered Jan 4 at 7:10
jmerryjmerry
2,521312
The key to this problem is that $x-2$ and $x-4$ are the factors of $x^2-4$.
Suppose we split $f$ into two parts through that division: $f(x) = (x^2-4)g(x) + ax+b$. What does it look like if we divide that by $x-2$? By $x+2$?
answered Jan 4 at 7:10
jmerryjmerry
2,521312
answered Jan 4 at 7:10
jmerryjmerry
2,521312
answered Jan 4 at 7:10
jmerryjmerry
2,521312
answered Jan 4 at 7:10
jmerryjmerry
2,521312
2,521312
I solved the problem and got a = 0 , b = 2 but it says that's wrong
– CosmoCrash
Jan 4 at 7:24
Indeed. That would be a remainder of $2$ when divided by $x-2$ and a remainder of $2$ when divided by $x+2$. We can't say more unless you say more about what you've tried to do.
– jmerry
Jan 4 at 8:00
add a comment |
I solved the problem and got a = 0 , b = 2 but it says that's wrong
– CosmoCrash
Jan 4 at 7:24
Indeed. That would be a remainder of $2$ when divided by $x-2$ and a remainder of $2$ when divided by $x+2$. We can't say more unless you say more about what you've tried to do.
– jmerry
Jan 4 at 8:00
I solved the problem and got a = 0 , b = 2 but it says that's wrong
– CosmoCrash
Jan 4 at 7:24
I solved the problem and got a = 0 , b = 2 but it says that's wrong
– CosmoCrash
Jan 4 at 7:24
Indeed. That would be a remainder of $2$ when divided by $x-2$ and a remainder of $2$ when divided by $x+2$. We can't say more unless you say more about what you've tried to do.
– jmerry
Jan 4 at 8:00
Indeed. That would be a remainder of $2$ when divided by $x-2$ and a remainder of $2$ when divided by $x+2$. We can't say more unless you say more about what you've tried to do.
– jmerry
Jan 4 at 8:00
add a comment |
See this answer for an easy way
– Bill Dubuque
Jan 4 at 19:32