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How can I factorize this: “$X^3 + X^2 + X - 3$â€
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up vote
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I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
beginalignX^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \
&= (X - 1)^2 + X(X - 1) \
&= (X-1)(X-1+X) \
&= (X - 1)(2X - 1)
endalign
But for a few months I have not been able to find a teacher around here who can factorize this:
$$X^3 + X^2 + X - 3$$
Do we have to solve it in this way?
$$X^3 + X^2 + X - 3 = X^2(X + 1) + X - 3$$
Or something else?
I'd appreciate your help with this.
polynomials factoring
edited Aug 22 at 10:54
Daniel Buck
2,5151625
asked Aug 22 at 8:12
AmirhoseinRiazi
1114
 |Â
show 1 more comment
up vote
1
down vote
favorite
I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
beginalignX^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \
&= (X - 1)^2 + X(X - 1) \
&= (X-1)(X-1+X) \
&= (X - 1)(2X - 1)
endalign
But for a few months I have not been able to find a teacher around here who can factorize this:
$$X^3 + X^2 + X - 3$$
Do we have to solve it in this way?
$$X^3 + X^2 + X - 3 = X^2(X + 1) + X - 3$$
Or something else?
I'd appreciate your help with this.
polynomials factoring
edited Aug 22 at 10:54
Daniel Buck
2,5151625
asked Aug 22 at 8:12
AmirhoseinRiazi
1114
Hint: $x=1$ is a root of your polynomial, meaning that the polynomial is divisible by $x-1$.
– Sobi
Aug 22 at 8:13
1
$X^3+X^2+X-3 = X^3-1 + X^2-1 + X-1 = (X-1)(X^2+X+1)+(X-1)(X+1)+(X-1) = (X-1)(X^2+2X+3)$, $X^2+2X+3$ you can't factorize when working with real numbers
– Rumpelstiltskin
Aug 22 at 8:15
@AmirhoseinRiazi You might want to look at this or this
– Rumpelstiltskin
Aug 22 at 8:29
Please use MathJax to format your posts.
– Chase Ryan Taylor
Aug 22 at 9:19
@ChaseRyanTaylor Excuse me, this question may seem strange or silly, but I do not know how to use this MathJox. I'm new to this, where should I start from?
– AmirhoseinRiazi
Aug 22 at 11:14
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
beginalignX^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \
&= (X - 1)^2 + X(X - 1) \
&= (X-1)(X-1+X) \
&= (X - 1)(2X - 1)
endalign
But for a few months I have not been able to find a teacher around here who can factorize this:
$$X^3 + X^2 + X - 3$$
Do we have to solve it in this way?
$$X^3 + X^2 + X - 3 = X^2(X + 1) + X - 3$$
Or something else?
I'd appreciate your help with this.
polynomials factoring
edited Aug 22 at 10:54
Daniel Buck
2,5151625
asked Aug 22 at 8:12
AmirhoseinRiazi
1114
I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
beginalignX^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \
&= (X - 1)^2 + X(X - 1) \
&= (X-1)(X-1+X) \
&= (X - 1)(2X - 1)
endalign
But for a few months I have not been able to find a teacher around here who can factorize this:
$$X^3 + X^2 + X - 3$$
Do we have to solve it in this way?
$$X^3 + X^2 + X - 3 = X^2(X + 1) + X - 3$$
Or something else?
I'd appreciate your help with this.
polynomials factoring
edited Aug 22 at 10:54
Daniel Buck
2,5151625
asked Aug 22 at 8:12
AmirhoseinRiazi
1114
edited Aug 22 at 10:54
Daniel Buck
2,5151625
edited Aug 22 at 10:54
Daniel Buck
2,5151625
edited Aug 22 at 10:54
Daniel Buck
2,5151625
2,5151625
asked Aug 22 at 8:12
AmirhoseinRiazi
1114
asked Aug 22 at 8:12
AmirhoseinRiazi
1114
asked Aug 22 at 8:12
AmirhoseinRiazi
1114
1114
Hint: $x=1$ is a root of your polynomial, meaning that the polynomial is divisible by $x-1$.
– Sobi
Aug 22 at 8:13
1
$X^3+X^2+X-3 = X^3-1 + X^2-1 + X-1 = (X-1)(X^2+X+1)+(X-1)(X+1)+(X-1) = (X-1)(X^2+2X+3)$, $X^2+2X+3$ you can't factorize when working with real numbers
– Rumpelstiltskin
Aug 22 at 8:15
@AmirhoseinRiazi You might want to look at this or this
– Rumpelstiltskin
Aug 22 at 8:29
Please use MathJax to format your posts.
– Chase Ryan Taylor
Aug 22 at 9:19
@ChaseRyanTaylor Excuse me, this question may seem strange or silly, but I do not know how to use this MathJox. I'm new to this, where should I start from?
– AmirhoseinRiazi
Aug 22 at 11:14
 |Â
show 1 more comment
Hint: $x=1$ is a root of your polynomial, meaning that the polynomial is divisible by $x-1$.
– Sobi
Aug 22 at 8:13
1
$X^3+X^2+X-3 = X^3-1 + X^2-1 + X-1 = (X-1)(X^2+X+1)+(X-1)(X+1)+(X-1) = (X-1)(X^2+2X+3)$, $X^2+2X+3$ you can't factorize when working with real numbers
– Rumpelstiltskin
Aug 22 at 8:15
@AmirhoseinRiazi You might want to look at this or this
– Rumpelstiltskin
Aug 22 at 8:29
Please use MathJax to format your posts.
– Chase Ryan Taylor
Aug 22 at 9:19
@ChaseRyanTaylor Excuse me, this question may seem strange or silly, but I do not know how to use this MathJox. I'm new to this, where should I start from?
– AmirhoseinRiazi
Aug 22 at 11:14
Hint: $x=1$ is a root of your polynomial, meaning that the polynomial is divisible by $x-1$.
– Sobi
Aug 22 at 8:13
Hint: $x=1$ is a root of your polynomial, meaning that the polynomial is divisible by $x-1$.
– Sobi
Aug 22 at 8:13
1
1
$X^3+X^2+X-3 = X^3-1 + X^2-1 + X-1 = (X-1)(X^2+X+1)+(X-1)(X+1)+(X-1) = (X-1)(X^2+2X+3)$, $X^2+2X+3$ you can't factorize when working with real numbers
– Rumpelstiltskin
Aug 22 at 8:15
$X^3+X^2+X-3 = X^3-1 + X^2-1 + X-1 = (X-1)(X^2+X+1)+(X-1)(X+1)+(X-1) = (X-1)(X^2+2X+3)$, $X^2+2X+3$ you can't factorize when working with real numbers
– Rumpelstiltskin
Aug 22 at 8:15
@AmirhoseinRiazi You might want to look at this or this
– Rumpelstiltskin
Aug 22 at 8:29
@AmirhoseinRiazi You might want to look at this or this
– Rumpelstiltskin
Aug 22 at 8:29
Please use MathJax to format your posts.
– Chase Ryan Taylor
Aug 22 at 9:19
Please use MathJax to format your posts.
– Chase Ryan Taylor
Aug 22 at 9:19
@ChaseRyanTaylor Excuse me, this question may seem strange or silly, but I do not know how to use this MathJox. I'm new to this, where should I start from?
– AmirhoseinRiazi
Aug 22 at 11:14
@ChaseRyanTaylor Excuse me, this question may seem strange or silly, but I do not know how to use this MathJox. I'm new to this, where should I start from?
– AmirhoseinRiazi
Aug 22 at 11:14
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
4
down vote
Surely $1$ is a root of $x^3+x^2+x-3$ therefore $x-1$ is a factor of it. We have $$x^3+x^2+x-3=(x-1)(x^2+x+1+x+1+1)=(x-1)(x^2+2x+3)$$
answered Aug 22 at 8:15
Mostafa Ayaz
10.1k3730
I will show this to my teacher, he maybe recall, thanks a lot!
– AmirhoseinRiazi
Aug 22 at 8:27
1
You're welcome! Good luck!
– Mostafa Ayaz
Aug 22 at 8:31
add a comment |Â
up vote
3
down vote
Hint: $(x^3+x^2+x-3):(x-1)=x^2+2x+3$
answered Aug 22 at 8:15
Fred
38.2k1238
add a comment |Â
up vote
3
down vote
For this kind of problem it is worth knowing about the factor theorem
This says if $a$ is a root of your polynomial $f(x)$ i.e. $f(a)=0$ then $x-a$ is a factor of, i.e. divides, the polynomial.
In these kinds of problems it is worth trying a few values such as $pm1, pm2$.
In your example $f(1)=0$ so $x-1$ divides your polynomial, allowing you to factorize it as $(x-1)(Ax^2+Bx+C)$ where you need to find $A,B,C$.
answered Aug 22 at 8:33
PM.
3,1982822
Useful, I can't adequately express my gratefulness!
– AmirhoseinRiazi
Aug 22 at 9:21
Thank-you. I am very pleased it helped you.
– PM.
Aug 22 at 9:25
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Surely $1$ is a root of $x^3+x^2+x-3$ therefore $x-1$ is a factor of it. We have $$x^3+x^2+x-3=(x-1)(x^2+x+1+x+1+1)=(x-1)(x^2+2x+3)$$
answered Aug 22 at 8:15
Mostafa Ayaz
10.1k3730
I will show this to my teacher, he maybe recall, thanks a lot!
– AmirhoseinRiazi
Aug 22 at 8:27
1
You're welcome! Good luck!
– Mostafa Ayaz
Aug 22 at 8:31
add a comment |Â
up vote
4
down vote
Surely $1$ is a root of $x^3+x^2+x-3$ therefore $x-1$ is a factor of it. We have $$x^3+x^2+x-3=(x-1)(x^2+x+1+x+1+1)=(x-1)(x^2+2x+3)$$
answered Aug 22 at 8:15
Mostafa Ayaz
10.1k3730
I will show this to my teacher, he maybe recall, thanks a lot!
– AmirhoseinRiazi
Aug 22 at 8:27
1
You're welcome! Good luck!
– Mostafa Ayaz
Aug 22 at 8:31
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Surely $1$ is a root of $x^3+x^2+x-3$ therefore $x-1$ is a factor of it. We have $$x^3+x^2+x-3=(x-1)(x^2+x+1+x+1+1)=(x-1)(x^2+2x+3)$$
answered Aug 22 at 8:15
Mostafa Ayaz
10.1k3730
Surely $1$ is a root of $x^3+x^2+x-3$ therefore $x-1$ is a factor of it. We have $$x^3+x^2+x-3=(x-1)(x^2+x+1+x+1+1)=(x-1)(x^2+2x+3)$$
answered Aug 22 at 8:15
Mostafa Ayaz
10.1k3730
answered Aug 22 at 8:15
Mostafa Ayaz
10.1k3730
answered Aug 22 at 8:15
Mostafa Ayaz
10.1k3730
answered Aug 22 at 8:15
Mostafa Ayaz
10.1k3730
10.1k3730
I will show this to my teacher, he maybe recall, thanks a lot!
– AmirhoseinRiazi
Aug 22 at 8:27
1
You're welcome! Good luck!
– Mostafa Ayaz
Aug 22 at 8:31
add a comment |Â
I will show this to my teacher, he maybe recall, thanks a lot!
– AmirhoseinRiazi
Aug 22 at 8:27
1
You're welcome! Good luck!
– Mostafa Ayaz
Aug 22 at 8:31
I will show this to my teacher, he maybe recall, thanks a lot!
– AmirhoseinRiazi
Aug 22 at 8:27
I will show this to my teacher, he maybe recall, thanks a lot!
– AmirhoseinRiazi
Aug 22 at 8:27
1
1
You're welcome! Good luck!
– Mostafa Ayaz
Aug 22 at 8:31
You're welcome! Good luck!
– Mostafa Ayaz
Aug 22 at 8:31
add a comment |Â
up vote
3
down vote
Hint: $(x^3+x^2+x-3):(x-1)=x^2+2x+3$
answered Aug 22 at 8:15
Fred
38.2k1238
add a comment |Â
up vote
3
down vote
Hint: $(x^3+x^2+x-3):(x-1)=x^2+2x+3$
answered Aug 22 at 8:15
Fred
38.2k1238
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hint: $(x^3+x^2+x-3):(x-1)=x^2+2x+3$
answered Aug 22 at 8:15
Fred
38.2k1238
Hint: $(x^3+x^2+x-3):(x-1)=x^2+2x+3$
answered Aug 22 at 8:15
Fred
38.2k1238
answered Aug 22 at 8:15
Fred
38.2k1238
answered Aug 22 at 8:15
Fred
38.2k1238
answered Aug 22 at 8:15
Fred
38.2k1238
38.2k1238
add a comment |Â
add a comment |Â
up vote
3
down vote
For this kind of problem it is worth knowing about the factor theorem
This says if $a$ is a root of your polynomial $f(x)$ i.e. $f(a)=0$ then $x-a$ is a factor of, i.e. divides, the polynomial.
In these kinds of problems it is worth trying a few values such as $pm1, pm2$.
In your example $f(1)=0$ so $x-1$ divides your polynomial, allowing you to factorize it as $(x-1)(Ax^2+Bx+C)$ where you need to find $A,B,C$.
answered Aug 22 at 8:33
PM.
3,1982822
Useful, I can't adequately express my gratefulness!
– AmirhoseinRiazi
Aug 22 at 9:21
Thank-you. I am very pleased it helped you.
– PM.
Aug 22 at 9:25
add a comment |Â
up vote
3
down vote
For this kind of problem it is worth knowing about the factor theorem
This says if $a$ is a root of your polynomial $f(x)$ i.e. $f(a)=0$ then $x-a$ is a factor of, i.e. divides, the polynomial.
In these kinds of problems it is worth trying a few values such as $pm1, pm2$.
In your example $f(1)=0$ so $x-1$ divides your polynomial, allowing you to factorize it as $(x-1)(Ax^2+Bx+C)$ where you need to find $A,B,C$.
answered Aug 22 at 8:33
PM.
3,1982822
Useful, I can't adequately express my gratefulness!
– AmirhoseinRiazi
Aug 22 at 9:21
Thank-you. I am very pleased it helped you.
– PM.
Aug 22 at 9:25
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For this kind of problem it is worth knowing about the factor theorem
This says if $a$ is a root of your polynomial $f(x)$ i.e. $f(a)=0$ then $x-a$ is a factor of, i.e. divides, the polynomial.
In these kinds of problems it is worth trying a few values such as $pm1, pm2$.
In your example $f(1)=0$ so $x-1$ divides your polynomial, allowing you to factorize it as $(x-1)(Ax^2+Bx+C)$ where you need to find $A,B,C$.
answered Aug 22 at 8:33
PM.
3,1982822
For this kind of problem it is worth knowing about the factor theorem
This says if $a$ is a root of your polynomial $f(x)$ i.e. $f(a)=0$ then $x-a$ is a factor of, i.e. divides, the polynomial.
In these kinds of problems it is worth trying a few values such as $pm1, pm2$.
In your example $f(1)=0$ so $x-1$ divides your polynomial, allowing you to factorize it as $(x-1)(Ax^2+Bx+C)$ where you need to find $A,B,C$.
answered Aug 22 at 8:33
PM.
3,1982822
edited Aug 22 at 9:25
answered Aug 22 at 8:33
PM.
3,1982822
answered Aug 22 at 8:33
PM.
3,1982822
answered Aug 22 at 8:33
PM.
3,1982822
3,1982822
Useful, I can't adequately express my gratefulness!
– AmirhoseinRiazi
Aug 22 at 9:21
Thank-you. I am very pleased it helped you.
– PM.
Aug 22 at 9:25
add a comment |Â
Useful, I can't adequately express my gratefulness!
– AmirhoseinRiazi
Aug 22 at 9:21
Thank-you. I am very pleased it helped you.
– PM.
Aug 22 at 9:25
Useful, I can't adequately express my gratefulness!
– AmirhoseinRiazi
Aug 22 at 9:21
Useful, I can't adequately express my gratefulness!
– AmirhoseinRiazi
Aug 22 at 9:21
Thank-you. I am very pleased it helped you.
– PM.
Aug 22 at 9:25
Thank-you. I am very pleased it helped you.
– PM.
Aug 22 at 9:25
add a comment |Â
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Hint: $x=1$ is a root of your polynomial, meaning that the polynomial is divisible by $x-1$.
– Sobi
Aug 22 at 8:13
1
$X^3+X^2+X-3 = X^3-1 + X^2-1 + X-1 = (X-1)(X^2+X+1)+(X-1)(X+1)+(X-1) = (X-1)(X^2+2X+3)$, $X^2+2X+3$ you can't factorize when working with real numbers
– Rumpelstiltskin
Aug 22 at 8:15
@AmirhoseinRiazi You might want to look at this or this
– Rumpelstiltskin
Aug 22 at 8:29
Please use MathJax to format your posts.
– Chase Ryan Taylor
Aug 22 at 9:19
@ChaseRyanTaylor Excuse me, this question may seem strange or silly, but I do not know how to use this MathJox. I'm new to this, where should I start from?
– AmirhoseinRiazi
Aug 22 at 11:14