How do I factor this complex polynomial?
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My math teacher offered my class extra credit if we could factor this equation. It is not factorable by grouping. If you put the values into a polynomial root solver the zeros are Integer.
Equation: $x(x^3 + 4x^2 -3x -18) = 0$
algebra-precalculus
edited Dec 13 '18 at 20:44
user376343
3,7883828
asked Dec 13 '18 at 20:43
xXNebulaNinjaXxxXNebulaNinjaXx
31
$endgroup$
add a comment |
$begingroup$
My math teacher offered my class extra credit if we could factor this equation. It is not factorable by grouping. If you put the values into a polynomial root solver the zeros are Integer.
Equation: $x(x^3 + 4x^2 -3x -18) = 0$
algebra-precalculus
edited Dec 13 '18 at 20:44
user376343
3,7883828
asked Dec 13 '18 at 20:43
xXNebulaNinjaXxxXNebulaNinjaXx
31
$endgroup$
7
$begingroup$
Are you familiar with the rational root theorem?
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– platty
Dec 13 '18 at 20:44
$begingroup$
there were both + - with the term 3x. What has to be there?
$endgroup$
– user376343
Dec 13 '18 at 20:45
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Why don’t you try to find small integers that are roots of the polynomial?
$endgroup$
– Mindlack
Dec 13 '18 at 20:48
$begingroup$
Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
$endgroup$
– R. Burton
Dec 13 '18 at 22:44
add a comment |
$begingroup$
My math teacher offered my class extra credit if we could factor this equation. It is not factorable by grouping. If you put the values into a polynomial root solver the zeros are Integer.
Equation: $x(x^3 + 4x^2 -3x -18) = 0$
algebra-precalculus
edited Dec 13 '18 at 20:44
user376343
3,7883828
asked Dec 13 '18 at 20:43
xXNebulaNinjaXxxXNebulaNinjaXx
31
$endgroup$
My math teacher offered my class extra credit if we could factor this equation. It is not factorable by grouping. If you put the values into a polynomial root solver the zeros are Integer.
Equation: $x(x^3 + 4x^2 -3x -18) = 0$
algebra-precalculus
algebra-precalculus
edited Dec 13 '18 at 20:44
user376343
3,7883828
asked Dec 13 '18 at 20:43
xXNebulaNinjaXxxXNebulaNinjaXx
31
edited Dec 13 '18 at 20:44
user376343
3,7883828
asked Dec 13 '18 at 20:43
xXNebulaNinjaXxxXNebulaNinjaXx
31
edited Dec 13 '18 at 20:44
user376343
3,7883828
edited Dec 13 '18 at 20:44
user376343
3,7883828
edited Dec 13 '18 at 20:44
user376343
3,7883828
3,7883828
asked Dec 13 '18 at 20:43
xXNebulaNinjaXxxXNebulaNinjaXx
31
asked Dec 13 '18 at 20:43
xXNebulaNinjaXxxXNebulaNinjaXx
31
asked Dec 13 '18 at 20:43
xXNebulaNinjaXxxXNebulaNinjaXx
31
31
7
$begingroup$
Are you familiar with the rational root theorem?
$endgroup$
– platty
Dec 13 '18 at 20:44
$begingroup$
there were both + - with the term 3x. What has to be there?
$endgroup$
– user376343
Dec 13 '18 at 20:45
$begingroup$
Why don’t you try to find small integers that are roots of the polynomial?
$endgroup$
– Mindlack
Dec 13 '18 at 20:48
$begingroup$
Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
$endgroup$
– R. Burton
Dec 13 '18 at 22:44
add a comment |
7
$begingroup$
Are you familiar with the rational root theorem?
$endgroup$
– platty
Dec 13 '18 at 20:44
$begingroup$
there were both + - with the term 3x. What has to be there?
$endgroup$
– user376343
Dec 13 '18 at 20:45
$begingroup$
Why don’t you try to find small integers that are roots of the polynomial?
$endgroup$
– Mindlack
Dec 13 '18 at 20:48
$begingroup$
Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
$endgroup$
– R. Burton
Dec 13 '18 at 22:44
7
7
$begingroup$
Are you familiar with the rational root theorem?
$endgroup$
– platty
Dec 13 '18 at 20:44
$begingroup$
Are you familiar with the rational root theorem?
$endgroup$
– platty
Dec 13 '18 at 20:44
$begingroup$
there were both + - with the term 3x. What has to be there?
$endgroup$
– user376343
Dec 13 '18 at 20:45
$begingroup$
there were both + - with the term 3x. What has to be there?
$endgroup$
– user376343
Dec 13 '18 at 20:45
$begingroup$
Why don’t you try to find small integers that are roots of the polynomial?
$endgroup$
– Mindlack
Dec 13 '18 at 20:48
$begingroup$
Why don’t you try to find small integers that are roots of the polynomial?
$endgroup$
– Mindlack
Dec 13 '18 at 20:48
$begingroup$
Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
$endgroup$
– R. Burton
Dec 13 '18 at 22:44
$begingroup$
Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
$endgroup$
– R. Burton
Dec 13 '18 at 22:44
add a comment |
2 Answers
2
active
oldest
votes
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Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$
answered Dec 14 '18 at 12:14
AryanSonwatikarAryanSonwatikar
453114
$endgroup$
1
$begingroup$
Corrected: $q(x)=(x+3)^2$
$endgroup$
– z100
Dec 14 '18 at 14:04
$begingroup$
@z100 Sorry, my bad.
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 15:21
add a comment |
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$x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$
answered Dec 14 '18 at 13:03
z100z100
48848
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$
answered Dec 14 '18 at 12:14
AryanSonwatikarAryanSonwatikar
453114
$endgroup$
1
$begingroup$
Corrected: $q(x)=(x+3)^2$
$endgroup$
– z100
Dec 14 '18 at 14:04
$begingroup$
@z100 Sorry, my bad.
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 15:21
add a comment |
$begingroup$
Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$
answered Dec 14 '18 at 12:14
AryanSonwatikarAryanSonwatikar
453114
$endgroup$
1
$begingroup$
Corrected: $q(x)=(x+3)^2$
$endgroup$
– z100
Dec 14 '18 at 14:04
$begingroup$
@z100 Sorry, my bad.
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 15:21
add a comment |
$begingroup$
Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$
answered Dec 14 '18 at 12:14
AryanSonwatikarAryanSonwatikar
453114
$endgroup$
Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$
answered Dec 14 '18 at 12:14
AryanSonwatikarAryanSonwatikar
453114
edited Dec 14 '18 at 15:20
answered Dec 14 '18 at 12:14
AryanSonwatikarAryanSonwatikar
453114
answered Dec 14 '18 at 12:14
AryanSonwatikarAryanSonwatikar
453114
answered Dec 14 '18 at 12:14
AryanSonwatikarAryanSonwatikar
453114
453114
1
$begingroup$
Corrected: $q(x)=(x+3)^2$
$endgroup$
– z100
Dec 14 '18 at 14:04
$begingroup$
@z100 Sorry, my bad.
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 15:21
add a comment |
1
$begingroup$
Corrected: $q(x)=(x+3)^2$
$endgroup$
– z100
Dec 14 '18 at 14:04
$begingroup$
@z100 Sorry, my bad.
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 15:21
1
1
$begingroup$
Corrected: $q(x)=(x+3)^2$
$endgroup$
– z100
Dec 14 '18 at 14:04
$begingroup$
Corrected: $q(x)=(x+3)^2$
$endgroup$
– z100
Dec 14 '18 at 14:04
$begingroup$
@z100 Sorry, my bad.
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 15:21
$begingroup$
@z100 Sorry, my bad.
$endgroup$
– AryanSonwatikar
Dec 14 '18 at 15:21
add a comment |
$begingroup$
$x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$
answered Dec 14 '18 at 13:03
z100z100
48848
$endgroup$
add a comment |
$begingroup$
$x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$
answered Dec 14 '18 at 13:03
z100z100
48848
$endgroup$
add a comment |
$begingroup$
$x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$
answered Dec 14 '18 at 13:03
z100z100
48848
$endgroup$
$x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$
answered Dec 14 '18 at 13:03
z100z100
48848
answered Dec 14 '18 at 13:03
z100z100
48848
answered Dec 14 '18 at 13:03
z100z100
48848
answered Dec 14 '18 at 13:03
z100z100
48848
48848
add a comment |
add a comment |
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7
$begingroup$
Are you familiar with the rational root theorem?
$endgroup$
– platty
Dec 13 '18 at 20:44
$begingroup$
there were both + - with the term 3x. What has to be there?
$endgroup$
– user376343
Dec 13 '18 at 20:45
$begingroup$
Why don’t you try to find small integers that are roots of the polynomial?
$endgroup$
– Mindlack
Dec 13 '18 at 20:48
$begingroup$
Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
$endgroup$
– R. Burton
Dec 13 '18 at 22:44