Mixing
Simplifying $frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$
$begingroup$
In simplifying
$$frac{2x^2-5x-3}{6x^3-2x^4}$$
I got this far
$$frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$$
but there aren't same brackets to cancel out.
algebra-precalculus rational-functions
edited Jan 31 at 21:58
Blue
49.5k870157
asked Jan 31 at 18:16
xx_Gcsemathstudent_xxxx_Gcsemathstudent_xx
356
$endgroup$
add a comment |
$begingroup$
In simplifying
$$frac{2x^2-5x-3}{6x^3-2x^4}$$
I got this far
$$frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$$
but there aren't same brackets to cancel out.
algebra-precalculus rational-functions
edited Jan 31 at 21:58
Blue
49.5k870157
asked Jan 31 at 18:16
xx_Gcsemathstudent_xxxx_Gcsemathstudent_xx
356
$endgroup$
add a comment |
$begingroup$
In simplifying
$$frac{2x^2-5x-3}{6x^3-2x^4}$$
I got this far
$$frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$$
but there aren't same brackets to cancel out.
algebra-precalculus rational-functions
edited Jan 31 at 21:58
Blue
49.5k870157
asked Jan 31 at 18:16
xx_Gcsemathstudent_xxxx_Gcsemathstudent_xx
356
$endgroup$
In simplifying
$$frac{2x^2-5x-3}{6x^3-2x^4}$$
I got this far
$$frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$$
but there aren't same brackets to cancel out.
algebra-precalculus rational-functions
algebra-precalculus rational-functions
edited Jan 31 at 21:58
Blue
49.5k870157
asked Jan 31 at 18:16
xx_Gcsemathstudent_xxxx_Gcsemathstudent_xx
356
edited Jan 31 at 21:58
Blue
49.5k870157
asked Jan 31 at 18:16
xx_Gcsemathstudent_xxxx_Gcsemathstudent_xx
356
edited Jan 31 at 21:58
Blue
49.5k870157
edited Jan 31 at 21:58
Blue
49.5k870157
edited Jan 31 at 21:58
Blue
49.5k870157
49.5k870157
asked Jan 31 at 18:16
xx_Gcsemathstudent_xxxx_Gcsemathstudent_xx
356
asked Jan 31 at 18:16
xx_Gcsemathstudent_xxxx_Gcsemathstudent_xx
356
asked Jan 31 at 18:16
xx_Gcsemathstudent_xxxx_Gcsemathstudent_xx
356
356
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$dfrac{2x^2-5x-3}{6x^3-2x^4}=dfrac{(2x+1)(x-3)}{2x^3(3-x)}=dfrac{(2x+1)(x-3)}{2x^3[-(x-3)]}=dfrac{(2x+1)(x-3)}{-2x^3(x-3)}=dfrac{2x+1}{-2x^3}$$
answered Jan 31 at 18:22
Key FlexKey Flex
8,58771233
$endgroup$
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
@xx_Gcsemathstudent_xx Your welcome!
$endgroup$
– Key Flex
Jan 31 at 18:23
$begingroup$
Don't forget to write the domain of the given term!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
add a comment |
$begingroup$
There are!
Don't forget that $(x-3)=-(3-x)$ so you can cancel to give:
$$frac{-(2x+1)}{2x^3}$$
Just remember that $xneq 3$ is a requirement else you're dividing by $0$ (Thanks Dr)
answered Jan 31 at 18:18
Rhys HughesRhys Hughes
7,0501630
$endgroup$
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
I have written that and got a $-1$, i think this is a joke!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
$begingroup$
You can only cancel if $xneq 3$!!!!!!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:14
add a comment |
$begingroup$
$frac{(2x+1)(x-3)}{2x^3(3-x)}= -frac{(2x+1)(x-3)}{2x^3(x-3)}= -frac{(2x+1)}{2x^3} $
answered Jan 31 at 18:18
Federico FalluccaFederico Fallucca
2,320210
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$dfrac{2x^2-5x-3}{6x^3-2x^4}=dfrac{(2x+1)(x-3)}{2x^3(3-x)}=dfrac{(2x+1)(x-3)}{2x^3[-(x-3)]}=dfrac{(2x+1)(x-3)}{-2x^3(x-3)}=dfrac{2x+1}{-2x^3}$$
answered Jan 31 at 18:22
Key FlexKey Flex
8,58771233
$endgroup$
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
@xx_Gcsemathstudent_xx Your welcome!
$endgroup$
– Key Flex
Jan 31 at 18:23
$begingroup$
Don't forget to write the domain of the given term!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
add a comment |
$begingroup$
$$dfrac{2x^2-5x-3}{6x^3-2x^4}=dfrac{(2x+1)(x-3)}{2x^3(3-x)}=dfrac{(2x+1)(x-3)}{2x^3[-(x-3)]}=dfrac{(2x+1)(x-3)}{-2x^3(x-3)}=dfrac{2x+1}{-2x^3}$$
answered Jan 31 at 18:22
Key FlexKey Flex
8,58771233
$endgroup$
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
@xx_Gcsemathstudent_xx Your welcome!
$endgroup$
– Key Flex
Jan 31 at 18:23
$begingroup$
Don't forget to write the domain of the given term!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
add a comment |
$begingroup$
$$dfrac{2x^2-5x-3}{6x^3-2x^4}=dfrac{(2x+1)(x-3)}{2x^3(3-x)}=dfrac{(2x+1)(x-3)}{2x^3[-(x-3)]}=dfrac{(2x+1)(x-3)}{-2x^3(x-3)}=dfrac{2x+1}{-2x^3}$$
answered Jan 31 at 18:22
Key FlexKey Flex
8,58771233
$endgroup$
$$dfrac{2x^2-5x-3}{6x^3-2x^4}=dfrac{(2x+1)(x-3)}{2x^3(3-x)}=dfrac{(2x+1)(x-3)}{2x^3[-(x-3)]}=dfrac{(2x+1)(x-3)}{-2x^3(x-3)}=dfrac{2x+1}{-2x^3}$$
answered Jan 31 at 18:22
Key FlexKey Flex
8,58771233
edited Jan 31 at 18:23
answered Jan 31 at 18:22
Key FlexKey Flex
8,58771233
answered Jan 31 at 18:22
Key FlexKey Flex
8,58771233
answered Jan 31 at 18:22
Key FlexKey Flex
8,58771233
8,58771233
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
@xx_Gcsemathstudent_xx Your welcome!
$endgroup$
– Key Flex
Jan 31 at 18:23
$begingroup$
Don't forget to write the domain of the given term!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
add a comment |
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
@xx_Gcsemathstudent_xx Your welcome!
$endgroup$
– Key Flex
Jan 31 at 18:23
$begingroup$
Don't forget to write the domain of the given term!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
@xx_Gcsemathstudent_xx Your welcome!
$endgroup$
– Key Flex
Jan 31 at 18:23
$begingroup$
@xx_Gcsemathstudent_xx Your welcome!
$endgroup$
– Key Flex
Jan 31 at 18:23
$begingroup$
Don't forget to write the domain of the given term!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
$begingroup$
Don't forget to write the domain of the given term!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
add a comment |
$begingroup$
There are!
Don't forget that $(x-3)=-(3-x)$ so you can cancel to give:
$$frac{-(2x+1)}{2x^3}$$
Just remember that $xneq 3$ is a requirement else you're dividing by $0$ (Thanks Dr)
answered Jan 31 at 18:18
Rhys HughesRhys Hughes
7,0501630
$endgroup$
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
I have written that and got a $-1$, i think this is a joke!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
$begingroup$
You can only cancel if $xneq 3$!!!!!!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:14
add a comment |
$begingroup$
There are!
Don't forget that $(x-3)=-(3-x)$ so you can cancel to give:
$$frac{-(2x+1)}{2x^3}$$
Just remember that $xneq 3$ is a requirement else you're dividing by $0$ (Thanks Dr)
answered Jan 31 at 18:18
Rhys HughesRhys Hughes
7,0501630
$endgroup$
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
I have written that and got a $-1$, i think this is a joke!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
$begingroup$
You can only cancel if $xneq 3$!!!!!!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:14
add a comment |
$begingroup$
There are!
Don't forget that $(x-3)=-(3-x)$ so you can cancel to give:
$$frac{-(2x+1)}{2x^3}$$
Just remember that $xneq 3$ is a requirement else you're dividing by $0$ (Thanks Dr)
answered Jan 31 at 18:18
Rhys HughesRhys Hughes
7,0501630
$endgroup$
There are!
Don't forget that $(x-3)=-(3-x)$ so you can cancel to give:
$$frac{-(2x+1)}{2x^3}$$
Just remember that $xneq 3$ is a requirement else you're dividing by $0$ (Thanks Dr)
answered Jan 31 at 18:18
Rhys HughesRhys Hughes
7,0501630
edited Jan 31 at 20:05
answered Jan 31 at 18:18
Rhys HughesRhys Hughes
7,0501630
answered Jan 31 at 18:18
Rhys HughesRhys Hughes
7,0501630
answered Jan 31 at 18:18
Rhys HughesRhys Hughes
7,0501630
7,0501630
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
I have written that and got a $-1$, i think this is a joke!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
$begingroup$
You can only cancel if $xneq 3$!!!!!!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:14
add a comment |
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
I have written that and got a $-1$, i think this is a joke!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
$begingroup$
You can only cancel if $xneq 3$!!!!!!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:14
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
Thank You For Helping Me!
$endgroup$
– xx_Gcsemathstudent_xx
Jan 31 at 18:23
$begingroup$
I have written that and got a $-1$, i think this is a joke!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
$begingroup$
I have written that and got a $-1$, i think this is a joke!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:13
$begingroup$
You can only cancel if $xneq 3$!!!!!!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:14
$begingroup$
You can only cancel if $xneq 3$!!!!!!
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:14
add a comment |
$begingroup$
$frac{(2x+1)(x-3)}{2x^3(3-x)}= -frac{(2x+1)(x-3)}{2x^3(x-3)}= -frac{(2x+1)}{2x^3} $
answered Jan 31 at 18:18
Federico FalluccaFederico Fallucca
2,320210
$endgroup$
add a comment |
$begingroup$
$frac{(2x+1)(x-3)}{2x^3(3-x)}= -frac{(2x+1)(x-3)}{2x^3(x-3)}= -frac{(2x+1)}{2x^3} $
answered Jan 31 at 18:18
Federico FalluccaFederico Fallucca
2,320210
$endgroup$
add a comment |
$begingroup$
$frac{(2x+1)(x-3)}{2x^3(3-x)}= -frac{(2x+1)(x-3)}{2x^3(x-3)}= -frac{(2x+1)}{2x^3} $
answered Jan 31 at 18:18
Federico FalluccaFederico Fallucca
2,320210
$endgroup$
$frac{(2x+1)(x-3)}{2x^3(3-x)}= -frac{(2x+1)(x-3)}{2x^3(x-3)}= -frac{(2x+1)}{2x^3} $
answered Jan 31 at 18:18
Federico FalluccaFederico Fallucca
2,320210
answered Jan 31 at 18:18
Federico FalluccaFederico Fallucca
2,320210
answered Jan 31 at 18:18
Federico FalluccaFederico Fallucca
2,320210
answered Jan 31 at 18:18
Federico FalluccaFederico Fallucca
2,320210
2,320210
add a comment |
add a comment |
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