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Factoring given expression: $3x^frac32-9x^frac12+6x^-frac12$
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Factor given expression
$$3x^frac32-9x^frac12+6x^-frac12$$
It seems that there's $x^frac 1 2 $ in common. Let's rewrite this expression
$$3x^3 frac12-9x^frac 1 2+6x^-frac 12 $$
Recalling that $x^frac 1 2 = t $
$$3t^3-9t+6t^-1$$
Where have I gone wrong so far?
algebra-precalculus factoring
asked yesterday
Maxwell
296
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up vote
0
down vote
favorite
Factor given expression
$$3x^frac32-9x^frac12+6x^-frac12$$
It seems that there's $x^frac 1 2 $ in common. Let's rewrite this expression
$$3x^3 frac12-9x^frac 1 2+6x^-frac 12 $$
Recalling that $x^frac 1 2 = t $
$$3t^3-9t+6t^-1$$
Where have I gone wrong so far?
algebra-precalculus factoring
asked yesterday
Maxwell
296
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Factor given expression
$$3x^frac32-9x^frac12+6x^-frac12$$
It seems that there's $x^frac 1 2 $ in common. Let's rewrite this expression
$$3x^3 frac12-9x^frac 1 2+6x^-frac 12 $$
Recalling that $x^frac 1 2 = t $
$$3t^3-9t+6t^-1$$
Where have I gone wrong so far?
algebra-precalculus factoring
asked yesterday
Maxwell
296
Factor given expression
$$3x^frac32-9x^frac12+6x^-frac12$$
It seems that there's $x^frac 1 2 $ in common. Let's rewrite this expression
$$3x^3 frac12-9x^frac 1 2+6x^-frac 12 $$
Recalling that $x^frac 1 2 = t $
$$3t^3-9t+6t^-1$$
Where have I gone wrong so far?
algebra-precalculus factoring
asked yesterday
Maxwell
296
asked yesterday
Maxwell
296
asked yesterday
Maxwell
296
asked yesterday
Maxwell
296
296
add a comment |Â
add a comment |Â
5 Answers
5
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1
down vote
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)= 3x^-frac12(x-2)(x-1)
$$ As you see, there is no need for substitution.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
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up vote
1
down vote
It seems that there's $x^frac12$ in common.
No, actually the smallest term $x^-frac12$ is a common term. Then:
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)=3x^-frac12(x-1)(x-2).$$
Now, if you want, you can use $x^-frac12=t$ to get rid of the fractional exponent.
answered yesterday
farruhota
13.4k2632
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0
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$$3t^3 -9t +6t^-1 =(3t^4 -9t^2 +6)t^-1 .$$
For the parenthesis, let $u=t^2$; can you factor it?
answered yesterday
Rócherz
2,1811417
I didn't get it. Why is that $3t^4$?
– Maxwell
yesterday
Because $3t^3$ equals $3t^4$ times $t^-1$.
– Rócherz
yesterday
add a comment |Â
up vote
0
down vote
Or $$frac3(t^4-3t^2+2)t$$ or
$$frac3(t-sqrt2)(t+sqrt2)(t-1)(t+1)t.$$
answered yesterday
Michael Rozenberg
86.9k1575178
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0
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let's solve it from where you left. Write the last expression as
$$frac3t^4-9t^2+6t$$
now factorize the numerator which gives us
$$3*(t^2-1)*(t^2-2)$$
thus expression becomes
$$frac3*(t^2-1)*(t^2-2)t$$
put $t=x^1/2$
answered yesterday
James
325113
Why is that $3t^4$?
– Maxwell
yesterday
multiply your last expression by t and divide by t you will get the same result
– James
yesterday
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)= 3x^-frac12(x-2)(x-1)
$$ As you see, there is no need for substitution.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
add a comment |Â
up vote
1
down vote
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)= 3x^-frac12(x-2)(x-1)
$$ As you see, there is no need for substitution.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)= 3x^-frac12(x-2)(x-1)
$$ As you see, there is no need for substitution.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)= 3x^-frac12(x-2)(x-1)
$$ As you see, there is no need for substitution.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
26.9k41849
add a comment |Â
add a comment |Â
up vote
1
down vote
It seems that there's $x^frac12$ in common.
No, actually the smallest term $x^-frac12$ is a common term. Then:
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)=3x^-frac12(x-1)(x-2).$$
Now, if you want, you can use $x^-frac12=t$ to get rid of the fractional exponent.
answered yesterday
farruhota
13.4k2632
add a comment |Â
up vote
1
down vote
It seems that there's $x^frac12$ in common.
No, actually the smallest term $x^-frac12$ is a common term. Then:
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)=3x^-frac12(x-1)(x-2).$$
Now, if you want, you can use $x^-frac12=t$ to get rid of the fractional exponent.
answered yesterday
farruhota
13.4k2632
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It seems that there's $x^frac12$ in common.
No, actually the smallest term $x^-frac12$ is a common term. Then:
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)=3x^-frac12(x-1)(x-2).$$
Now, if you want, you can use $x^-frac12=t$ to get rid of the fractional exponent.
answered yesterday
farruhota
13.4k2632
It seems that there's $x^frac12$ in common.
No, actually the smallest term $x^-frac12$ is a common term. Then:
$$3x^frac32-9x^frac12+6x^-frac12=3x^-frac12(x^2-3x+2)=3x^-frac12(x-1)(x-2).$$
Now, if you want, you can use $x^-frac12=t$ to get rid of the fractional exponent.
answered yesterday
farruhota
13.4k2632
answered yesterday
farruhota
13.4k2632
answered yesterday
farruhota
13.4k2632
answered yesterday
farruhota
13.4k2632
13.4k2632
add a comment |Â
add a comment |Â
up vote
0
down vote
$$3t^3 -9t +6t^-1 =(3t^4 -9t^2 +6)t^-1 .$$
For the parenthesis, let $u=t^2$; can you factor it?
answered yesterday
Rócherz
2,1811417
I didn't get it. Why is that $3t^4$?
– Maxwell
yesterday
Because $3t^3$ equals $3t^4$ times $t^-1$.
– Rócherz
yesterday
add a comment |Â
up vote
0
down vote
$$3t^3 -9t +6t^-1 =(3t^4 -9t^2 +6)t^-1 .$$
For the parenthesis, let $u=t^2$; can you factor it?
answered yesterday
Rócherz
2,1811417
I didn't get it. Why is that $3t^4$?
– Maxwell
yesterday
Because $3t^3$ equals $3t^4$ times $t^-1$.
– Rócherz
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$3t^3 -9t +6t^-1 =(3t^4 -9t^2 +6)t^-1 .$$
For the parenthesis, let $u=t^2$; can you factor it?
answered yesterday
Rócherz
2,1811417
$$3t^3 -9t +6t^-1 =(3t^4 -9t^2 +6)t^-1 .$$
For the parenthesis, let $u=t^2$; can you factor it?
answered yesterday
Rócherz
2,1811417
answered yesterday
Rócherz
2,1811417
answered yesterday
Rócherz
2,1811417
answered yesterday
Rócherz
2,1811417
2,1811417
I didn't get it. Why is that $3t^4$?
– Maxwell
yesterday
Because $3t^3$ equals $3t^4$ times $t^-1$.
– Rócherz
yesterday
add a comment |Â
I didn't get it. Why is that $3t^4$?
– Maxwell
yesterday
Because $3t^3$ equals $3t^4$ times $t^-1$.
– Rócherz
yesterday
I didn't get it. Why is that $3t^4$?
– Maxwell
yesterday
I didn't get it. Why is that $3t^4$?
– Maxwell
yesterday
Because $3t^3$ equals $3t^4$ times $t^-1$.
– Rócherz
yesterday
Because $3t^3$ equals $3t^4$ times $t^-1$.
– Rócherz
yesterday
add a comment |Â
up vote
0
down vote
Or $$frac3(t^4-3t^2+2)t$$ or
$$frac3(t-sqrt2)(t+sqrt2)(t-1)(t+1)t.$$
answered yesterday
Michael Rozenberg
86.9k1575178
add a comment |Â
up vote
0
down vote
Or $$frac3(t^4-3t^2+2)t$$ or
$$frac3(t-sqrt2)(t+sqrt2)(t-1)(t+1)t.$$
answered yesterday
Michael Rozenberg
86.9k1575178
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Or $$frac3(t^4-3t^2+2)t$$ or
$$frac3(t-sqrt2)(t+sqrt2)(t-1)(t+1)t.$$
answered yesterday
Michael Rozenberg
86.9k1575178
Or $$frac3(t^4-3t^2+2)t$$ or
$$frac3(t-sqrt2)(t+sqrt2)(t-1)(t+1)t.$$
answered yesterday
Michael Rozenberg
86.9k1575178
answered yesterday
Michael Rozenberg
86.9k1575178
answered yesterday
Michael Rozenberg
86.9k1575178
answered yesterday
Michael Rozenberg
86.9k1575178
86.9k1575178
add a comment |Â
add a comment |Â
up vote
0
down vote
let's solve it from where you left. Write the last expression as
$$frac3t^4-9t^2+6t$$
now factorize the numerator which gives us
$$3*(t^2-1)*(t^2-2)$$
thus expression becomes
$$frac3*(t^2-1)*(t^2-2)t$$
put $t=x^1/2$
answered yesterday
James
325113
Why is that $3t^4$?
– Maxwell
yesterday
multiply your last expression by t and divide by t you will get the same result
– James
yesterday
add a comment |Â
up vote
0
down vote
let's solve it from where you left. Write the last expression as
$$frac3t^4-9t^2+6t$$
now factorize the numerator which gives us
$$3*(t^2-1)*(t^2-2)$$
thus expression becomes
$$frac3*(t^2-1)*(t^2-2)t$$
put $t=x^1/2$
answered yesterday
James
325113
Why is that $3t^4$?
– Maxwell
yesterday
multiply your last expression by t and divide by t you will get the same result
– James
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
let's solve it from where you left. Write the last expression as
$$frac3t^4-9t^2+6t$$
now factorize the numerator which gives us
$$3*(t^2-1)*(t^2-2)$$
thus expression becomes
$$frac3*(t^2-1)*(t^2-2)t$$
put $t=x^1/2$
answered yesterday
James
325113
let's solve it from where you left. Write the last expression as
$$frac3t^4-9t^2+6t$$
now factorize the numerator which gives us
$$3*(t^2-1)*(t^2-2)$$
thus expression becomes
$$frac3*(t^2-1)*(t^2-2)t$$
put $t=x^1/2$
answered yesterday
James
325113
edited yesterday
answered yesterday
James
325113
answered yesterday
James
325113
answered yesterday
James
325113
325113
Why is that $3t^4$?
– Maxwell
yesterday
multiply your last expression by t and divide by t you will get the same result
– James
yesterday
add a comment |Â
Why is that $3t^4$?
– Maxwell
yesterday
multiply your last expression by t and divide by t you will get the same result
– James
yesterday
Why is that $3t^4$?
– Maxwell
yesterday
Why is that $3t^4$?
– Maxwell
yesterday
multiply your last expression by t and divide by t you will get the same result
– James
yesterday
multiply your last expression by t and divide by t you will get the same result
– James
yesterday
add a comment |Â
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