Mixing
Simplify this expression: $25x^2 + 30x^3 - 35x^6$ [closed]
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I was thinking about simplifying this expression:
$$25x^2 + 30x^3 - 35x^6$$
Note that I don't mean just factoring $5x^2$ from it!
algebra-precalculus factoring
edited Jan 14 at 16:10
Key Flex
8,27261233
asked Jan 14 at 16:02
amiraliamirali
12
$endgroup$
closed as unclear what you're asking by amWhy, Omnomnomnom, John Douma, José Carlos Santos, Alexander Gruber♦ Jan 15 at 11:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
I was thinking about simplifying this expression:
$$25x^2 + 30x^3 - 35x^6$$
Note that I don't mean just factoring $5x^2$ from it!
algebra-precalculus factoring
edited Jan 14 at 16:10
Key Flex
8,27261233
asked Jan 14 at 16:02
amiraliamirali
12
$endgroup$
closed as unclear what you're asking by amWhy, Omnomnomnom, John Douma, José Carlos Santos, Alexander Gruber♦ Jan 15 at 11:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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How exactly do you solve an expression?
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– KM101
Jan 14 at 16:05
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What do you mean by "solving"? That's just an expression.
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– Ramiro Scorolli
Jan 14 at 16:06
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@KM101 I mean simplify it.
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– amirali
Jan 14 at 16:06
2
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"Simplifying" is vague and subjective (although not completely arbitrary; $2$ is clearly simpler than $frac{105+95}{2cdot 50}$). Do you mean "factor"?
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– Arthur
Jan 14 at 16:08
2
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@amirali Yes, that's factoring. This one is awful, though. Like, really, unimaginably awful. Are you sure you want it factored?
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– user3482749
Jan 14 at 16:14
|
show 5 more comments
$begingroup$
I was thinking about simplifying this expression:
$$25x^2 + 30x^3 - 35x^6$$
Note that I don't mean just factoring $5x^2$ from it!
algebra-precalculus factoring
edited Jan 14 at 16:10
Key Flex
8,27261233
asked Jan 14 at 16:02
amiraliamirali
12
$endgroup$
I was thinking about simplifying this expression:
$$25x^2 + 30x^3 - 35x^6$$
Note that I don't mean just factoring $5x^2$ from it!
algebra-precalculus factoring
algebra-precalculus factoring
edited Jan 14 at 16:10
Key Flex
8,27261233
asked Jan 14 at 16:02
amiraliamirali
12
edited Jan 14 at 16:10
Key Flex
8,27261233
asked Jan 14 at 16:02
amiraliamirali
12
edited Jan 14 at 16:10
Key Flex
8,27261233
edited Jan 14 at 16:10
Key Flex
8,27261233
edited Jan 14 at 16:10
Key Flex
8,27261233
8,27261233
asked Jan 14 at 16:02
amiraliamirali
12
asked Jan 14 at 16:02
amiraliamirali
12
asked Jan 14 at 16:02
amiraliamirali
12
12
closed as unclear what you're asking by amWhy, Omnomnomnom, John Douma, José Carlos Santos, Alexander Gruber♦ Jan 15 at 11:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by amWhy, Omnomnomnom, John Douma, José Carlos Santos, Alexander Gruber♦ Jan 15 at 11:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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How exactly do you solve an expression?
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– KM101
Jan 14 at 16:05
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What do you mean by "solving"? That's just an expression.
$endgroup$
– Ramiro Scorolli
Jan 14 at 16:06
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@KM101 I mean simplify it.
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– amirali
Jan 14 at 16:06
2
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"Simplifying" is vague and subjective (although not completely arbitrary; $2$ is clearly simpler than $frac{105+95}{2cdot 50}$). Do you mean "factor"?
$endgroup$
– Arthur
Jan 14 at 16:08
2
$begingroup$
@amirali Yes, that's factoring. This one is awful, though. Like, really, unimaginably awful. Are you sure you want it factored?
$endgroup$
– user3482749
Jan 14 at 16:14
|
show 5 more comments
$begingroup$
How exactly do you solve an expression?
$endgroup$
– KM101
Jan 14 at 16:05
$begingroup$
What do you mean by "solving"? That's just an expression.
$endgroup$
– Ramiro Scorolli
Jan 14 at 16:06
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@KM101 I mean simplify it.
$endgroup$
– amirali
Jan 14 at 16:06
2
$begingroup$
"Simplifying" is vague and subjective (although not completely arbitrary; $2$ is clearly simpler than $frac{105+95}{2cdot 50}$). Do you mean "factor"?
$endgroup$
– Arthur
Jan 14 at 16:08
2
$begingroup$
@amirali Yes, that's factoring. This one is awful, though. Like, really, unimaginably awful. Are you sure you want it factored?
$endgroup$
– user3482749
Jan 14 at 16:14
$begingroup$
How exactly do you solve an expression?
$endgroup$
– KM101
Jan 14 at 16:05
$begingroup$
How exactly do you solve an expression?
$endgroup$
– KM101
Jan 14 at 16:05
$begingroup$
What do you mean by "solving"? That's just an expression.
$endgroup$
– Ramiro Scorolli
Jan 14 at 16:06
$begingroup$
What do you mean by "solving"? That's just an expression.
$endgroup$
– Ramiro Scorolli
Jan 14 at 16:06
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@KM101 I mean simplify it.
$endgroup$
– amirali
Jan 14 at 16:06
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@KM101 I mean simplify it.
$endgroup$
– amirali
Jan 14 at 16:06
2
2
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"Simplifying" is vague and subjective (although not completely arbitrary; $2$ is clearly simpler than $frac{105+95}{2cdot 50}$). Do you mean "factor"?
$endgroup$
– Arthur
Jan 14 at 16:08
$begingroup$
"Simplifying" is vague and subjective (although not completely arbitrary; $2$ is clearly simpler than $frac{105+95}{2cdot 50}$). Do you mean "factor"?
$endgroup$
– Arthur
Jan 14 at 16:08
2
2
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@amirali Yes, that's factoring. This one is awful, though. Like, really, unimaginably awful. Are you sure you want it factored?
$endgroup$
– user3482749
Jan 14 at 16:14
$begingroup$
@amirali Yes, that's factoring. This one is awful, though. Like, really, unimaginably awful. Are you sure you want it factored?
$endgroup$
– user3482749
Jan 14 at 16:14
|
show 5 more comments
1 Answer
1
active
oldest
votes
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Your expression factors as $frac{-5}{7}x^2a(x)b(x)c(x)$, where $a(x)$, $b(x)$, and $c(x)$ are, respectively,
$$left(!!x! -!! sqrt{!sqrt[3]{left(81!+!sqrt{48561}right)^2}! -! 10sqrt[3]{42}}!+!frac{!!!!!sqrt{!10sqrt[3]{!42}! -!! sqrt[3]{!left(81!+!sqrt{48561}right)^2}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2}! - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}!!right)$$
and
$$left(!!x! -!! sqrt{!!sqrt[3]{!left(!81!!+!!sqrt{!48561}!right)^2}!! -!! 10sqrt[3]{42}}!-!!frac{!!!!!sqrt{!10sqrt[3]{!42} !-!! sqrt[3]{!!left(81!!+!!sqrt{!48561}right)^2}!!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2} - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}right)$$
and
$$left(x^2 -alpha x+betaright),$$
where $$alpha=2sqrt{sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{41}}$$
and $$beta!=!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}! +!! frac{!!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{42}}}}{sqrt[3]{2^5}sqrt[3]{21^2}sqrt[3]{81+sqrt{48561}}}.$$
answered Jan 14 at 16:49
user3482749user3482749
4,266919
$endgroup$
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You have warned OP above, so the shock was mitigated.....
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– Andreas
Jan 14 at 16:54
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Is this an output of a program?
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– kelalaka
Jan 14 at 17:01
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@kelalaka The initial factorisation was, yes. I LaTeXed it manually.
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– user3482749
Jan 14 at 17:01
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+1 for the latex
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– kelalaka
Jan 14 at 17:02
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Could you please share your solution?
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– amirali
Jan 14 at 18:06
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your expression factors as $frac{-5}{7}x^2a(x)b(x)c(x)$, where $a(x)$, $b(x)$, and $c(x)$ are, respectively,
$$left(!!x! -!! sqrt{!sqrt[3]{left(81!+!sqrt{48561}right)^2}! -! 10sqrt[3]{42}}!+!frac{!!!!!sqrt{!10sqrt[3]{!42}! -!! sqrt[3]{!left(81!+!sqrt{48561}right)^2}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2}! - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}!!right)$$
and
$$left(!!x! -!! sqrt{!!sqrt[3]{!left(!81!!+!!sqrt{!48561}!right)^2}!! -!! 10sqrt[3]{42}}!-!!frac{!!!!!sqrt{!10sqrt[3]{!42} !-!! sqrt[3]{!!left(81!!+!!sqrt{!48561}right)^2}!!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2} - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}right)$$
and
$$left(x^2 -alpha x+betaright),$$
where $$alpha=2sqrt{sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{41}}$$
and $$beta!=!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}! +!! frac{!!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{42}}}}{sqrt[3]{2^5}sqrt[3]{21^2}sqrt[3]{81+sqrt{48561}}}.$$
answered Jan 14 at 16:49
user3482749user3482749
4,266919
$endgroup$
$begingroup$
You have warned OP above, so the shock was mitigated.....
$endgroup$
– Andreas
Jan 14 at 16:54
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Is this an output of a program?
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– kelalaka
Jan 14 at 17:01
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@kelalaka The initial factorisation was, yes. I LaTeXed it manually.
$endgroup$
– user3482749
Jan 14 at 17:01
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+1 for the latex
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– kelalaka
Jan 14 at 17:02
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Could you please share your solution?
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– amirali
Jan 14 at 18:06
|
show 1 more comment
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Your expression factors as $frac{-5}{7}x^2a(x)b(x)c(x)$, where $a(x)$, $b(x)$, and $c(x)$ are, respectively,
$$left(!!x! -!! sqrt{!sqrt[3]{left(81!+!sqrt{48561}right)^2}! -! 10sqrt[3]{42}}!+!frac{!!!!!sqrt{!10sqrt[3]{!42}! -!! sqrt[3]{!left(81!+!sqrt{48561}right)^2}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2}! - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}!!right)$$
and
$$left(!!x! -!! sqrt{!!sqrt[3]{!left(!81!!+!!sqrt{!48561}!right)^2}!! -!! 10sqrt[3]{42}}!-!!frac{!!!!!sqrt{!10sqrt[3]{!42} !-!! sqrt[3]{!!left(81!!+!!sqrt{!48561}right)^2}!!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2} - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}right)$$
and
$$left(x^2 -alpha x+betaright),$$
where $$alpha=2sqrt{sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{41}}$$
and $$beta!=!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}! +!! frac{!!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{42}}}}{sqrt[3]{2^5}sqrt[3]{21^2}sqrt[3]{81+sqrt{48561}}}.$$
answered Jan 14 at 16:49
user3482749user3482749
4,266919
$endgroup$
$begingroup$
You have warned OP above, so the shock was mitigated.....
$endgroup$
– Andreas
Jan 14 at 16:54
$begingroup$
Is this an output of a program?
$endgroup$
– kelalaka
Jan 14 at 17:01
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@kelalaka The initial factorisation was, yes. I LaTeXed it manually.
$endgroup$
– user3482749
Jan 14 at 17:01
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+1 for the latex
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– kelalaka
Jan 14 at 17:02
$begingroup$
Could you please share your solution?
$endgroup$
– amirali
Jan 14 at 18:06
|
show 1 more comment
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Your expression factors as $frac{-5}{7}x^2a(x)b(x)c(x)$, where $a(x)$, $b(x)$, and $c(x)$ are, respectively,
$$left(!!x! -!! sqrt{!sqrt[3]{left(81!+!sqrt{48561}right)^2}! -! 10sqrt[3]{42}}!+!frac{!!!!!sqrt{!10sqrt[3]{!42}! -!! sqrt[3]{!left(81!+!sqrt{48561}right)^2}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2}! - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}!!right)$$
and
$$left(!!x! -!! sqrt{!!sqrt[3]{!left(!81!!+!!sqrt{!48561}!right)^2}!! -!! 10sqrt[3]{42}}!-!!frac{!!!!!sqrt{!10sqrt[3]{!42} !-!! sqrt[3]{!!left(81!!+!!sqrt{!48561}right)^2}!!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2} - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}right)$$
and
$$left(x^2 -alpha x+betaright),$$
where $$alpha=2sqrt{sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{41}}$$
and $$beta!=!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}! +!! frac{!!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{42}}}}{sqrt[3]{2^5}sqrt[3]{21^2}sqrt[3]{81+sqrt{48561}}}.$$
answered Jan 14 at 16:49
user3482749user3482749
4,266919
$endgroup$
Your expression factors as $frac{-5}{7}x^2a(x)b(x)c(x)$, where $a(x)$, $b(x)$, and $c(x)$ are, respectively,
$$left(!!x! -!! sqrt{!sqrt[3]{left(81!+!sqrt{48561}right)^2}! -! 10sqrt[3]{42}}!+!frac{!!!!!sqrt{!10sqrt[3]{!42}! -!! sqrt[3]{!left(81!+!sqrt{48561}right)^2}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2}! - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}!!right)$$
and
$$left(!!x! -!! sqrt{!!sqrt[3]{!left(!81!!+!!sqrt{!48561}!right)^2}!! -!! 10sqrt[3]{42}}!-!!frac{!!!!!sqrt{!10sqrt[3]{!42} !-!! sqrt[3]{!!left(81!!+!!sqrt{!48561}right)^2}!!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{!left(81+sqrt{48561}right)^2} - 10sqrt[3]{42}}}}}{sqrt[6]{2^5}sqrt[3]{21}sqrt[6]{81+sqrt{48561}}}right)$$
and
$$left(x^2 -alpha x+betaright),$$
where $$alpha=2sqrt{sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{41}}$$
and $$beta!=!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}! +!! frac{!!!sqrt[3]{!left(81!!+!!sqrt{!48561}right)^2}!!-!!10sqrt[3]{42}!+!!18sqrt{!!frac{2left(81+sqrt{48561}right)}{!sqrt[3]{left(81+sqrt{48561}right)^2}-10sqrt[3]{42}}}}{sqrt[3]{2^5}sqrt[3]{21^2}sqrt[3]{81+sqrt{48561}}}.$$
answered Jan 14 at 16:49
user3482749user3482749
4,266919
edited Jan 14 at 21:03
answered Jan 14 at 16:49
user3482749user3482749
4,266919
answered Jan 14 at 16:49
user3482749user3482749
4,266919
answered Jan 14 at 16:49
user3482749user3482749
4,266919
4,266919
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You have warned OP above, so the shock was mitigated.....
$endgroup$
– Andreas
Jan 14 at 16:54
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Is this an output of a program?
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– kelalaka
Jan 14 at 17:01
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@kelalaka The initial factorisation was, yes. I LaTeXed it manually.
$endgroup$
– user3482749
Jan 14 at 17:01
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+1 for the latex
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– kelalaka
Jan 14 at 17:02
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Could you please share your solution?
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– amirali
Jan 14 at 18:06
|
show 1 more comment
$begingroup$
You have warned OP above, so the shock was mitigated.....
$endgroup$
– Andreas
Jan 14 at 16:54
$begingroup$
Is this an output of a program?
$endgroup$
– kelalaka
Jan 14 at 17:01
$begingroup$
@kelalaka The initial factorisation was, yes. I LaTeXed it manually.
$endgroup$
– user3482749
Jan 14 at 17:01
$begingroup$
+1 for the latex
$endgroup$
– kelalaka
Jan 14 at 17:02
$begingroup$
Could you please share your solution?
$endgroup$
– amirali
Jan 14 at 18:06
$begingroup$
You have warned OP above, so the shock was mitigated.....
$endgroup$
– Andreas
Jan 14 at 16:54
$begingroup$
You have warned OP above, so the shock was mitigated.....
$endgroup$
– Andreas
Jan 14 at 16:54
$begingroup$
Is this an output of a program?
$endgroup$
– kelalaka
Jan 14 at 17:01
$begingroup$
Is this an output of a program?
$endgroup$
– kelalaka
Jan 14 at 17:01
$begingroup$
@kelalaka The initial factorisation was, yes. I LaTeXed it manually.
$endgroup$
– user3482749
Jan 14 at 17:01
$begingroup$
@kelalaka The initial factorisation was, yes. I LaTeXed it manually.
$endgroup$
– user3482749
Jan 14 at 17:01
$begingroup$
+1 for the latex
$endgroup$
– kelalaka
Jan 14 at 17:02
$begingroup$
+1 for the latex
$endgroup$
– kelalaka
Jan 14 at 17:02
$begingroup$
Could you please share your solution?
$endgroup$
– amirali
Jan 14 at 18:06
$begingroup$
Could you please share your solution?
$endgroup$
– amirali
Jan 14 at 18:06
|
show 1 more comment
$begingroup$
How exactly do you solve an expression?
$endgroup$
– KM101
Jan 14 at 16:05
$begingroup$
What do you mean by "solving"? That's just an expression.
$endgroup$
– Ramiro Scorolli
Jan 14 at 16:06
$begingroup$
@KM101 I mean simplify it.
$endgroup$
– amirali
Jan 14 at 16:06
2
$begingroup$
"Simplifying" is vague and subjective (although not completely arbitrary; $2$ is clearly simpler than $frac{105+95}{2cdot 50}$). Do you mean "factor"?
$endgroup$
– Arthur
Jan 14 at 16:08
2
$begingroup$
@amirali Yes, that's factoring. This one is awful, though. Like, really, unimaginably awful. Are you sure you want it factored?
$endgroup$
– user3482749
Jan 14 at 16:14