Mixing
A polynomial equation using synthetic division
$begingroup$
$$frac{left(4g^4+3g^2-6g^3-g+12right)}{left(4g-4right)^{-1}}$$
I was trying to solve this polynomial equation, only to be met by frustration.
At first I applied the negative exponent to the second polynomial, and got this;
$(frac{1}{g}-1)$
But then I realized g is still not 0, which makes it impossible to use the synthetic division.
What do I have to do?
roots
edited Jan 14 at 9:32
José Carlos Santos
161k22127232
asked Nov 21 '16 at 21:36
whitedevilwhitedevil
1123
$endgroup$
|
show 7 more comments
$begingroup$
$$frac{left(4g^4+3g^2-6g^3-g+12right)}{left(4g-4right)^{-1}}$$
I was trying to solve this polynomial equation, only to be met by frustration.
At first I applied the negative exponent to the second polynomial, and got this;
$(frac{1}{g}-1)$
But then I realized g is still not 0, which makes it impossible to use the synthetic division.
What do I have to do?
roots
edited Jan 14 at 9:32
José Carlos Santos
161k22127232
asked Nov 21 '16 at 21:36
whitedevilwhitedevil
1123
$endgroup$
1
$begingroup$
There are two problems. I think the slash / and the negative exponent are redundant. Dividing by crud$^{-1}$ is to multiply by crud; no division necessary. Second, the exponent does not distribute over subtraction. You can't get $(1/g - 1).$
$endgroup$
– B. Goddard
Nov 21 '16 at 21:42
$begingroup$
What you have is $left(4g^4+3g^2-6g^3-g+12right) (4g-4)$. (No division), As BGoddard said. Did you intend to divide $left(4g^4+3g^2-6g^3-g+12right)$ by $left(4g-4right)$?
$endgroup$
– amWhy
Nov 21 '16 at 21:44
$begingroup$
@B.Goddard Thanks you. But I thought that the exponent was just negative, not subtraction. Can you please explain further?
$endgroup$
– whitedevil
Nov 21 '16 at 21:54
$begingroup$
All you can do is to simplify your expression (it is not, at the time of this comment, an equation.)
$endgroup$
– amWhy
Nov 21 '16 at 21:56
1
$begingroup$
@dxis In case you missed it (1) "Did you intend to divide (4g4+3g2−6g3−g+12) by (4g−4)? – amWhy"; reply " I was trying to divide (4g4+3g2−6g3−g+12) by (4g−4)^1" Me "then see my answer." OP: "at amWhy: Oh now I just got it. I don't know what I was thinking. Thanks! – whitedevil".
$endgroup$
– amWhy
Nov 21 '16 at 23:13
|
show 7 more comments
$begingroup$
$$frac{left(4g^4+3g^2-6g^3-g+12right)}{left(4g-4right)^{-1}}$$
I was trying to solve this polynomial equation, only to be met by frustration.
At first I applied the negative exponent to the second polynomial, and got this;
$(frac{1}{g}-1)$
But then I realized g is still not 0, which makes it impossible to use the synthetic division.
What do I have to do?
roots
edited Jan 14 at 9:32
José Carlos Santos
161k22127232
asked Nov 21 '16 at 21:36
whitedevilwhitedevil
1123
$endgroup$
$$frac{left(4g^4+3g^2-6g^3-g+12right)}{left(4g-4right)^{-1}}$$
I was trying to solve this polynomial equation, only to be met by frustration.
At first I applied the negative exponent to the second polynomial, and got this;
$(frac{1}{g}-1)$
But then I realized g is still not 0, which makes it impossible to use the synthetic division.
What do I have to do?
roots
roots
edited Jan 14 at 9:32
José Carlos Santos
161k22127232
asked Nov 21 '16 at 21:36
whitedevilwhitedevil
1123
edited Jan 14 at 9:32
José Carlos Santos
161k22127232
asked Nov 21 '16 at 21:36
whitedevilwhitedevil
1123
edited Jan 14 at 9:32
José Carlos Santos
161k22127232
edited Jan 14 at 9:32
José Carlos Santos
161k22127232
edited Jan 14 at 9:32
José Carlos Santos
161k22127232
161k22127232
asked Nov 21 '16 at 21:36
whitedevilwhitedevil
1123
asked Nov 21 '16 at 21:36
whitedevilwhitedevil
1123
asked Nov 21 '16 at 21:36
whitedevilwhitedevil
1123
1123
1
$begingroup$
There are two problems. I think the slash / and the negative exponent are redundant. Dividing by crud$^{-1}$ is to multiply by crud; no division necessary. Second, the exponent does not distribute over subtraction. You can't get $(1/g - 1).$
$endgroup$
– B. Goddard
Nov 21 '16 at 21:42
$begingroup$
What you have is $left(4g^4+3g^2-6g^3-g+12right) (4g-4)$. (No division), As BGoddard said. Did you intend to divide $left(4g^4+3g^2-6g^3-g+12right)$ by $left(4g-4right)$?
$endgroup$
– amWhy
Nov 21 '16 at 21:44
$begingroup$
@B.Goddard Thanks you. But I thought that the exponent was just negative, not subtraction. Can you please explain further?
$endgroup$
– whitedevil
Nov 21 '16 at 21:54
$begingroup$
All you can do is to simplify your expression (it is not, at the time of this comment, an equation.)
$endgroup$
– amWhy
Nov 21 '16 at 21:56
1
$begingroup$
@dxis In case you missed it (1) "Did you intend to divide (4g4+3g2−6g3−g+12) by (4g−4)? – amWhy"; reply " I was trying to divide (4g4+3g2−6g3−g+12) by (4g−4)^1" Me "then see my answer." OP: "at amWhy: Oh now I just got it. I don't know what I was thinking. Thanks! – whitedevil".
$endgroup$
– amWhy
Nov 21 '16 at 23:13
|
show 7 more comments
1
$begingroup$
There are two problems. I think the slash / and the negative exponent are redundant. Dividing by crud$^{-1}$ is to multiply by crud; no division necessary. Second, the exponent does not distribute over subtraction. You can't get $(1/g - 1).$
$endgroup$
– B. Goddard
Nov 21 '16 at 21:42
$begingroup$
What you have is $left(4g^4+3g^2-6g^3-g+12right) (4g-4)$. (No division), As BGoddard said. Did you intend to divide $left(4g^4+3g^2-6g^3-g+12right)$ by $left(4g-4right)$?
$endgroup$
– amWhy
Nov 21 '16 at 21:44
$begingroup$
@B.Goddard Thanks you. But I thought that the exponent was just negative, not subtraction. Can you please explain further?
$endgroup$
– whitedevil
Nov 21 '16 at 21:54
$begingroup$
All you can do is to simplify your expression (it is not, at the time of this comment, an equation.)
$endgroup$
– amWhy
Nov 21 '16 at 21:56
1
$begingroup$
@dxis In case you missed it (1) "Did you intend to divide (4g4+3g2−6g3−g+12) by (4g−4)? – amWhy"; reply " I was trying to divide (4g4+3g2−6g3−g+12) by (4g−4)^1" Me "then see my answer." OP: "at amWhy: Oh now I just got it. I don't know what I was thinking. Thanks! – whitedevil".
$endgroup$
– amWhy
Nov 21 '16 at 23:13
1
1
$begingroup$
There are two problems. I think the slash / and the negative exponent are redundant. Dividing by crud$^{-1}$ is to multiply by crud; no division necessary. Second, the exponent does not distribute over subtraction. You can't get $(1/g - 1).$
$endgroup$
– B. Goddard
Nov 21 '16 at 21:42
$begingroup$
There are two problems. I think the slash / and the negative exponent are redundant. Dividing by crud$^{-1}$ is to multiply by crud; no division necessary. Second, the exponent does not distribute over subtraction. You can't get $(1/g - 1).$
$endgroup$
– B. Goddard
Nov 21 '16 at 21:42
$begingroup$
What you have is $left(4g^4+3g^2-6g^3-g+12right) (4g-4)$. (No division), As BGoddard said. Did you intend to divide $left(4g^4+3g^2-6g^3-g+12right)$ by $left(4g-4right)$?
$endgroup$
– amWhy
Nov 21 '16 at 21:44
$begingroup$
What you have is $left(4g^4+3g^2-6g^3-g+12right) (4g-4)$. (No division), As BGoddard said. Did you intend to divide $left(4g^4+3g^2-6g^3-g+12right)$ by $left(4g-4right)$?
$endgroup$
– amWhy
Nov 21 '16 at 21:44
$begingroup$
@B.Goddard Thanks you. But I thought that the exponent was just negative, not subtraction. Can you please explain further?
$endgroup$
– whitedevil
Nov 21 '16 at 21:54
$begingroup$
@B.Goddard Thanks you. But I thought that the exponent was just negative, not subtraction. Can you please explain further?
$endgroup$
– whitedevil
Nov 21 '16 at 21:54
$begingroup$
All you can do is to simplify your expression (it is not, at the time of this comment, an equation.)
$endgroup$
– amWhy
Nov 21 '16 at 21:56
$begingroup$
All you can do is to simplify your expression (it is not, at the time of this comment, an equation.)
$endgroup$
– amWhy
Nov 21 '16 at 21:56
1
1
$begingroup$
@dxis In case you missed it (1) "Did you intend to divide (4g4+3g2−6g3−g+12) by (4g−4)? – amWhy"; reply " I was trying to divide (4g4+3g2−6g3−g+12) by (4g−4)^1" Me "then see my answer." OP: "at amWhy: Oh now I just got it. I don't know what I was thinking. Thanks! – whitedevil".
$endgroup$
– amWhy
Nov 21 '16 at 23:13
$begingroup$
@dxis In case you missed it (1) "Did you intend to divide (4g4+3g2−6g3−g+12) by (4g−4)? – amWhy"; reply " I was trying to divide (4g4+3g2−6g3−g+12) by (4g−4)^1" Me "then see my answer." OP: "at amWhy: Oh now I just got it. I don't know what I was thinking. Thanks! – whitedevil".
$endgroup$
– amWhy
Nov 21 '16 at 23:13
|
show 7 more comments
1 Answer
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$begingroup$
$(4g-4)^{-1} = dfrac 1{4(g-1)}$. You've got your numerator, being divided by $dfrac 1{4g-4}$, which, as I said earlier is equal to $$frac{left(4g^4+3g^2-6g^3-g+12right)}{dfrac 1{4g-4}} = (4g^4+3g^2-6g^3-g+12)(4g-4)tag{*}$$
(E.g. $dfrac x{y^{-1}} = dfrac{x}{frac 1{y}}$. Then multiply numerator and denominator by $y$ to get $dfrac{xy}{y/y} = xy$)
There is no division required. See if you can factor $$(4g^4+3g^2-6g^3-g+12)$$
Unfortunately, $4g^4+3g^2 -6g^3-g+12$ is irreducible, so $(*)$ cannot be factored any further (in the real numbers).
answered Nov 21 '16 at 22:04
amWhyamWhy
1
$endgroup$
add a comment |
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$begingroup$
$(4g-4)^{-1} = dfrac 1{4(g-1)}$. You've got your numerator, being divided by $dfrac 1{4g-4}$, which, as I said earlier is equal to $$frac{left(4g^4+3g^2-6g^3-g+12right)}{dfrac 1{4g-4}} = (4g^4+3g^2-6g^3-g+12)(4g-4)tag{*}$$
(E.g. $dfrac x{y^{-1}} = dfrac{x}{frac 1{y}}$. Then multiply numerator and denominator by $y$ to get $dfrac{xy}{y/y} = xy$)
There is no division required. See if you can factor $$(4g^4+3g^2-6g^3-g+12)$$
Unfortunately, $4g^4+3g^2 -6g^3-g+12$ is irreducible, so $(*)$ cannot be factored any further (in the real numbers).
answered Nov 21 '16 at 22:04
amWhyamWhy
1
$endgroup$
add a comment |
$begingroup$
$(4g-4)^{-1} = dfrac 1{4(g-1)}$. You've got your numerator, being divided by $dfrac 1{4g-4}$, which, as I said earlier is equal to $$frac{left(4g^4+3g^2-6g^3-g+12right)}{dfrac 1{4g-4}} = (4g^4+3g^2-6g^3-g+12)(4g-4)tag{*}$$
(E.g. $dfrac x{y^{-1}} = dfrac{x}{frac 1{y}}$. Then multiply numerator and denominator by $y$ to get $dfrac{xy}{y/y} = xy$)
There is no division required. See if you can factor $$(4g^4+3g^2-6g^3-g+12)$$
Unfortunately, $4g^4+3g^2 -6g^3-g+12$ is irreducible, so $(*)$ cannot be factored any further (in the real numbers).
answered Nov 21 '16 at 22:04
amWhyamWhy
1
$endgroup$
add a comment |
$begingroup$
$(4g-4)^{-1} = dfrac 1{4(g-1)}$. You've got your numerator, being divided by $dfrac 1{4g-4}$, which, as I said earlier is equal to $$frac{left(4g^4+3g^2-6g^3-g+12right)}{dfrac 1{4g-4}} = (4g^4+3g^2-6g^3-g+12)(4g-4)tag{*}$$
(E.g. $dfrac x{y^{-1}} = dfrac{x}{frac 1{y}}$. Then multiply numerator and denominator by $y$ to get $dfrac{xy}{y/y} = xy$)
There is no division required. See if you can factor $$(4g^4+3g^2-6g^3-g+12)$$
Unfortunately, $4g^4+3g^2 -6g^3-g+12$ is irreducible, so $(*)$ cannot be factored any further (in the real numbers).
answered Nov 21 '16 at 22:04
amWhyamWhy
1
$endgroup$
$(4g-4)^{-1} = dfrac 1{4(g-1)}$. You've got your numerator, being divided by $dfrac 1{4g-4}$, which, as I said earlier is equal to $$frac{left(4g^4+3g^2-6g^3-g+12right)}{dfrac 1{4g-4}} = (4g^4+3g^2-6g^3-g+12)(4g-4)tag{*}$$
(E.g. $dfrac x{y^{-1}} = dfrac{x}{frac 1{y}}$. Then multiply numerator and denominator by $y$ to get $dfrac{xy}{y/y} = xy$)
There is no division required. See if you can factor $$(4g^4+3g^2-6g^3-g+12)$$
Unfortunately, $4g^4+3g^2 -6g^3-g+12$ is irreducible, so $(*)$ cannot be factored any further (in the real numbers).
answered Nov 21 '16 at 22:04
amWhyamWhy
1
edited Nov 21 '16 at 23:00
answered Nov 21 '16 at 22:04
amWhyamWhy
1
answered Nov 21 '16 at 22:04
amWhyamWhy
1
answered Nov 21 '16 at 22:04
amWhyamWhy
1
1
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$begingroup$
There are two problems. I think the slash / and the negative exponent are redundant. Dividing by crud$^{-1}$ is to multiply by crud; no division necessary. Second, the exponent does not distribute over subtraction. You can't get $(1/g - 1).$
$endgroup$
– B. Goddard
Nov 21 '16 at 21:42
$begingroup$
What you have is $left(4g^4+3g^2-6g^3-g+12right) (4g-4)$. (No division), As BGoddard said. Did you intend to divide $left(4g^4+3g^2-6g^3-g+12right)$ by $left(4g-4right)$?
$endgroup$
– amWhy
Nov 21 '16 at 21:44
$begingroup$
@B.Goddard Thanks you. But I thought that the exponent was just negative, not subtraction. Can you please explain further?
$endgroup$
– whitedevil
Nov 21 '16 at 21:54
$begingroup$
All you can do is to simplify your expression (it is not, at the time of this comment, an equation.)
$endgroup$
– amWhy
Nov 21 '16 at 21:56
1
$begingroup$
@dxis In case you missed it (1) "Did you intend to divide (4g4+3g2−6g3−g+12) by (4g−4)? – amWhy"; reply " I was trying to divide (4g4+3g2−6g3−g+12) by (4g−4)^1" Me "then see my answer." OP: "at amWhy: Oh now I just got it. I don't know what I was thinking. Thanks! – whitedevil".
$endgroup$
– amWhy
Nov 21 '16 at 23:13