Mixing
How is this step acheived? $h(x)= 27x^6+26x^3-1 $
$begingroup$
Find all real roots of $h(x)$.
Solution
I have solved the question by letting $u = x^3$ and then using the quadratic formula to solve $27u^2+26u-1 = 0$.
However I don't have a clue as to how they've achieved $27x^3 -1x^3+1=0$ as one of the steps. I have highlighted the section in the image linked Solution.
functions roots quadratics
edited Jan 17 at 15:17
N. F. Taussig
44.4k93357
asked Jan 17 at 0:15
rootroot
133
$endgroup$
add a comment |
$begingroup$
Find all real roots of $h(x)$.
Solution
I have solved the question by letting $u = x^3$ and then using the quadratic formula to solve $27u^2+26u-1 = 0$.
However I don't have a clue as to how they've achieved $27x^3 -1x^3+1=0$ as one of the steps. I have highlighted the section in the image linked Solution.
functions roots quadratics
edited Jan 17 at 15:17
N. F. Taussig
44.4k93357
asked Jan 17 at 0:15
rootroot
133
$endgroup$
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 0:22
1
$begingroup$
$(27u-1)(u+1) = 27 u^2 +26 u - 1$
$endgroup$
– Will Jagy
Jan 17 at 0:25
add a comment |
$begingroup$
Find all real roots of $h(x)$.
Solution
I have solved the question by letting $u = x^3$ and then using the quadratic formula to solve $27u^2+26u-1 = 0$.
However I don't have a clue as to how they've achieved $27x^3 -1x^3+1=0$ as one of the steps. I have highlighted the section in the image linked Solution.
functions roots quadratics
edited Jan 17 at 15:17
N. F. Taussig
44.4k93357
asked Jan 17 at 0:15
rootroot
133
$endgroup$
Find all real roots of $h(x)$.
Solution
I have solved the question by letting $u = x^3$ and then using the quadratic formula to solve $27u^2+26u-1 = 0$.
However I don't have a clue as to how they've achieved $27x^3 -1x^3+1=0$ as one of the steps. I have highlighted the section in the image linked Solution.
functions roots quadratics
functions roots quadratics
edited Jan 17 at 15:17
N. F. Taussig
44.4k93357
asked Jan 17 at 0:15
rootroot
133
edited Jan 17 at 15:17
N. F. Taussig
44.4k93357
asked Jan 17 at 0:15
rootroot
133
edited Jan 17 at 15:17
N. F. Taussig
44.4k93357
edited Jan 17 at 15:17
N. F. Taussig
44.4k93357
edited Jan 17 at 15:17
N. F. Taussig
44.4k93357
44.4k93357
asked Jan 17 at 0:15
rootroot
133
asked Jan 17 at 0:15
rootroot
133
asked Jan 17 at 0:15
rootroot
133
133
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 0:22
1
$begingroup$
$(27u-1)(u+1) = 27 u^2 +26 u - 1$
$endgroup$
– Will Jagy
Jan 17 at 0:25
add a comment |
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 0:22
1
$begingroup$
$(27u-1)(u+1) = 27 u^2 +26 u - 1$
$endgroup$
– Will Jagy
Jan 17 at 0:25
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 0:22
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 0:22
1
1
$begingroup$
$(27u-1)(u+1) = 27 u^2 +26 u - 1$
$endgroup$
– Will Jagy
Jan 17 at 0:25
$begingroup$
$(27u-1)(u+1) = 27 u^2 +26 u - 1$
$endgroup$
– Will Jagy
Jan 17 at 0:25
add a comment |
1 Answer
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$begingroup$
It's a typo, it should read
$$ begin{align}&27(x^3)^2 + 26 x^3 - 1\ = &(27x^3-1)(x^3+1) end{align}$$
Remark $ $ You could have inferred this from what follows, i.e. they reduce it to
$,x^3 = 1/27,$ and $,x^3 = -1,$ implying that it has factors $,27x^3-1,$ and $,x^3+1$
Worth mention is that we can use the AC method to factor such polynomials, i.e.
$$begin{eqnarray} rm: a:f(x):! ,=,:! a:(a:x^2 + b:x + c) &,=,&!!rm: color{#c00}{X^2} + b:X + !!!!!!!!!!!!!!!!!!!!!{overbrace{ac,}^{rmqquad {bflarge AC-method}}}!!!!!!!!!!!!!!!! X = a:x \
end{eqnarray}$$ In our case $$ {begin{eqnarray}
f , &,=,& , 27 x^2+ 26 x, - 1\
Rightarrow, 27f, &,=,&!, (27x)^2! +26(27x)-1\
&,=,& color{#c00}{X^2}+, 26 X, - 1,, X, =, 27x\
&,=,& (X+27) (X-1)\
&,=,& (27x+27),(27x-1)\
Rightarrow f:=: color{#0a0}{27^{-1}},(27f), &,=,& (x+ 1) (27x-1)\
end{eqnarray}}$$
answered Jan 17 at 0:24
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
add a comment |
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1 Answer
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$begingroup$
It's a typo, it should read
$$ begin{align}&27(x^3)^2 + 26 x^3 - 1\ = &(27x^3-1)(x^3+1) end{align}$$
Remark $ $ You could have inferred this from what follows, i.e. they reduce it to
$,x^3 = 1/27,$ and $,x^3 = -1,$ implying that it has factors $,27x^3-1,$ and $,x^3+1$
Worth mention is that we can use the AC method to factor such polynomials, i.e.
$$begin{eqnarray} rm: a:f(x):! ,=,:! a:(a:x^2 + b:x + c) &,=,&!!rm: color{#c00}{X^2} + b:X + !!!!!!!!!!!!!!!!!!!!!{overbrace{ac,}^{rmqquad {bflarge AC-method}}}!!!!!!!!!!!!!!!! X = a:x \
end{eqnarray}$$ In our case $$ {begin{eqnarray}
f , &,=,& , 27 x^2+ 26 x, - 1\
Rightarrow, 27f, &,=,&!, (27x)^2! +26(27x)-1\
&,=,& color{#c00}{X^2}+, 26 X, - 1,, X, =, 27x\
&,=,& (X+27) (X-1)\
&,=,& (27x+27),(27x-1)\
Rightarrow f:=: color{#0a0}{27^{-1}},(27f), &,=,& (x+ 1) (27x-1)\
end{eqnarray}}$$
answered Jan 17 at 0:24
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
add a comment |
$begingroup$
It's a typo, it should read
$$ begin{align}&27(x^3)^2 + 26 x^3 - 1\ = &(27x^3-1)(x^3+1) end{align}$$
Remark $ $ You could have inferred this from what follows, i.e. they reduce it to
$,x^3 = 1/27,$ and $,x^3 = -1,$ implying that it has factors $,27x^3-1,$ and $,x^3+1$
Worth mention is that we can use the AC method to factor such polynomials, i.e.
$$begin{eqnarray} rm: a:f(x):! ,=,:! a:(a:x^2 + b:x + c) &,=,&!!rm: color{#c00}{X^2} + b:X + !!!!!!!!!!!!!!!!!!!!!{overbrace{ac,}^{rmqquad {bflarge AC-method}}}!!!!!!!!!!!!!!!! X = a:x \
end{eqnarray}$$ In our case $$ {begin{eqnarray}
f , &,=,& , 27 x^2+ 26 x, - 1\
Rightarrow, 27f, &,=,&!, (27x)^2! +26(27x)-1\
&,=,& color{#c00}{X^2}+, 26 X, - 1,, X, =, 27x\
&,=,& (X+27) (X-1)\
&,=,& (27x+27),(27x-1)\
Rightarrow f:=: color{#0a0}{27^{-1}},(27f), &,=,& (x+ 1) (27x-1)\
end{eqnarray}}$$
answered Jan 17 at 0:24
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
add a comment |
$begingroup$
It's a typo, it should read
$$ begin{align}&27(x^3)^2 + 26 x^3 - 1\ = &(27x^3-1)(x^3+1) end{align}$$
Remark $ $ You could have inferred this from what follows, i.e. they reduce it to
$,x^3 = 1/27,$ and $,x^3 = -1,$ implying that it has factors $,27x^3-1,$ and $,x^3+1$
Worth mention is that we can use the AC method to factor such polynomials, i.e.
$$begin{eqnarray} rm: a:f(x):! ,=,:! a:(a:x^2 + b:x + c) &,=,&!!rm: color{#c00}{X^2} + b:X + !!!!!!!!!!!!!!!!!!!!!{overbrace{ac,}^{rmqquad {bflarge AC-method}}}!!!!!!!!!!!!!!!! X = a:x \
end{eqnarray}$$ In our case $$ {begin{eqnarray}
f , &,=,& , 27 x^2+ 26 x, - 1\
Rightarrow, 27f, &,=,&!, (27x)^2! +26(27x)-1\
&,=,& color{#c00}{X^2}+, 26 X, - 1,, X, =, 27x\
&,=,& (X+27) (X-1)\
&,=,& (27x+27),(27x-1)\
Rightarrow f:=: color{#0a0}{27^{-1}},(27f), &,=,& (x+ 1) (27x-1)\
end{eqnarray}}$$
answered Jan 17 at 0:24
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
It's a typo, it should read
$$ begin{align}&27(x^3)^2 + 26 x^3 - 1\ = &(27x^3-1)(x^3+1) end{align}$$
Remark $ $ You could have inferred this from what follows, i.e. they reduce it to
$,x^3 = 1/27,$ and $,x^3 = -1,$ implying that it has factors $,27x^3-1,$ and $,x^3+1$
Worth mention is that we can use the AC method to factor such polynomials, i.e.
$$begin{eqnarray} rm: a:f(x):! ,=,:! a:(a:x^2 + b:x + c) &,=,&!!rm: color{#c00}{X^2} + b:X + !!!!!!!!!!!!!!!!!!!!!{overbrace{ac,}^{rmqquad {bflarge AC-method}}}!!!!!!!!!!!!!!!! X = a:x \
end{eqnarray}$$ In our case $$ {begin{eqnarray}
f , &,=,& , 27 x^2+ 26 x, - 1\
Rightarrow, 27f, &,=,&!, (27x)^2! +26(27x)-1\
&,=,& color{#c00}{X^2}+, 26 X, - 1,, X, =, 27x\
&,=,& (X+27) (X-1)\
&,=,& (27x+27),(27x-1)\
Rightarrow f:=: color{#0a0}{27^{-1}},(27f), &,=,& (x+ 1) (27x-1)\
end{eqnarray}}$$
answered Jan 17 at 0:24
Bill DubuqueBill Dubuque
211k29193646
edited Jan 17 at 0:42
answered Jan 17 at 0:24
Bill DubuqueBill Dubuque
211k29193646
answered Jan 17 at 0:24
Bill DubuqueBill Dubuque
211k29193646
answered Jan 17 at 0:24
Bill DubuqueBill Dubuque
211k29193646
211k29193646
add a comment |
add a comment |
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$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 0:22
1
$begingroup$
$(27u-1)(u+1) = 27 u^2 +26 u - 1$
$endgroup$
– Will Jagy
Jan 17 at 0:25