Simplifying a fraction of polynomials
$begingroup$
This is a really simple question. All I want to know is how it goes from that first line of the equation to the second line, what is it they're doing. I'm just missing something, can someone explain.
$$
begin{split}
H(z) &= frac{0.3249(z+1)}{(z-1)+0.3249z+0.3249}\
H(z) &= frac{0.3249(z+1)}{1.3249z-0.6751}
end{split}
$$
algebra-precalculus polynomials
edited Jan 8 at 17:16
gt6989b
35.2k22557
asked Jan 8 at 17:13
controlled_controlled_
111
$endgroup$
add a comment |
$begingroup$
This is a really simple question. All I want to know is how it goes from that first line of the equation to the second line, what is it they're doing. I'm just missing something, can someone explain.
$$
begin{split}
H(z) &= frac{0.3249(z+1)}{(z-1)+0.3249z+0.3249}\
H(z) &= frac{0.3249(z+1)}{1.3249z-0.6751}
end{split}
$$
algebra-precalculus polynomials
edited Jan 8 at 17:16
gt6989b
35.2k22557
asked Jan 8 at 17:13
controlled_controlled_
111
$endgroup$
$begingroup$
$begin{align}{rm The denominator } &= color{#c00}z!-!1 + color{#c00}{az}!+!a\[.2em] &= color{#c00}{(a!+!1)z} +a!-!1 {rm for} a = 0.3249end{align}$
$endgroup$
– Bill Dubuque
Jan 8 at 17:34
add a comment |
$begingroup$
This is a really simple question. All I want to know is how it goes from that first line of the equation to the second line, what is it they're doing. I'm just missing something, can someone explain.
$$
begin{split}
H(z) &= frac{0.3249(z+1)}{(z-1)+0.3249z+0.3249}\
H(z) &= frac{0.3249(z+1)}{1.3249z-0.6751}
end{split}
$$
algebra-precalculus polynomials
edited Jan 8 at 17:16
gt6989b
35.2k22557
asked Jan 8 at 17:13
controlled_controlled_
111
$endgroup$
This is a really simple question. All I want to know is how it goes from that first line of the equation to the second line, what is it they're doing. I'm just missing something, can someone explain.
$$
begin{split}
H(z) &= frac{0.3249(z+1)}{(z-1)+0.3249z+0.3249}\
H(z) &= frac{0.3249(z+1)}{1.3249z-0.6751}
end{split}
$$
algebra-precalculus polynomials
algebra-precalculus polynomials
edited Jan 8 at 17:16
gt6989b
35.2k22557
asked Jan 8 at 17:13
controlled_controlled_
111
edited Jan 8 at 17:16
gt6989b
35.2k22557
asked Jan 8 at 17:13
controlled_controlled_
111
edited Jan 8 at 17:16
gt6989b
35.2k22557
edited Jan 8 at 17:16
gt6989b
35.2k22557
edited Jan 8 at 17:16
gt6989b
35.2k22557
35.2k22557
asked Jan 8 at 17:13
controlled_controlled_
111
asked Jan 8 at 17:13
controlled_controlled_
111
asked Jan 8 at 17:13
controlled_controlled_
111
111
$begingroup$
$begin{align}{rm The denominator } &= color{#c00}z!-!1 + color{#c00}{az}!+!a\[.2em] &= color{#c00}{(a!+!1)z} +a!-!1 {rm for} a = 0.3249end{align}$
$endgroup$
– Bill Dubuque
Jan 8 at 17:34
add a comment |
$begingroup$
$begin{align}{rm The denominator } &= color{#c00}z!-!1 + color{#c00}{az}!+!a\[.2em] &= color{#c00}{(a!+!1)z} +a!-!1 {rm for} a = 0.3249end{align}$
$endgroup$
– Bill Dubuque
Jan 8 at 17:34
$begingroup$
$begin{align}{rm The denominator } &= color{#c00}z!-!1 + color{#c00}{az}!+!a\[.2em] &= color{#c00}{(a!+!1)z} +a!-!1 {rm for} a = 0.3249end{align}$
$endgroup$
– Bill Dubuque
Jan 8 at 17:34
$begingroup$
$begin{align}{rm The denominator } &= color{#c00}z!-!1 + color{#c00}{az}!+!a\[.2em] &= color{#c00}{(a!+!1)z} +a!-!1 {rm for} a = 0.3249end{align}$
$endgroup$
– Bill Dubuque
Jan 8 at 17:34
add a comment |
1 Answer
1
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$begingroup$
Note that the denominator can be simplified:
$$
(z-1) + 0.3249z + 0.3249 = (z + 0.3249z) + (0.3249-1) = 1.3249z - 0.6751.
$$
answered Jan 8 at 17:14
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
Note that the denominator can be simplified:
$$
(z-1) + 0.3249z + 0.3249 = (z + 0.3249z) + (0.3249-1) = 1.3249z - 0.6751.
$$
answered Jan 8 at 17:14
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
$begingroup$
Note that the denominator can be simplified:
$$
(z-1) + 0.3249z + 0.3249 = (z + 0.3249z) + (0.3249-1) = 1.3249z - 0.6751.
$$
answered Jan 8 at 17:14
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
$begingroup$
Note that the denominator can be simplified:
$$
(z-1) + 0.3249z + 0.3249 = (z + 0.3249z) + (0.3249-1) = 1.3249z - 0.6751.
$$
answered Jan 8 at 17:14
gt6989bgt6989b
35.2k22557
$endgroup$
Note that the denominator can be simplified:
$$
(z-1) + 0.3249z + 0.3249 = (z + 0.3249z) + (0.3249-1) = 1.3249z - 0.6751.
$$
answered Jan 8 at 17:14
gt6989bgt6989b
35.2k22557
answered Jan 8 at 17:14
gt6989bgt6989b
35.2k22557
answered Jan 8 at 17:14
gt6989bgt6989b
35.2k22557
answered Jan 8 at 17:14
gt6989bgt6989b
35.2k22557
35.2k22557
add a comment |
add a comment |
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$begingroup$
$begin{align}{rm The denominator } &= color{#c00}z!-!1 + color{#c00}{az}!+!a\[.2em] &= color{#c00}{(a!+!1)z} +a!-!1 {rm for} a = 0.3249end{align}$
$endgroup$
– Bill Dubuque
Jan 8 at 17:34