Weltideepq

2021-11-06

For polynomial arithmetic with coefficients in ${Z}_{11}$, perform the following calculations.
a. $\left({x}^{2}+2x+9\right)\left({x}^{3}+11{x}^{2}+x+7\right)$
b.$\left(8{x}^{2}+3x+2\right)\left(5{x}^{2}+6\right)$

### Answer & Explanation

Todd Williams

a.
$\left({x}^{2}+2x+9\right)\left({x}^{3}+11{x}^{2}+x+7\right)$
$={x}^{5}+11{x}^{4}+{x}^{3}+7{x}^{2}+2{x}^{4}+22{x}^{3}+2{x}^{2}+14x+9{x}^{3}+99{x}^{2}+9x+63$
$={x}^{5}+13{x}^{4}+32{x}^{3}+108{x}^{2}+23x+63$
In coefficients ${Z}_{11}$
$={x}^{5}+2{x}^{4}+10{x}^{3}+9{x}^{2}+x+8$
$\left[\text{oti:-1\%11=1, 13\%11=2, 32\%11=10, 108\%11=9, 23\%11=1, 63\%11=8\right}oti:-1\mathrm{%}11=1,13\mathrm{%}11=2,32\mathrm{%}11=10,108\mathrm{%}11=9,23\mathrm{%}11=1,63\mathrm{%}11=8\right]$
b.
$\left(8{x}^{2}+3x+2\right)\left(5{x}^{2}+6\right)$
$=40{x}^{4}+58{x}^{2}+15{x}^{3}+18x+12$
In${Z}_{11}$ coefficients,
$=7{x}^{4}+4{x}^{3}+3{x}^{2}+7x+1$
$\left[\text{oti:-40\%11=7, 58\%11=3, 15\%11=4, 18\%11=7, 12\%11=1\right}oti:-40\mathrm{%}11=7,58\mathrm{%}11=3,15\mathrm{%}11=4,18\mathrm{%}11=7,12\mathrm{%}11=1\right]$

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