Substract 9x^2y^2-3x^2-5 from the sum of -6x^2y^2-x^2-1 and 5x^2y^2+2x^2-1

Amari Flowers 2021-09-06 Answered
Substract \(\displaystyle{9}{x}^{{2}}{y}^{{2}}-{3}{x}^{{2}}-{5}\) from the sum of \(\displaystyle-{6}{x}^{{2}}{y}^{{2}}-{x}^{{2}}-{1}\) and \(\displaystyle{5}{x}^{{2}}{y}^{{2}}+{2}{x}^{{2}}-{1}\)

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Expert Answer

diskusje5
Answered 2021-09-07 Author has 12405 answers
First, find the sum of the polynomials \(\displaystyle-{6}{x}^{{2}}{y}^{{2}}-{x}^{{2}}-{1}\) and \(\displaystyle{5}{x}^{{2}}{y}^{{2}}+{2}{x}^{{2}}-{1}\)
Then,
\(\displaystyle{\left(-{6}{x}^{{2}}{y}^{{2}}-{x}^{{2}}-{1}\right)}+{\left({5}{x}^{{2}}{y}^{{2}}+{2}{x}^{{2}}-{1}\right)}=-{6}{x}^{{2}}{y}^{{2}}+{5}{x}^{{2}}{y}^{{2}}-{x}^{{2}}+{2}{x}^{{2}}-{1}-{1}\)
\(\displaystyle={\left(-{6}+{5}\right)}{x}^{{2}}{y}^{{2}}+{\left(-{1}+{2}\right)}{x}^{{2}}-{2}\)
\(\displaystyle=-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}\)
Therefore, the sum of the polynomials \(\displaystyle-{6}{x}^{{2}}{y}^{{2}}-{x}^{{2}}-{1}\) and \(\displaystyle{5}{x}^{{2}}{y}^{{2}}+{2}{x}^{{2}}-{1}\) is \(\displaystyle-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}\)
Now, subtract \(\displaystyle{9}{x}^{{2}}{y}^{{2}}-{3}{x}^{{2}}-{5}\) from \(\displaystyle-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}\)
So,
\(\displaystyle{\left(-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}\right)}-{\left({9}{x}^{{2}}{y}^{{2}}-{3}{x}^{{2}}-{5}\right)}=-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}-{9}{x}^{{2}}{y}^{{2}}+{3}{x}^{{2}}+{5}\)
\(\displaystyle=-{x}^{{2}}{y}^{{2}}-{9}{x}^{{2}}{y}^{{2}}+{x}^{{2}}+{3}{x}^{{2}}-{2}+{5}\)
\(\displaystyle={\left(-{1}-{9}\right)}{x}^{{2}}{y}^{{2}}+{\left({1}+{3}\right)}{x}^{{2}}+{5}-{2}\)
\(\displaystyle=-{10}{x}^{{2}}{y}^{{2}}+{4}{x}^{{2}}+{3}\)
Hence, the required polynomial is \(\displaystyle-{10}{x}^{{2}}{y}^{{2}}+{4}{x}^{{2}}+{3}\)
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