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Question

# Find the following product and verify the result for x = − 1, y = − 2: (x2y − 1) (3 − 2x2y)

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Solution

## To multiply, we will use distributive law as follows: $\left({x}^{2}y-1\right)\left(3-2{x}^{2}y\right)\phantom{\rule{0ex}{0ex}}={x}^{2}y\left(3-2{x}^{2}y\right)-1×\left(3-2{x}^{2}y\right)\phantom{\rule{0ex}{0ex}}=3{x}^{2}y-2{x}^{4}{y}^{2}-3+2{x}^{2}y\phantom{\rule{0ex}{0ex}}=5{x}^{2}y-2{x}^{4}{y}^{2}-3$ $\therefore$ $\left({x}^{2}y-1\right)\left(3-2{x}^{2}y\right)=5{x}^{2}y-2{x}^{4}{y}^{2}-3$ Now, we put x = $-$1 and y = $-$2 on both sides to verify the result. $\text{LHS =}\left({x}^{2}y-1\right)\left(3-2{x}^{2}y\right)\phantom{\rule{0ex}{0ex}}=\left[{\left(-1\right)}^{2}\left(-2\right)-1\right]\left[3-2{\left(-1\right)}^{2}\left(-2\right)\right]\phantom{\rule{0ex}{0ex}}=\left[1×\left(-2\right)-1\right]\left[3-2×1×\left(-2\right)\right]\phantom{\rule{0ex}{0ex}}=\left(-2-1\right)\left(3+4\right)\phantom{\rule{0ex}{0ex}}=-3×7\phantom{\rule{0ex}{0ex}}=-21$ $\text{RHS}=5{x}^{2}y-2{x}^{4}{y}^{2}-3\phantom{\rule{0ex}{0ex}}=5{\left(-1\right)}^{2}\left(-2\right)-2{\left(-1\right)}^{4}{\left(-2\right)}^{2}-3\phantom{\rule{0ex}{0ex}}=\left[5×1×\left(-2\right)\right]-\left[2×1×4\right]-3\phantom{\rule{0ex}{0ex}}=-10-8-3\phantom{\rule{0ex}{0ex}}=-21$ Because LHS is equal to RHS, the result is verified. Thus, the answer is $5{x}^{2}y-2{x}^{4}{y}^{2}-3$.

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