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प्रश्न
Find the following product and verify the result for x = − 1, y = − 2:
(3x − 5y) (x + y)
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उत्तर
To multiply, we will use distributive law as follows:
\[\left( 3x - 5y \right)\left( x + y \right)\]
\[ = 3x\left( x + y \right) - 5y\left( x + y \right)\]
\[ = 3 x^2 + 3xy - 5xy - 5 y^2 \]
\[ = 3 x^2 - 2xy - 5 y^2\]
\[\therefore\] \[\left( 3x - 5y \right)\left( x + y \right) = 3 x^2 - 2xy - 5 y^2\].
Now, we put x = \[-\] 1 and y = \[-\] 2 on both sides to verify the result.
\[\text { LHS } = \left( 3x - 5y \right)\left( x + y \right)\]
\[ = \left\{ 3\left( - 1 \right) - 5\left( - 2 \right) \right\}\left\{ - 1 + \left( - 2 \right) \right\}\]
\[ = \left( - 3 + 10 \right)\left( - 3 \right)\]
\[ = \left( 7 \right)\left( - 3 \right)\]
\[ = - 21\]
\[\text { RHS } = 3 x^2 - 2xy - 5 y^2 \]
\[ = 3 \left( - 1 \right)^2 - 2\left( - 1 \right)\left( - 2 \right) - 5 \left( - 2 \right)^2 \]
\[ = 3 \times 1 - 4 - 5 \times 4\]
\[ = 3 - 4 - 20\]
\[ = - 21\]
Because LHS is equal to RHS, the result is verified.
Thus, the answer is \[3 x^2 - 2xy - 5 y^2\].
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संबंधित प्रश्न
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Express each of the following product as a monomials and verify the result in each case for x = 1:
(4x2) × (−3x) × \[\left( \frac{4}{5} x^3 \right)\]
Find the value of (5x6) × (−1.5x2y3) × (−12xy2) when x = 1, y = 0.5.
Evaluate each of the following when x = 2, y = −1.
\[(2xy) \times \left( \frac{x^2 y}{4} \right) \times \left( x^2 \right) \times \left( y^2 \right)\]
Simplify: x(x + 4) + 3x(2x2 − 1) + 4x2 + 4
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(a − 1) by (0.1a2 + 3)
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(3x2y − 5xy2) by \[\left( \frac{1}{5} x^2 + \frac{1}{3} y^2 \right)\].
Find the following product and verify the result for x = − 1, y = − 2: \[\left( \frac{1}{3}x - \frac{y^2}{5} \right)\left( \frac{1}{3}x + \frac{y^2}{5} \right)\]
Simplify:
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