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Mathematics

Multiply :

3 - 23\dfrac{2}{3} xy + 57\dfrac{5}{7} xy2 - 1621\dfrac{16}{21} x2y by - 21x2y2

Algebraic Expressions

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Answer

(323xy+57xy21621x2y)×(21x2y2)=(3×(21x2y2)23xy×(21x2y2)+57xy2×(21x2y2)1621x2y×(21x2y2))=(63x2y22×(21)3x1+2y1+2+5×(21)7x1+2y2+216×(21)21x2+2y1+2)=63x2y2423x3y3+1057x3y4+16×(21)21x4y3=63x2y2+14x3y315x3y4+16x4y3\Big(3 - \dfrac{2}{3}xy + \dfrac{5}{7}xy^2 - \dfrac{16}{21}x^2y\Big) \times (-21x^2y^2)\\[1em] = \Big(3 \times (-21x^2y^2) - \dfrac{2}{3}xy \times (-21x^2y^2) + \dfrac{5}{7}xy^2 \times (-21x^2y^2) - \dfrac{16}{21}x^2y \times (-21x^2y^2)\Big)\\[1em] = \Big(-63x^2y^2 - \dfrac{2 \times (-21)}{3}x^{1+2}y^{1+2} + \dfrac{5\times (-21)}{7}x^{1+2}y^{2+2} - \dfrac{16 \times (-21)}{21}x^{2+2}y^{1+2}\Big)\\[1em] = -63x^2y^2 - \dfrac{-42}{3}x^3y^3 + \dfrac{-105}{7}x^3y^4 + \dfrac{16 \times \cancel{(21)}}{\cancel{21}}x^4y^3\\[1em] = -63x^2y^2 + 14x^3y^3 - 15x^3y^4 + 16x^4y^3

Hence, (3 - 23xy+57xy21621\dfrac{2}{3} xy + \dfrac{5}{7} xy^2 - \dfrac{16}{21} x2y) x (- 21x2y2) = -63x2y2 + 14x3y3 - 15x3y4 + 16x4y3

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