Math, asked by chandrai3786, 4 months ago

Find HCF of the given equation:21py2–56py

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Answered by bhawanixe9

Answer:

Factorise the following (1 to 8) polynomials: 1. ( i) 8xy3 ... 2. (i) 21 py2 – 56py (ii) 4x3 – 6x2. Solution: (i) 21 py2 – 56py = 7py (3y – 8). (ii) 4x 3 ...

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Find HCF of the given equation:21py2–56py​

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Find HCF of the given equation:21py2–56py