4\Xd7(3\Xd7-5), 3\Xd7y\Xb2(2\Xd7 + Y - 1)

4×(3×-5)
3×y²(2× + y - 1)

Answer:

4x(3x−5)

=(4x)(3x+−5)

=(4x)(3x)+(4x)(−5)

=12x²−20x

3xy²(2x+y−1)

=(3xy²)(2x+y−1)

=(3xy²)(2x)+(3xy²)(y)+(3xy²)(−1)

=6x²y²+3xy³−3xy²


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