How do you multiply $3x ^ { 3} y ^ { 4} \cdot - 2x ^ { 4} y ^ { 3} \cdot - y x ^ { 2}$ ?

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Answer 1 · 2017-06-10T16:46:46

$3x^3y^4*-2x^4y^3*-yx^2=6x^9y^8$

As there are two negatives in the product of three monomials, the product would be positive.

Further on multiplication, powers of same base are added as $a^mxxa^n=a^((m+n))$

Hence $3x^3y^4*-2x^4y^3*-yx^2$

= $(3xx2xx1)x^((3+4+2))y^((4+3+1))$

= $6x^9y^8$

How do you multiply 3x ^ { 3} y ^ { 4} \cdot - 2x ^ { 4} y ^ { 3} \cdot - y x ^ { 2}3x ^ { 3} y ^ { 4} \cdot - 2x ^ { 4} y ^ { 3} \cdot - y x ^ { 2}?