How do you find the product $2 a b (7 a^{4} b^{2} + a^{5} b - 2 a)$ $2ab(7a^4b^2+a^5b-2a)$ ?

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Answer 1 · 2017-08-22T07:09:06

See explanation.

To find the product you have to multiply each term in the brackets by $2 a b$ $2ab$ :

$2 a b \times (7 a^{4} b^{2} + a^{5} b - 2 a) =$ $2abxx(7a^4b^2+a^5b-2a)=$

$= 2 a b \times 7 a^{4} b + 2 a b \times a^{5} b + 2 a b \times (- 2 a) =$ $=2abxx7a^4b+2abxxa^5b+2abxx(-2a)=$

$= 14 a^{5} b^{2} + 2 a^{6} b^{2} + (- 4 a^{2} b) =$ $=14a^5b^2+2a^6b^2+(-4a^2b)=$

$= 14 a^{5} b^{2} + 2 a^{6} b^{2} - 4 a^{2} b$ $=14a^5b^2+2a^6b^2-4a^2b$

How do you find the product 2ab(7a4b2+a5b−2a)2ab(7a^4b^2+a^5b-2a)?