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Find the sum of 6(1-a^2), -2(3-a+2a^2) and 5(-2a+3a^2)​

1 Answer

6 votes

Answer:


5a^2-8a

Explanation:


6(1-a^2)-2(3-a+2a^2)+5(-2a+3a^2)


(6)(1)+(6)(-a^2)+(-2)(3)+(-2)(-a)+(-2)(2a^2)+(5)(-2a)+(5)(3a^2)


6+-6a^2+-6+2a+-4a^2+-10a+15a^2


(-6a^2+-4a^2+15a^2)+(2a+-10a)+(6+-6)


5a^2-8a

User Shaun Ryan
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