Question

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

Answer 1

Answer:

$5a^2-8a$

Step-by-step explanation:

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

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

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

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

$5a^2-8a$