Find the product of 3a+5 and 2a^(2)+4a-2

1 answer:

6 0

The answer is 6a3+22a2+14a−10
My steps:
(3a+5)(2a^2+4a−2)=(3a+5)(2a^2+4a+−2)=(3a)(2a^2)+(3a)(4a)+(3a)(−2)+(5)(2a^2)+(5)(4a)+(5)(−2)=6a^3+12a^2−6a+10a^2+20a−10=6a^3+22a^2+14a−10

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$x^3+(3+b)x^2+(3b-2)x-6\\\downarrow\qquad\ \ \downarrow\qquad\qquad\ \ \ \ \downarrow\\x^3\ +\ \ 6x^2\ \ \ \ +\ \ \ \ \ 7x\ \ -\ 6\\\\x^3\to1=1\\x^2\to3+b=6\\x\to3b-2=7\\constans\to-6=-6\\\\3+b=6\ \wedge\ 3b-2=7\\\\3+b=6\ \ \ |-3\\b=3\\\\3b-2=\ \ \ \ |+2\\3b=9\ \ \ \ |:3\\b=3\\\\Answer:\ \boxed{b=3}\to\boxed{B.}$

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