SOLUTION: (a+2)(3a2-a+5) in (3a2-a+5)(3a is to the second power) Please work this problem out so i can see what i am doing work. Thanks

Click here to see ALL problems on Polynomials-and-rational-expressions

Question 425972: (a+2)(3a2-a+5)
in (3a2-a+5)(3a is to the second power)
Please work this problem out so i can see what i am doing work. Thanks

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
3a squared would be shown as 3a^2

the ^ is exponent symbol.

your problem is (a+2) * (3a^2 - a + 5)

in polynomial multiplication, every element in the each factor is multiplied by every element in each other factor.

your problem is equivalent to:

(a * (3a^2 - a + 5)) + (2 * (3a^2 - a + 5))

this gets you:

3a^3 - a^2 + 5a + 6a^2 - 2a + 10

combine like terms to get:

3a^3 + 5a^2 + 3a + 10

this works with any number of factors.

you can use it in place of foil.

(a + 2) * (a - 2) = a * (a - 2) + 2 * (a - 2) = a^2 - 2a + 2a + 4 = a^2 + 4

the general form of this would be:

(a + b) * (c + d) = a * (c + d) + b * (c + d)

in your problem, the general form would be:

(a + b) * (c + d + e) = a * (c + d + e) + b * (c + d + e)