Question

Rewrite the expression -2a(a+b-5)+3(-5a+2b)+b(6a+b-8)

Answer 1

We distribute the constant to each of the terms in the parenthesis.  in other words, we rewrite the expression as:
(-2a)(a) + (-2a)b - (-2a)(5) + 3(-5a) + 3(2b) + (b)(6a) + (b)(b) - b(8)
-2$a^{2}$ - 2ab + 10a - 15a + 6b + 6ab + $b ^{2}$ - 8b
now we can combine like terms
-2$a^{2}$ + 4ab + 5a -2b + $b ^{2}$ and that's your answer.

whether you want to read the next part is optional but i want to show you why distributive property works.  consider the expression ab + ac.  multiplication is repeated addition so this expression states that we first add a, b times, and then add a, c times.  that means in total we added a, (b+c) times so the expression can be rewritten as a(b+c)
since ab + ac = a(b+c) we can reverse the equality and get a(b+c) = ab + ac.  this can be extended to any number of terms (in the case of this problem, you had to distribute a number to 3 terms).  

now consider ab - ac.  this expression states that first add a, b times and then take away c of the a's we added.  therefore we added a, (b - c) times.  that means a(b-c) = ab - ac.  the point of showing you the subtraction example seperately is to remind you to keep the signs when you distribute and why you need to do that.  when most students see a(b-c) they forget the negative sign in front of the c and write ab + ac, but that's wrong.

let me know if you have any questions!

Answer 2

First separately:
$-2a(a+b-5) = -2a^{2}-2ab+10a$
$3(-5a+2b) = -15a + 6b$
$b(6a+b+8) = 6ab+b^{2}+8b$

Then you add then all together and get:
$-2a^{2}-2ab+10a-15a + 6b+ 6ab+b^{2}+8b =$
$-2a^{2}-5a+4ab+14b+b^{2}$