Question 1040299
multiply each term in the first by each term in the second.  There are four terms.
For the first, it would be 2a*2+2a(12b)-3b(a)-3b(12b)=2a^2-3ab+24ab-36b^2
That is 2a^2+21ab-36b^2. Remember, ab and ba are the same.
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The second is 3m^2-11mn-42n^2.
(-m+6n)(-3m-7n)=(-m)(-3m)+(-m)(-7n)+(6n)(-3m)+(6n)(-7n)=3m^2+7mn-18mn-42n^2.  There are four terms from a 2 x 2 binomial.  Each term in the first is multiplied by each term in the second.  There will be a m^2 and n^2 and two mn terms.  The others are done the same way.
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The third is -8p^2-38pq-9q^2
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The fourth is done by putting the -3 in last.  (5h^2-60h+55), so it becomes -15h^2+180h-165
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The fifth is (x+5)(x-2), because the parentheses don't change anything when there is addition.  We can add in any order we choose.
That is foiled out to x^2+3x-10.
The long way is (x+3)^2-5(x+3)+2(x=3)-10
That is (x+3)^2-3(x+3)-10
That is x^2+6x+9-3x-9-10
That is x^2+3x-10