Replace ∗ with a monomial so that the derived expression may be represented as a square of a binomial: *−42pq+49q^2

36p²

Step-by-step explanation:

The square of a binomial can be written as (a + b)².  If you expand this formula, you get

(a + b)(a + b).  Use F.O.I.L. to multiply this out..

F - stands for 'Firsts' (the first values in each set of parenthesis)

O - stands for 'Outsides'  ( the first value in the first set of parenthesis, and the

                                           second value in the second set of parenthesis)

I - stands for 'Insides'  ( the second value in the first set of parenthesis, and

                                       the first value in the second set of parenthesis)

L - stands for 'Lasts'   ( the second value in each set of parenthesis)

This is the order you multiply them in...so we get

F -  a²     (a times a)

O - ab    

I  -  ba    (which we rewrite as ab, since order doesn't matter when multiplying)

L  -  b²

           We add them together to get   a² + ab + ab + b²

which simplifies to a² + 2ab + b²

Read this as, the square of the first term in the parenthesis (a²), plus twice the product of the terms (2ab is twice the product of a and b), plus the square of the last term in the parenthesis)

So to solve this we need to know what makes *−42pq+49q^2 factor into some form of (a + b)².

Look at the bold paragraph above and work backwards.

Take the square root of the term 49q², which is 7q.  That goes in the second value of the parenthesis, so we have

    (c - 7q)²    (there is a subtraction sign because it's -42pq)

We know that 42pq is twice the product of the two terms, so we divide by 2 to see the product of the two terms.

  42pq/2  = 42pq,

We know one term, so divide 42pq by the second term, 7q to find the first term.

  42pq/7q = -6p

So the first term in our set of parenthesis is -6p, so we have

   (6p - 7q)²  

To get the missing value, square the first term, (6p)² = 36p²

So 36p² is our missing term.

16x^2+24xy + 9y² = (4x + 3y)²