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