1)Factor x⁴ + 4y⁴.

That's the sum of two squares.  But we can only factor the
difference of two squares, so we see if it is possible to
change it to the difference of two squares by adding and
subtracting a square, like this

x⁴ + ___ + 4y⁴ - ___

We ask ourselves this question:

What term would have to be placed in those blanks 
so that the first three terms would factor as the 
square of the sum of their square roots, x² and 2y²?

To find out we multiply out the square of the sum of their
square roots: (x² + 2y²)² = x⁴ + 4x²y² + 4y⁴

We see that the term that must be placed in the two blanks
would be 4x²y², which does happen to be a square.  So we 
place 4x²y² in the two blanks:

x⁴ + 4x²y² + 4y⁴ - 4x²y²

Then we factor the first three terms:

(x²+2y²)² - 4x²y²

and that is the difference of two squares,
and factors as

[(x²+2y²) - 2xy][(x+2y²) + 2xy]

Removing the parentheses inside the brackets:

[x² + 2y² - 2xy][x² + 2y² + 2xy]

Then we can change the brackets to parentheses
and rearrange the trinomials in descending order
of x and ascending order of y:

(x² - 2xy + 2y²)(x² + 2xy + 2y²)


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2)What is the coefficient of x in (x⁴ + x³ + x² + x + 1)⁴?

We write the parentheses as multiplied 4 times:

(x⁴ + x³ + x² + x + 1)(x⁴ + x³ + x² + x + 1)(x⁴ + x³ + x² + x + 1)(x⁴ + x³ + x² + x + 1)

If we were to go to the trouble to multiply that all the way out,
we would have 5*5*5*5 = 625 terms before we started collecting
like terms.  The only like terms in x among those 625 would be
when we multiplied x from one of those 4 parenthetical expressions
times 1's from the other 3.  We can choose the x any of 4 ways, so
the coefficient would be 4. 


3)What is the coefficient of x³ in (x⁴ + x³ + x² + x + 1)⁴?

As in the preceding problem, We write the parentheses as 
multiplied 4 times:

(x⁴ + x³ + x² + x + 1)(x⁴ + x³ + x² + x + 1)(x⁴ + x³ + x² + x + 1)(x⁴ + x³ + x² + x + 1)

The only terms among the 625 that would give us x³ is when we choose

1.  x³ from one of the 4 to multiply by 1's from the other three, which 
    we would do in 4 ways.

or

2. x² from one of the 4 to multiply by x from another one of the
   other 3, and 1s from the remaining 2, which we would do in 4*3 or 
   12 ways.

or

3. x from 3 of them and 1 from the 4th, which we would do in 4 ways
 

So among the 625 multiplications there are total of 4+12+4=20 ways to get
x³. So the coefficient of x³ is 20.  

Edwin