SOLUTION: Factor {{{21y^4+49y^3+14y^2}}}

Factor

21y4 + 49y3 +14y2

First we look for a number to take out,
which is 7, because 7 is the largest integer
that is a factor of 21, 49 and 14

Now we look for a letter.  The smallest power of
y that occurs in every term is y².

So we factor out 7y²

7y²(3y² + 7y + 2)

7y²(3y² + 7y + 2)

Multiply the red 3 by the blue 2, which gives
6, Now since the green sign is " + ", we 
think of two integers which have product equal
to 6 and which have sum equal to the purple 7.

Such integers are 6 and 1. So we use these to
rewrite the 7y as 6y + 1y

7y²(3y² + 6y + 1y + 2)

We change the parentheses to brackets so we
can put parentheses inside them:

7y²[3y² + 6y + 1y + 2]

Now we factor 3y out of the first two terms
in the brackets

7y²[3y(y + 2) + 1y + 2]

Now we factor 1 out of the last two terms in
the brackets:

7y²[3y(y + 2) + 1(y + 2)]

Now notice that inside the brackets there is
a common factor of (y + 2)

So within the bracets take out the common
factor (y + 2) and put the factors 3y and +1
inside parentheses on the right:

7y²[(y + 2)(3y + 1)]

Now we can erase the brackets and get

7y²(y + 2)(3y + 1)

Edwin