We have to solve the inequality u^2 -2u + 1 < 16.

u^2 -2u + 1 < 16

=> u^2 - 2u - 15 < 0

=> u^2 - 5u + 3u - 15 < 0

=> u( u - 5) + 3( u -5) < 0

=> (u +3) (u - 5)< 0

Now for this to be true either of the terms u + 3 and u-5 should be negative.

=> u+3> 0 and u-5 < 0

=> u > -3 and u < 5

This gives us that u lies in (-3 , 5)

u+3 < 0 and u - 5> 0

=> u < -3 and u > 5 which is not possible.

**So the only valid value of u lie in ( -3, 5)**

Given the inequality:

u^2 - 2u + 1 < 16

First we will factor the quadratic equation.

We know that (u-1)^2 = u62 - 2u + 1

=> (u-1)^2 < 16

Now we will take the square root for both sides.

==> +-(u-1) < +-4

==> l u-1 l < l 4 l

==> -4 < u-1 < 4

We will add 1 to both sides.

==> -4 +1 < u < 4+1

==> -3 < u < 5

**Then, the values of u belong to the interval (-3, 5)**

**==> u = ( -3, 5)**

To find the values of u for which u^2-2u+1 < 16.

We notice that let side is a perfect square of (u-1)^2 and the right is a perfect square.

So (u-1)^2 < 16 .

We take the square root of both sides:

u-1 < 4, or u-1 > -4.

u < 5, or u > -4+1 = -3.

Therefore -3 < u < 5. Or u takes values from the open interval (-3 , 5).