Factor completely x2 − 10x + 25. (2 points)

(x-5)(x-5)

Step-by-step explanation:

Ok... This ones actually pretty simple.

So all you have to do is find 2 numbers that add up to -10, and multiply to 25. You will quickly realize that the only 2 numbers that do that are -5 and -5. Then, all you have to do is write the now factored equation. (x-5)(x-5)

There are 2 ways to check your work. The first is to FOIL (first, inside, outside, last). This should get you back to your original equation

The second way to do it is to plug it into a graphing program. If you graph your factored equation and your original equation, they should be the same!

I hope this helps!

If you have any other questions just feel free to ask me.

For this case we must factor the following expression:

For this, we must find two numbers that when multiplied give 25 and when added to -10.

These numbers are -5 and -5:

Then, the factorization is given by:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

there still no table