Can somebody solve this: x^4+2x^3-25x^2-26x+120

 (x + 5) • (x + 3) • (x - 2) • (x - 4)

Step-by-step explanation:

Step-1 : Multiply the coefficient of the first term by the constant   1 • 15 = 15

Step-2 : Find two factors of  15  whose sum equals the coefficient of the middle term, which is   8 .

     -15    +    -1    =    -16

     -5    +    -3    =    -8

     -3    +    -5    =    -8

     -1    +    -15    =    -16

     1    +    15    =    16

     3    +    5    =    8    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  3  and  5

                    x2 + 3x + 5x + 15

Step-4 : Add up the first 2 terms, pulling out like factors :

                   x • (x+3)

             Add up the last 2 terms, pulling out common factors :

                   5 • (x+3)

Step-5 : Add up the four terms of step 4 :

                   (x+5)  •  (x+3)

            Which is the desired factorization

 (x + 5) • (x + 3) • (x - 2) • (x - 4)

there are 25 people coming to the party

step-by-step explanation:

the text says that there will be 22 children at this party, 1 teacher, and 2 parents. so if you add them all together (22+1+2) it will equal 25. (=25)

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.