PL HELP! Find all real zeros and complex of each solution x^3-5x^2+7x-35=0. SHOW WORK!!!

Hey there! :)

x = 5, x = -i√7, x = i√7

Step-by-step explanation:

We can factor the equation x³ - 5x² + 7x - 35 = 0 by grouping:

x²(x - 5) + 7(x -5) = 0

We get:

(x² + 7)(x - 5) = 0

or

(x + i√7)(x - i√7)(x - 5) = 0

Use the Zero Product Property to solve for the solutions:

x - 5 = 0

x = 5.

x² + 7 = 0

x² = -7 (We will have imaginary solutions in this instance)

x = ±i√7

Therefore, the zeros of this equation are:

x = 5, x = -i√7, x = i√7

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.

1st part is 64,000,000

2nd part 500,000

step-by-step explanation: