given that (x-1)(x+2) are factors of x3-px2-qx+6=0 find the value of p and q and determine d zero of the polynomial​

points are handed out whenever you post answers.

brainliest answers are handed out by the person asking the question, so whether your answer is picked depends entirely on whether the asker bothers to pick

if you're asking about your rank, it depends on having a certain number of points and a certain number of brainliest answers. for example, at the time of this writing, i have 44338 points out of 15000 needed for the next (nearly triple the requirement! ) but i only have 24 out of 50 required brainliest answers. this number decays, meaning after a certain amount of time (i think about a month), the total number of brainliest answers you've earned doesn't count toward this total. (i've been stuck at the same rank for a long

the lesson to learn here is that you shouldn't worry about points or rank. chances are, no matter how good an answer you may provide, your efforts will go mostly unnoticed unless you focus on the newest/most active questions that get posted.

actually your concept is wrong

step-by-step explanation:

150% of x =300

150x=300×100

x=(300×100)÷150

x=200

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.