Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mkuma36

  • 2 years ago

x^3+px^2-7x-6 can be factorised into three simple factors only when p is equal to ?

  • This Question is Closed
  1. moneybird
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what does simple factors mean?

  2. mkuma36
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    meaning (x+1) (x+2)(x+4) something like this

  3. moneybird
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it can't be (2x+1), right?

  4. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    simple had me too, i wanna say integers but simple can mean nothing in math :)

  5. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (x^3-7x)+(px^2-6) x(x^2 - 7) + p(x^2 - 6/p) p = 6/7 if i see it right

  6. mkuma36
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let us say p=0 the equation will be x^3-7x-6 = x^2(x+1)-x(x+1)-6(x+1)=(x-3)(x+2)(x+1).

  7. mkuma36
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but this is not concrete solution ..this is hit and trial...As choices were 0,1,2,3

  8. moneybird
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    try vieta formula

  9. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (x+a)(x+b)(x+c) (x^2 + 2abx + ab)(x+c) x^3 + 2abx^2 + abx +cx^2 +2abcx + abc -------------------------- x^3 + p x^2 + 7x - 6 if thats workable :)

  10. moneybird
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    possible rational roots are pm 1, pm 2, pm 3, pm 6

  11. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    0 is it according to the trial and error method :)

  12. dturpin53
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[x^3+px^2-7x-6=x^x(x+p)-(7x+6)=(x^2-1)(x+p) =(x^2-1)(7x+6)\] hence x+p=7x+6 p=6x+6

  13. Chlorophyll
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    when p= 0 -> f(x) = x^3 - 7x - 6 = ( x + 1) ( x^2 - x - 6) => x = -1, x = -2, x = 3

  14. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.