anonymous
  • anonymous
x^3+px^2-7x-6 can be factorised into three simple factors only when p is equal to ?
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
what does simple factors mean?
anonymous
  • anonymous
meaning (x+1) (x+2)(x+4) something like this
anonymous
  • anonymous
it can't be (2x+1), right?

Looking for something else?

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

More answers

amistre64
  • amistre64
simple had me too, i wanna say integers but simple can mean nothing in math :)
amistre64
  • amistre64
(x^3-7x)+(px^2-6) x(x^2 - 7) + p(x^2 - 6/p) p = 6/7 if i see it right
anonymous
  • anonymous
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).
anonymous
  • anonymous
but this is not concrete solution ..this is hit and trial...As choices were 0,1,2,3
anonymous
  • anonymous
try vieta formula
amistre64
  • amistre64
(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 :)
anonymous
  • anonymous
possible rational roots are pm 1, pm 2, pm 3, pm 6
amistre64
  • amistre64
0 is it according to the trial and error method :)
anonymous
  • anonymous
\[x^3+px^2-7x-6=x^x(x+p)-(7x+6)=(x^2-1)(x+p) =(x^2-1)(7x+6)\] hence x+p=7x+6 p=6x+6
anonymous
  • anonymous
when p= 0 -> f(x) = x^3 - 7x - 6 = ( x + 1) ( x^2 - x - 6) => x = -1, x = -2, x = 3

Looking for something else?

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