Plz help with this math problem ?
回答 (6)
2009-05-02 12:30 pm
✔ 最佳答案
Hi,Problem : x^2 + (p + 1/p)x + 1
= x^2 + px + x/p + 1
= x(x + p) + 1/p(x + p)
= ( x+ p)(x + 1/p)<==FACTORED
Hope i helped u :)
2009-05-02 3:34 pm
x^2 + (p + 1/p)x + 1
= x^2 + px + x/p + 1
= (x^2 + px) + (x/p + 1)
= x(x + p) + 1/p(x + p)
= (x + p)(x + 1/p)
= x^2 + px + x/p + 1
= (x^2 + px) + (x/p + 1)
= x(x + p) + 1/p(x + p)
= (x + p)(x + 1/p)
2009-05-02 8:48 pm
hey it is simple....
x^2 + (p + 1/p)x + 1
= x^2 + px + x/p + 1
= (x^2 + px) + (x/p + 1)
= x(x + p) + 1/p(x + p)
= (x + p)(x + 1/p)
so dats d answer.
x^2 + (p + 1/p)x + 1
= x^2 + px + x/p + 1
= (x^2 + px) + (x/p + 1)
= x(x + p) + 1/p(x + p)
= (x + p)(x + 1/p)
so dats d answer.
2009-05-02 3:37 pm
Factor of what? Or you mean factorise?
It is = (x+p)(x+(1/p))
It is = (x+p)(x+(1/p))
2009-05-02 3:48 pm
x^2 + (p + 1/p)x + 1 =
Distribute the x into the numerator (p+1), the denominator remains p
x^2 + (xp + x)/p+ 1 =
Divide (xp + x) by p
x^2 + x + x/p + 1 =
Multiply by p to get p out of the denominator
p(x^2 + x + x/p + 1) =
x^2p + xp + x + 1=
Factor
(x + 1)(xp + 1)
Great problem. I hope I helped.Good luck.
Distribute the x into the numerator (p+1), the denominator remains p
x^2 + (xp + x)/p+ 1 =
Divide (xp + x) by p
x^2 + x + x/p + 1 =
Multiply by p to get p out of the denominator
p(x^2 + x + x/p + 1) =
x^2p + xp + x + 1=
Factor
(x + 1)(xp + 1)
Great problem. I hope I helped.Good luck.
2009-05-02 3:41 pm
x^2 + (p + 1/p)x + 1
x^2+px+x/p+1
x(x+p)+1/p(x+p)
(x+1/p)(x+p)
x= -p,-1/p
x^2+px+x/p+1
x(x+p)+1/p(x+p)
(x+1/p)(x+p)
x= -p,-1/p
收錄日期: 2021-05-01 12:32:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090502003035AAu7jkD