help me solve this math question?

Question

Could you tell me step by step on how do i solve or find x in the problem below:

n^3-8n^2+n+42

what if i write the equaiont so that it equals 0.

n^3-8x^2+n+42=42

Answer 1

your last three answers all correct.but they have their own way . i think no need me to explain it more.and i think choose happy_ min as the best answer, she is nice girl...

Good Luck.

Answer 2

You either use trial and error to find 1 real root of the equation, or input that equation into a program for graphs like mathcad.

By trial and error, assuming that n^3-8n^2 + n + 42 = 0

n = -2

Substitute n = -2 into equation to get 0, so ( n + 2) is a real root of the equation.

From then on, use long division, dividing the equation by ( n + 2) to get n^2 -10 n + 21,
factorise this expression to get ( n - 7 ) ( n - 3)

Therefore the equation can be expressed as ( n+2)(n-7)(n-3) = 0

n= -2, 3, or 7.

Answer 3

First some big assumptions:
1) when you say "find x" you really mean find "n"
2) this equation is equal to 0.

If those assumptions are valid the answers are -2, 3, and 7

The way to solve this is to factor thusly:
(n+2)(n-3)(n-7)=0
So, at least one of these values in parenthesis has to equal to zero, then take turns setting each to 0 to solve and you get the answer.
How did I get the original factors? Blind luck sorta. I figure that there had to be three factors to come up with n^3 so...
2-3-7 = -8 and (2)(-3)(-7) = 42.
I plugged those numbers in and it worked.

Clear as mud?

Answer 4

I think u've typed wrong, eqn. is : n^3-8n^2+n+42=0.
Use vanishing method. n=(-2) satisfies the eqn.
so, n^2.(n+2)-10n.(n+2)+21.(n+2)=0.
=> (n+2).(n^2-10n+21)=0.
=> (n+2).(n^2-3n-7n+21)=0.
=> (n+2).[n.(n-3)-7.(n-3)]=0.
=> (n+2).(n-3).(n-7)=0.
Therefore, n=(-2),3,7.

Plz, choose my answer the best.

Answer 5

There is no way to simplify this term.

Answer 6

impossible

Answer 7

www.quickmath.com should help you out.

Answer 8

you can't simplify this question