Thread: show equation and find the factor

1. show equation and find the factor

2a)given that f(x) =x^n-nx+n-1 for the integer n>1. by considering f(x) and f'(x), show that f(x)=(X-1)^2 X g (x) is true for all polynomial g(x) with interger coefficients.

hence , or otherwise
1) show that 3^2n - 8n -1 is divisiblke by 64 for all intergers n>1
2) show that the equation X^4 -4x+3 = 0

. question2

Moderator,

i am learning abt factor . example 4x^4-4x^3-9x^2+x+2 . find the factor , factorise it! ( without using trial , any other method provided?

2. Originally Posted by qweiop90
2a)given that f(x) =x^n-nx+n-1 for the integer n>1. by considering f(x) and f'(x), show that f(x)=(X-1)^2 X g (x) is true for all polynomial g(x) with interger coefficients.

[snip]
The result you have stated is wrong. The correct statement would be something like:

Show that $f(x) = (x-1)^2 g(x)$ for some polynomial g(x) with integer coefficients.

First note that x = 1 is a root of f(x) and so x - 1 is a factor of f(x). Then $f(x) = (x - 1) \, h(x)$.

Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of $f'(x) = h(x) + (x - 1) h'(x)$.

It follows that (x - 1) is a factor of h(x). Therefore $h(x) = (x - 1) g(x)$ and the result follows.

3. Originally Posted by qweiop90
2a)given that f(x) =x^n-nx+n-1 for the integer n>1. by considering f(x) and f'(x), show that f(x)=(X-1)^2 X g (x) is true for all polynomial g(x) with interger coefficients.

hence , or otherwise
1) show that 3^2n - 8n -1 is divisiblke by 64 for all intergers n>1

Mr F says: The result is simple to show using proof by induction.

2) show that the equation X^4 -4x+3 = 0

Mr F says: You probably mean

Solve the equation x^4 - 4x + 3 = 0.

Note that n = 4.

[snip]
..

4. Originally Posted by qweiop90
2a)given that f(x) =x^n-nx+n-1 for the integer n>1. by considering f(x) and f'(x), show that f(x)=(X-1)^2 X g (x) is true for all polynomial g(x) with interger coefficients.

hence , or otherwise
1) show that 3^2n - 8n -1 is divisiblke by 64 for all intergers n>1
[snip]
As an alternative to proof by induction, you can note that $3^{2n} = 9^n$. So substitute x = 9 into the previous result:

$f(9) = (9-1)^2 g(9) = 64 g(9)$.

Also $f(9) = 9^n - 9n + n - 1 = 3^{2n} - 8n - 1$.

Therefore $64 g(9) = 3^{2n} - 8n - 1 ~ .....$

5. Originally Posted by mr fantastic
The result you have stated is wrong. The correct statement would be something like:

Show that $f(x) = (x-1)^2 g(x)$ for some polynomial g(x) with integer coefficients.

First note that x = 1 is a root of f(x) and so x - 1 is a factor of f(x). Then $f(x) = (x - 1) \, h(x)$.

Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of $f'(x) = h(x) + (x - 1) h'(x)$.

It follows that (x - 1) is a factor of h(x). Therefore $h(x) = (x - 1) g(x)$ and the result follows.

dont really understand? mind to show how the solution can be done?

6. Originally Posted by qweiop90
dont really understand? mind to show how the solution can be done?
What part don't you understand?

7. er show that . f(x) = (x-1)^2 g(x) . where is ^2 come frm? and how to prove?

8. Originally Posted by qweiop90
er show that . f(x) = (x-1)^2 g(x) . where is ^2 come frm? and how to prove?
What part of my reply (post #2) don't you understand.

9. Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of .

It follows that (x - 1) is a factor of h(x). Therefore and the result follows.

. how it comes frm?

10. Originally Posted by qweiop90
Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of .

It follows that (x - 1) is a factor of h(x). Therefore and the result follows.

. how it comes frm?
Substitute into $f(x) = (x - 1) \, h(x)$.

11. First note that x = 1 is a root of f(x) and so x - 1 is a factor of f(x). Then .<---1

Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of <---2

why it become like this?

12. Originally Posted by qweiop90
First note that x = 1 is a root of f(x) and so x - 1 is a factor of f(x). Then .<---1

Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of <---2

why it become like this?
I got f(x) = (x-1) h(x) by noting that (x-1) is a factor of f(x). At this point I think it would be best if you went back and thoroughly reviewed the factorising of polynomials of degree greater than 2.

I differentiated f(x) =x^n-nx+n-1 in the usual way to get $f'(x) = n x^{n-1} - n$. Clearly x = 1 is a root of f'(x) and therefore (x - 1) is a factor of f'(x).

Then I differentiated f(x) = (x-1) h(x) using the product rule. I used the result to show that if (x-1) is a factor of f'(x) then it must be a factor of h(x).

I have no idea what your mathematics background. It might be that you don't know the product rule. In which case you have to accept as a rule that if f(x) and f'(x) both have a factor of (x - 1), then (x - 1) is a repeated factor of f(x), that is, (x-1)^2 is a factor of f(x).

13. dont understand . y h(x) + (x-1)h'(x) . how 2 exist?

f(x) = (x-1) h(x) using the product rule. I used the result to show that if (x-1) is a factor of f'(x) then it must be a factor of h(x).

at here. i understood that h(x) is quadrant .

14. Originally Posted by qweiop90
[snip]

dont understand . y h(x) + (x-1)h'(x) . how 2 exist?

[snip]
In post #12 I said I used the product rule to get that result. You clearly have not learnt this rule yet. So forget all of that. Please recall:

Originally Posted by mr fantastic
It might be that you don't know the product rule. In which case you have to accept as a rule that if f(x) and f'(x) both have a factor of (x - 1), then (x - 1) is a repeated factor of f(x), that is, (x-1)^2 is a factor of f(x).