# Thread: polynomial stuff

1. ## polynomial stuff

determine the polynomial f(x) that possesses the following characteristic :

1) f(x) is a polynomial of degree4
2) (x-1) is a factor of f(x) and f ' (x)
3) f(0) =3 and f'(0)= -5
4) the ramiander when f(x) is divided by (x-2) is 13

Find the set of value of x such tht f'(x) >0. find all four roots of f(x) = 0

STUCK at

f(x) = ax^4 + bx^3 + cx^2 +dx +e
f(1) = a+B+C+D+E = 0
f '(X) = 4ax^3+3b^2 +2cx + d = 0
f " (1) = 4a+3b+2c+d= 0
f(0)= e =3
f'(0) =d =-5

then wht to do?

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

i am not sure whether my step is correct. kinda blur with the question

a)i substitue n= 2 into f(2) = X^2 -2x+1 is identity to (X-1)^2 g(x)
g(x) = 1 <--- interger coefficients ?

1)do not have any idea
2) not sure. by substituting the value of 1 into the f(1) = 1 -4+3 =0, then i differentiate 4x^3-4 = 0 then again substitue to show the value?

2. Originally Posted by qweiop90
determine the polynomial f(x) that possesses the following characteristic :

1) f(x) is a polynomial of degree4
2) (x-1) is a factor of f(x) and f ' (x)
3) f(0) =3 and f'(0)= -5
4) the ramiander when f(x) is divided by (x-2) is 13

Find the set of value of x such tht f'(x) >0. find all four roots of f(x) = 0

STUCK at

f(x) = ax^4 + bx^3 + cx^2 +dx +e
f(1) = a+B+C+D+E = 0
f '(X) = 4ax^3+3b^2 +2cx + d = 0
f " (1) = 4a+3b+2c+d= 0
f(0)= e =3
f'(0) =d =-5

then wht to do?
you are going to solve this by setting up 3 simultaneous equations.

you missed the last clue. if the remainder when f(x) is divided by (x - 2) is 13, then f(2) = 13. use this fact to come up with the last equation. you already know d and e, so now, just solve the system for a, b and c

can you continue?

3. ops. sry for the mistake . btw . i have calculated a=2, b = -3 c=3 d= -5 e=3.

so can you help to chk? and the question also mention that you need to find 4 of the roots? <- not sure . f '(X) >0

4. Originally Posted by qweiop90
ops. sry for the mistake . btw . i have calculated a=2, b = -3 c=3 d= -5 e=3.

so can you help to chk? and the question also mention that you need to find 4 of the roots? <- not sure . f '(X) >0
$2x^4-3x^3+3x^2-5x+3\underbrace{=}_{\text{factors}}(x-1)^2(2x^2+x+3)$

So you have the real answer $x=1\leftarrow\text{multiplicity of two}$
and the two imaginary answers

$x\underbrace{=}_{\text{quadractic formula}}\frac{-1\pm\sqrt{1-4(2)(3)}}{4}=\frac{-1\pm{i\sqrt{23}}}{4}$

5. anyway, f ' (X) > 0 , means?

Without using trial and error, isnt any method to solve the equation with higher degree?

6. Originally Posted by qweiop90
anyway, f ' (X) > 0 , means?

Without using trial and error, isnt any method to solve the equation with higher degree?
$f'(x)>0\Rightarrow\text{if }x_0

And not really, its kind of an art.

7. not really understand . zz x0 is , x1 is???

sry in lower 6, we havent learn this kind of things yet..

2)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

. ANYWAY, i had solve num1 , but for num2 , the question asked me to prove? so, how? i should let n = 4 ?

8. Originally Posted by qweiop90
not really understand . zz x0 is , x1 is???

sry in lower 6, we havent learn this kind of things yet..

2)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

. ANYWAY, i had solve num1 , but for num2 , the question asked me to prove? so, how? i should let n = 4 ?
Post new questions in new threads, and are you saying your in sixth grade and your learning about derivatives? This just means that since the derivative is always greater than zero that the function is always increasing.