1. ## factor theorem

Given that (x-3) and (2x+1) are factors of
$\displaystyle f(x)=ax^4+bx^3+13x^2=30x+9$, find the values of a and b . With these values of a and b , show that $\displaystyle f(x)\geq0$ for all x belongs to real numbers .

My working :

i found that a = 4 and b = -20

$\displaystyle 4x^4-20x^3+13x^2+30x+9$
$\displaystyle (x-3)(2x+1)(2x^2-5x-3)$

I am wondering how can i show that $\displaystyle f(x)\geq0$ from here .
Thanks for any help .

Given that (x-3) and (2x+1) are factors of
$\displaystyle f(x)=ax^4+bx^3+13x^2=30x+9$, find the values of a and b . With these values of a and b , show that $\displaystyle f(x)\geq0$ for all x belongs to real numbers .

My working :

i found that a = 4 and b = -20

$\displaystyle 4x^4-20x^3+13x^2+30x+9$
$\displaystyle (x-3)(2x+1)(2x^2-5x-3)$

I am wondering how can i show that $\displaystyle f(x)\geq0$ from here .
Thanks for any help .
I suggest you find the coordinates and nature of the turning points of y = f(x) (note that $\displaystyle 16x^3 -60x^2 + 26x + 30 = 2(x - 3)(2x + 1)(4x-5)$ .... )

3. I think f(x) not >=0 with all real numbers .Maybe the problem is wrong .

4. Originally Posted by fanfan1609
I think f(x) not >=0 with all real numbers .Maybe the problem is wrong .
You're wrong. On what grounds do you make this statement?

Given that (x-3) and (2x+1) are factors of
$\displaystyle f(x)=ax^4+bx^3+13x^2=30x+9$, find the values of a and b . With these values of a and b , show that $\displaystyle f(x)\geq0$ for all x belongs to real numbers .

My working :

i found that a = 4 and b = -20

$\displaystyle 4x^4-20x^3+13x^2+30x+9$
$\displaystyle (x-3)(2x+1)(2x^2-5x-3)$

I am wondering how can i show that $\displaystyle f(x)\geq0$ from here .
Thanks for any help .
Try factorising $\displaystyle 2x^2-5x-3$.

6. Originally Posted by Opalg
Try factorising $\displaystyle 2x^2-5x-3$.
Well, you only do that if you want to do it the easy way .....

7. Originally Posted by mr fantastic
You're wrong. On what grounds do you make this statement?
After factorising . I have $\displaystyle f(x)=(x-3).(2x+1).(2x-3).(x+1)$
If x belongs (3/2,3) ,f(x) <0 .Example x=2 ,I have f(2)=(2-3).(2.2+1).(2.2-3).(2+1)=(-1).5.1.3=-15
I thinks so that .

8. Originally Posted by fanfan1609
After factorising . I have $\displaystyle f(x)=(x-3).(2x+1).(2x-3).(x+1)$
If x belongs (3/2,3) ,f(x) <0 .Example x=2 ,I have f(2)=(2-3).(2.2+1).(2.2-3).(2+1)=(-1).5.1.3=-15
I thinks so that .
$\displaystyle 2x^2 - 5x - 3 \neq (2x - 3)(x+1)$.

If you had expanded as a check you would have noticed this.

9. Originally Posted by mr fantastic
$\displaystyle 2x^2 - 5x - 3 \neq (2x - 3)(x+1)$.

If you had expanded as a check you would have noticed this.
$\displaystyle 2x^2 -5x -3 = (2x+1)(x-3)$ If that ,$\displaystyle f(x)=(2x+1)^2(x-3)^2$ .so that f(x) >= 0 with all x belongs real numbers .