# factor theorem

• Jan 5th 2009, 05:18 AM
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 .
• Jan 5th 2009, 05:27 AM
mr fantastic
Quote:

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)\$ .... )
• Jan 5th 2009, 08:15 AM
fanfan1609
I think f(x) not >=0 with all real numbers .Maybe the problem is wrong .
• Jan 5th 2009, 11:30 AM
mr fantastic
Quote:

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?
• Jan 5th 2009, 12:21 PM
Opalg
Quote:

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\$.
• Jan 5th 2009, 02:16 PM
mr fantastic
Quote:

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 ..... (Rofl)
• Jan 5th 2009, 02:30 PM
fanfan1609
Quote:

Originally Posted by mr fantastic
You're wrong. On what grounds do you make this statement?

After factorising http://www.mathhelpforum.com/math-he...5aac9f30-1.gif. 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 .
• Jan 5th 2009, 03:04 PM
mr fantastic
Quote:

Originally Posted by fanfan1609
After factorising http://www.mathhelpforum.com/math-he...5aac9f30-1.gif. 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.
• Jan 5th 2009, 06:43 PM
fanfan1609
Quote:

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.