1. Cubic Functions

Show that (x+3) is a factor of the following cubic function,

2x cubed + x squared - 13x + 6 = 0,

Hence solve the cubic function completely

2. Have you learned how to use synthetic division or at least polynomial long division? Either method can demonstrate the problem.

You may have also learned the remainder theorem / factor theorem, which basically ask you to plug in the zero value that comes from your factor and check to see if the remainder is zero.

3. Workink a little with the polynom you will find that

$2x^3+x^2-13x+6 = (x+3)(2x^2-5x+2)
$

4. Originally Posted by reddevil
Show that (x+3) is a factor of the following cubic function,

2x cubed + x squared - 13x + 6 = 0,

Hence solve the cubic function completely
The remainder theorem tells you that when a polynomial $P(x)$ is divided by a linear factor $x-a$ the remainder is $P(a).$

Hence $(x+3)$ is a factor of $P(x)$ if and only if $P(-3)=0.$

CB

5. In other words, all you are asked to do is to show that if $f(x)= 2x^3+ x^2- 13x+ 6$, then f(-3)= 0. surely you can do that!

6. Originally Posted by reddevil
Show that (x+3) is a factor of... 2x cubed + x squared - 13x + 6 = 0. Hence solve the cubic function completely
It's already been pointed out that you can do the "showing" part using polynomial long division or else synthetic division, such as:

[HTML] 2x^2 - 5x
-----------------------
x + 3 ) 2x^3 + 1x^2 - 13x + 6
2x^3 + 6x^2
-----------------------
- 5x^2 - 13x + 6
- 5x^2 - 15x
-----------------------
2x + 6

-----------------------[/HTML]
...and so forth.

Once you have done the actual division, you will be left with a quadratic, being the quotient across the top of the long division above (once you've completed the process).

The exercise is unclear, since it gives the cubic "equal to zero", so you should be trying to find the zeroes, but then it asks you to show the factors (which would be factors of just "y = 2x^3 + x^2 - 13x + 6", without the "equals zero" stipulation). So I'm not sure if, for the "hence" part, you're supposed to complete the exercise by factoring the quadratic or by solving for the zeroes, such as with the Quadratic Formula.

You might want to do both.

7. Personally, what I would do is sit down with the problem for 10 or 15 minutes, and try to factor it. Once you get decent at factoring, it should be second nature, although factoring cubics isn't the easiest thing ever.