# Thread: Setting polynomial to zero

1. ## Setting polynomial to zero

I need help on how to set the following to zero;

12x^2 + 6x - 6

Can someone walk me through this?

2. Originally Posted by Humbled_by_math
I need help on how to set the following to zero;

12x^2 + 6x - 6

Can someone walk me through this?
if all you want to do is set it to zero, then you simply do the folowing

$\displaystyle 12x^2 + 6x - 6 = 0$

we're done!

on the other hand (since it is hardly likely that what you asked for was what you were actually asked to do--you should type the question in its entirety!), if you want to solve for $\displaystyle x$, that is a different story. try dividing through by 6 then factorizing. can you proceed?

3. Originally Posted by Humbled_by_math
I need help on how to set the following to zero;

12x^2 + 6x - 6

Can someone walk me through this?
In case the instructions want you also to solve, the first step will be to divide through by the common factor of 6. Then you'll need to factor the quadratic, set the factors equal to zero, and then solve the resulting linear equations; or else you'll need to apply the Quadratic Formula.

4. $\displaystyle 12x^2+6x-6=0 \Rightarrow \ 6(2x^2+x-1)=0 \Rightarrow \ 6(2x-1)(x+1)=0$

Now we can find the roots of the quadratic!

The roots are the values for x for which the quadratic is equal to 0.

This is when $\displaystyle x=-1$ and $\displaystyle x=\frac{1}{2}$.

5. You also have the following formula for solving this kind of equations:

$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where a,b,c are the coefficients of equation you have to solve:

$\displaystyle ax^2+bx+c=0 \Rightarrow 12x^2+6x-6=0$

$\displaystyle \Rightarrow$$\displaystyle a=12$$\displaystyle b=-6$$\displaystyle c=-6$

From here it's easy ,

Have a nice day!