# Thread: Trying to solve polynomial equations - need help

1. ## Trying to solve polynomial equations - need help

Ok so I'm in grade 12 taking math. Were working on polynomial functions and factoring. It's got to be the hardest stuff yet but looking forward at the curriculum that comment will look like a joke. Anyways here's the question that I'm stuck on.

-4x^4 + 60x^3 = 200x^2

This is my first post so I apologize if I'm not posting in the correct sub forum. I searched the forum and this seemed to be the most appropriate area. Any help would be great. Thanks.

2. Originally Posted by Arise01
Ok so I'm in grade 12 taking math. Were working on polynomial functions and factoring. It's got to be the hardest stuff yet but looking forward at the curriculum that comment will look like a joke. Anyways here's the question that I'm stuck on.

$-4x^4 + 60x^3 = 200x^2$

This is my first post so I apologize if I'm not posting in the correct sub forum. I searched the forum and this seemed to be the most appropriate area. Any help would be great. Thanks.
We have: $-4x^4 + 60x^3 = 200x^2$

Get everything on one side, starting with the highest power of x (for clarity). I added to the right hand side (it makes the working easier):

$0=4x^4-60x^3+200x^2$

Every term on the right is divisible by 4. So I take this out as a "common factor" as such, by 'dividing' each term by the common factor, which is left outside the bracket:

$0=4(x^4-15x^3+50x^2)$

There is also a common factor of $x^2$ as $x^3=x^2\times x$ and $x^4 = x^2\times x^2$. Again, out goes the common factor:

$0=4x^2(x^2-15x+50)$

Which means that either $4x^2=0$, or $x^2-15x+50=0$

Can you finish up?

...And yes, when you've finished the year, you will look back over this and laugh at how simple it seems.

3. Move all terms to one side then do some factoring.

Hint $x^2$ and -4 look like factors.

4. @quacky - how come you have down -600? shouldn't it be - 60?

I'm in the midst right now of moving everything to one side, than I'll start factoring.

5. Originally Posted by Arise01
@quacky - how come you have down -600? shouldn't it be - 60?

I'm in the midst right now of moving everything to one side, than I'll start factoring.
Yes, I made a typo which I carried through. It isn't THAT easy!

6. Ok so I got ...

4x^2(x^2-15x+50)

I'm not sure what to do from there.

7. Factor this part... $\displaystyle x^2-15x+50$

8. Originally Posted by Arise01
Ok so I got ...

$4x^2(x^2-15x+50)$

I'm not sure what to do from there.
You have turned your equation into an expression! What you meant to say was that:

$4x^2(x^2-15x+50)=0$

This can only be true if $4x^2=0$ or $x^2-15x+50=0$

If $4x^2=0$
Then $x^2=0$
And $x=0$

Similarly, if $x^2-15x+50=0$, then...

9. So 4x^2 = 0 and now I need to factor x^2 - 15x+50?

10. so in order to factor x^2-15x+50 ...

I need to find 2 #'s of the product of 50 and sum of 15?

11. Originally Posted by Arise01
so in order to factor x^2-15x+50 ...

I need to find 2 #'s of the product of 50 and sum of 15?
Literally speaking they have to sum -15.

Alternatively, you could use the formula, but factoring is far easier here.

12. Is this correct? The two numbers that are to be used are... -5,-10... so...

4x^2(x-5)(x-10)

Am I on the right track?

13. Originally Posted by Arise01
Is this correct? The two numbers that are to be used are... -5,-10... so...

4x^2(x-5)(x-10)

Am I on the right track?
Absolutely! Almost finished; remember that it is equal to 0, so either $4x^2=0$, $x-5=0$ or $x-10=0$. Therefore, there are three possible values for $x$ - what are they?

14. 4x^2, 5, 10? Are those the three possible values?

15. Originally Posted by Arise01
4x^2, 5, 10? Are those the three possible values?

Not $4x^2$, but you're thinking along the right lines.

If $4x^2=0$
Then divide both sides by 4 to obtain:
$x^2=0$

Then square root both sides:
$x=0$

x=0,5 or 10

For future reference, you can check by substituting those values back into the original equation.

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