Trying to solve polynomial equations - need help

**Arise01** · May 5th 2011, 01:08 PM

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.

**Quacky** · May 5th 2011, 01:18 PM

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.

**pickslides** · May 5th 2011, 01:18 PM

Move all terms to one side then do some factoring.

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

**Arise01** · May 5th 2011, 01:26 PM

@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.

**Quacky** · May 5th 2011, 01:28 PM

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!

**Arise01** · May 5th 2011, 01:30 PM

Ok so I got ...

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

I'm not sure what to do from there.

**pickslides** · May 5th 2011, 01:33 PM

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

**Quacky** · May 5th 2011, 01:33 PM

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...

**Arise01** · May 5th 2011, 01:37 PM

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

**Arise01** · May 5th 2011, 01:41 PM

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

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

**Quacky** · May 5th 2011, 01:45 PM

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.

**Arise01** · May 5th 2011, 01:50 PM

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?

**Quacky** · May 5th 2011, 01:57 PM

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?

**Arise01** · May 5th 2011, 02:00 PM

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

**Quacky** · May 5th 2011, 02:09 PM

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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