# Problems with factorization in regard to defining a polynomial

• Mar 24th 2011, 12:56 PM
Lia85
Problems with factorization in regard to defining a polynomial
Hi

Following cordinate-points are given to define a function:

(½,o), (-2,0) and (1,-3) with the parable roots being ½ and -2. Since d>0, I used the formula: f(x)= a(x-r1)(x-r2), where r1= first root and r2= second root.

So this is what I end up with:

f(x) = a(x-½)(x+2) <=> f(x)= a(x^2+ 2x- ½x- 1) <=> f(x)= ax^2+ 1½x-1

Then I isolate a, by inserting the x and y values of the third coordinate-point (1,-3) and I get:

-3= a(1-0.5)(1+2) <=> -3= a(1+2-0.5-1) <=> -3= a(1.5) <=> -3/1.5 = a <=> a= -2

So the final definition of the function is:

f(x) = -2ax^2 + 1½x - 1

But in the answer list in my text-book, it says the definition should be:

f(x)= 2x^2 + 3x- 2

Can someone please explain to me where I went wrong?

Regards
Lia
• Mar 24th 2011, 01:02 PM
topsquark
Quote:

Originally Posted by Lia85
Hi

Following cordinate-points are given to define a function:

(½,o), (-2,0) and (1,-3) with the parable roots being ½ and -2. Since d>0, I used the formula: f(x)= a(x-r1)(x-r2), where r1= first root and r2= second root.

So this is what I end up with:

f(x) = a(x-½)(x+2) <=> f(x)= a(x^2+ 2x- ½x- 1) <=> f(x)= ax^2+ 1½x-1

First, you should probably leave the $\displaystyle 1 \frac{1}{2}$ as 3/2.

Second in the line above the final entery should be $\displaystyle ax^2 + \frac{3}{2}ax - a$.

-Dan
• Mar 25th 2011, 03:26 AM
Lia85
Hi Dan

If a is the slope, then why does it appear thrice? The general formula for a quadratic is ax^2 + bx + c. But maybe it was just a typo? Or maybe there's something I'm just not seeing?

I'll be sure to rephrase 1½x to 3/2x, but even so, it still doesn't solve my problem and i still end up with the same expression: f(x) = -2ax^2 + 3/2x - 1, which is somewhat of a far cry from what the answer in my textbook is, namely
f(x)= 2x^2 + 3x- 2

(Worried)
• Mar 25th 2011, 04:13 AM
e^(i*pi)
When you used the formula $\displaystyle f(x) = a(x-r_1)(x-r_2)$ the a is out front so you need to expand it out. Indeed, it is not a slope, only straight lines have slope.

Do you understand that $\displaystyle f(x) = -2(x^2+1.5x-1) = (-2 \cdot x^2) + (-2 \cdot 1.5) + (-2 \cdot -1)$

Once simplified that is a factor of -1 out but I see no way to reconcile that (perhaps it's an even function?)
• Mar 25th 2011, 07:15 AM
Lia85
Hi e^(i*pi)

Quote:

Originally Posted by e^(i*pi)
When you used the formula $\displaystyle f(x) = a(x-r_1)(x-r_2)$ the a is out front so you need to expand it out. Indeed, it is not a slope, only straight lines have slope.

Yes you're right, a in this case is just a coefficient that appears infront of the quadratic term. I just got thrown away by assigning the same arbitrary letter to all of the constants. Sorry for the mix-up.

Quote:

Originally Posted by e^(i*pi)
Do you understand that $\displaystyle f(x) = -2(x^2+1.5x-1) = (-2 \cdot x^2) + (-2 \cdot 1.5) + (-2 \cdot -1)$

Yes I understand the point you're trying to make, but I don't know if it is applicable in this case.

The value of f(x)= -2x^2 + 3/2x - 1 is not conserved if you rephrase the expression to f(x)= (-2*x^2) + (-2*1.5x) + (-2* -1), because the coefficient -2 is only to be multiplied with the quadratic term, as seen in the original expression I ended up with. Had the expression been f(x)= -2(x^2 +1.5x -1), then the above method would be right.

The only way I can see how you might have arrived at f(x)= -2(x^2 +1.5x -1) is if maybe you extracted it from the formula f(x)= a(x-r1)(x-r2). If so could you please show me how, step by step?

I apologise in advance for my limited math-abilities, I just want to understand, that's all.

Thanks for the contributions
• Mar 26th 2011, 03:57 AM
Lia85
I got it now...
Having to do it the other way around, i.e. breaking down a predefined functions into factors, made me understand what both of you were trying to say. Thanks for the help... And yeah I guess the problem is solved, well except the negative sign infront of a... Oh well maybe it's an error in the textbook.