Need help factoring

**newbie43** · November 19th 2009, 06:58 PM

Hi
I was wondering if anyone could help me with this equation i have, im trying to factor this in my ti92+ calculator and im not seeing the textbook result my book shows. Im told to factor this

1/9*y^2 - 1/6*y*x^2 + 1/16*x^4

the answer the book gives is

(1/3*y - 1/4*x^2)^2

is there anything special i need to do to factor this equation.
The calculator gives me this as the answer

(3*x^2 - 8)^2 / 144

I know if i factor it by hand i can get the proper answer but i have alot of these to do and im not sure what im doing wrong. Ive also tried this in the sage math application and it didnt show the proper form either.

Any help would be appreciated thanks.

**Amer** · November 19th 2009, 07:22 PM

Originally Posted by newbie43

Hi
I was wondering if anyone could help me with this equation i have, im trying to factor this in my ti92+ calculator and im not seeing the textbook result my book shows. Im told to factor this

1/9*y^2 - 1/6*y*x^2 + 1/16*x^4

the answer the book gives is

(1/3*y - 1/4*x^2)^2

is there anything special i need to do to factor this equation.
The calculator gives me this as the answer

(3*x^2 - 8)^2 / 144

I know if i factor it by hand i can get the proper answer but i have alot of these to do and im not sure what im doing wrong. Ive also tried this in the sage math application and it didnt show the proper form either.

Any help would be appreciated thanks.

1/9*y^2 - 1/6*y*x^2 + 1/16*x^4

$\frac{y^2}{9} - \frac{yx^2}{6} + \frac{x^4}{16}$

$\left(\frac{1}{9}\right)\left(y^2 - \frac{yx^2 \cdot 9}{6} + \frac{9x^4}{16}\right)$

you can consider that x is a constant and apply the general law (I do not know what is the mathematics name)

$y = \frac{-b \mp \sqrt{b^2 -4ac}}{2a}$

you have
$a=1 , b=\frac{-9x^2}{6} , c=\frac{9x^4}{16}$

$y= \frac{\frac{9x^2}{6} \mp \sqrt{\left(\frac{9x^2}{6}\right)^2-4(1)\left(\frac{9x^4}{16}\right)}}{2}$

$y=\frac{\frac{9x^2}{6}\mp 0 }{2} = \frac{\frac{3x^2}{2}}{2}$

$y=\frac{3x^2}{4}$

so

$y^2 - \frac{yx^2 \cdot 9}{6} + \frac{9x^4}{16} = \left(y - \frac{3x^2}{4} \right)^2$

but

$\left(\frac{1}{9}\right)\left(y^2 - \frac{yx^2 \cdot 9}{6} + \frac{9x^4}{16}\right) = \left(\frac{1}{9} \right)\left(y-\frac{3x^3}{4}\right)$

$\left(\frac{1}{9}\right)\left(y^2 - \frac{yx^2 \cdot 9}{6} + \frac{9x^4}{16}\right) = \left(\frac{y}{3} - \frac{x^2}{4}\right)^2$

**newbie43** · November 20th 2009, 06:39 AM

Thanks very much for the help.

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