# Factor

• Sep 15th 2011, 02:17 PM
habibixox
Factor
3x^2+5x-7

I want to make sure this is an error in the book.
You can't factor this right?
• Sep 15th 2011, 02:19 PM
skeeter
Re: Factor
Quote:

Originally Posted by habibixox
3x^2+5x-7

I want to make sure this is an error in the book.
You can't factor this right?

$\displaystyle b^2 - 4ac = 25 - 4(3)(-7) = 109$ ... not a perfect square, therefore will not factor over the rationals.
• Sep 15th 2011, 02:26 PM
habibixox
Re: Factor
thanks :)
• Sep 15th 2011, 02:34 PM
Plato
Re: Factor
Quote:

Originally Posted by habibixox
3x^2+5x-7
I want to make sure this is an error in the book.
You can't factor this right?

I can factor it,
$\displaystyle \left( {x - \left[ {\frac{{ - 5}}{6} + \frac{{\sqrt {109} }}{6}} \right]} \right)\left( {x - \left[ {\frac{{ - 5}}{6} - \frac{{\sqrt {109} }}{6}} \right]} \right)$
• Sep 16th 2011, 05:43 AM
Prove It
Re: Factor
Quote:

Originally Posted by Plato
I can factor it,
$\displaystyle \left( {x - \left[ {\frac{{ - 5}}{6} + \frac{{\sqrt {109} }}{6}} \right]} \right)\left( {x - \left[ {\frac{{ - 5}}{6} - \frac{{\sqrt {109} }}{6}} \right]} \right)$

I certainly hope you mean $\displaystyle \displaystyle 3\left[x - \left(-\frac{5}{6} + \frac{\sqrt{109}}{6}\right)\right]\left[x - \left(-\frac{5}{6} - \frac{\sqrt{109}}{6}\right)\right]$...

\displaystyle \displaystyle \begin{align*} 3x^2 + 5x - 7 &= 3 \left( x^2 + \frac{5}{3}x - \frac{7}{3} \right) \\ &= 3 \left[ x^2 + \frac{5}{3}x + \left( \frac{5}{6} \right)^2 - \left(\frac{5}{6}\right)^2 - \frac{7}{3}\right] \\ &= 3\left[\left(x + \frac{5}{6}\right)^2 - \frac{25}{36} - \frac{84}{36}\right] \\ &= 3\left[\left(x + \frac{5}{6}\right)^2 - \frac{109}{36}\right] \\ &= 3\left[\left(x + \frac{5}{6}\right)^2 - \left(\frac{\sqrt{109}}{6}\right)^2\right] \\ &= 3\left(x + \frac{5}{6} - \frac{\sqrt{109}}{6}\right)\left(x + \frac{5}{6} + \frac{\sqrt{109}}{6}\right) \end{align*}