# Am I crazy or are these quadratics really not factorable??

• Feb 2nd 2010, 12:46 PM
dcd5105
Am I crazy or are these quadratics really not factorable??
My professor insists these quadratic equations can be factored, she’s asking us to graph them as well. Maybe its just me...

$\displaystyle 2x^2-3x-5=0$

&

$\displaystyle x^2-4x+2=0$

Any suggestions?
• Feb 2nd 2010, 12:55 PM
e^(i*pi)
Quote:

Originally Posted by dcd5105
My professor insists these quadratic equations can be factored, she’s asking us to graph them as well. Maybe its just me...

$\displaystyle 2x^2-3x-5=0$

&

$\displaystyle x^2-4x+2=0$

Any suggestions?

To find out if a quadratic can be factored calculate the discriminant, $\displaystyle \Delta$.

$\displaystyle \Delta = b^2-4ac$

If $\displaystyle \Delta$ is a perfect square it can be factorised.

-------------------------

$\displaystyle (-3)^2-(4)(2)(-5) = 49$ ------- This can be factored

$\displaystyle (-4)^2-(4)(1)(2) = 8$ --------- This cannot be factored
• Feb 3rd 2010, 04:39 AM
HallsofIvy
Quote:

Originally Posted by dcd5105
My professor insists these quadratic equations can be factored, she’s asking us to graph them as well. Maybe its just me...

$\displaystyle 2x^2-3x-5=0$

&

$\displaystyle x^2-4x+2=0$

Any suggestions?

Any quadratic "can" be factored, just not necessarily with integer coefficients.
For $\displaystyle 2x^2- 3x- 5$, by the quadratic formula, its zeros are $\displaystyle \frac{3\pm\sqrt{9+ 40}}{6}= \frac{3\pm 7}{4}$ or 5/2 and -1. $\displaystyle 2x^2- 3x- 5= 2(x- 5/2)(x+ 1)= (2x- 5)(x+ 1)$

For $\displaystyle x^2- 4x+ 2$ we can complete the square: $\displaystyle x^2- 4x+ 2= x^2- 4x+ 4- 4+ 2= (x- 2)^2- 2= (x-2+ \sqrt{2})(x-2-\sqrt{2})$. the last step is factoring the "difference of two squares", (x-2)^2 and $\displaystyle (\sqrt{2})^2$