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

**dcd5105** · February 2nd 2010, 12:46 PM

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

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

&

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

Any suggestions?

**e^(i*pi)** · February 2nd 2010, 12:55 PM

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...

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

&

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

Any suggestions?

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

$\Delta = b^2-4ac$

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

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

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

$(-4)^2-(4)(1)(2) = 8$ --------- This cannot be factored

**HallsofIvy** · February 3rd 2010, 04:39 AM

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...

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

&

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

Any suggestions?

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

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

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

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