Question

How do I factor x^2-3x-1?

Answer 1

not factorable over the integers

Answer 2

hey myininaya is right

Answer 3

so then how do I do x^2-3x-1/4x^2-9?

Answer 4

you can always complete the square

Answer 5

Because my original question from the book is:

2x^2+3x+1 x^2-3x-1
____________ - ________________
4x^2-9 -(4x^2-9)

Answer 6

I did:

(2x+1)(x+1)
____________ + ______
4x^2-9 4x^2-9

Answer 7

...

Answer 8

but it can still be factored
\[b=(-\frac{3}{2}+z)+(-\frac{3}{2}-z)\]
\[a*c=(\frac{-3}{2}+z)(\frac{-3}{2}-z)=-1\]
so what is z
\[\frac{9}{4}-z^2=-1 =>\frac{9}{4}+1=z^2=>z^2=\frac{13}{4}=>z=\pm \frac{\sqrt{13}}{2}\]
so we have
\[x^2+(\frac{-3}{2}+ \frac{\sqrt{13}}{2})x+(\frac{-3}{2}- \frac{\sqrt{13}}{2})x-1\]
\[x(x+\frac{-3 + \sqrt{13}}{2})+\frac{-3- \sqrt{13}}{2}(x-1 \cdot \frac{2}{-3- \sqrt{13}})\]
\[x(x+\frac{-3 + \sqrt{13}}{2})+\frac{-3- \sqrt{13}}{2}(x-\frac{2}{-3- \sqrt{13}} \cdot \frac{-3 + \sqrt{13}}{-3+ \sqrt{13}})\]
\[x(x+\frac{-3+ \sqrt{13}}{2})+\frac{-3 - \sqrt{13}}{2}(x-\frac{2(-3 + \sqrt{13})}{9-13})\]
\[x(x+\frac{-3+ \sqrt{13}}{2})+\frac{-3 - \sqrt{13}}{2}(x-\frac{2(-3 + \sqrt{13})}{-4})\]
\[x(x+\frac{-3 + \sqrt{13}}{2})+\frac{-3- \sqrt{13}}{2}(x+\frac{-3+ \sqrt{13}}{2})\]
\[(x+\frac{-3+\sqrt{13}}{2})(x+\frac{-3-\sqrt{13}}{2})\]

Answer 9

2x^2+3x+1 x^2-3x-1
____________ - ________________
4x^2-9 -(4x^2-9)

\[\frac{2x^2+3x+1}{4x^2-9}+\frac{x^2-3x-1}{4x^2-9}\]

Answer 10

you have same denominator
add the numerators

Answer 11

ohh, okay.