2x^2+5x+1

can this be factorised?

Printable View

- Jan 21st 2006, 01:39 PMx-disturbed-xfactorising
2x^2+5x+1

can this be factorised? - Jan 21st 2006, 01:43 PMTD!
Sure it is, but since you're asking I'm assuming that you mean over the reals and probably without using radicals. In that case, no.

- Jan 21st 2006, 01:51 PMx-disturbed-x
it must! there's a remainder of 3 from the expression i divided by (x-1)

but it's not part of the quadratic.. - Jan 21st 2006, 01:52 PMCaptainBlackQuote:

Originally Posted by**x-disturbed-x**

$\displaystyle 2x^2+5x+1=2(x-a)(x-b)$,

for some $\displaystyle a$ and $\displaystyle b$. But when $\displaystyle x=a$, or $\displaystyle x=b$

the RHS of the last equation is zero.

Therefore $\displaystyle b$ and $\displaystyle c$ are the roots of the quadratic

on the LHS.

The quadratic formula will give the roots of $\displaystyle 2x^2+5x+1$,

which will then allow you to factorise this quadratic.

RonL - Jan 21st 2006, 02:08 PMx-disturbed-x
captainblack! you scare me with talk of RHS. there must be an easier way :eek:

- Jan 21st 2006, 02:11 PMMathGuru
Actually it's pretty simple simply plug in the values of

A=2

B=5

C=1

into the quadratic equation - Jan 21st 2006, 02:11 PMTD!
A quadratic equation always has 2 roots, if the discriminant is positive, you have two distinct real zeroes, let's call them a and b. You can then always factor (x-a)(x-b). You can find a and b by using the quadratic formula, as CaptainBlack suggested.