# factorising

• Jan 21st 2006, 01:39 PM
x-disturbed-x
factorising
2x^2+5x+1

can this be factorised?
• Jan 21st 2006, 01:43 PM
TD!
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 PM
x-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 PM
CaptainBlack
Quote:

Originally Posted by x-disturbed-x
2x^2+5x+1

can this be factorised?

Suppose it can be factorised then we would have:

\$\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 PM
x-disturbed-x
captainblack! you scare me with talk of RHS. there must be an easier way :eek:
• Jan 21st 2006, 02:11 PM
MathGuru
Actually it's pretty simple simply plug in the values of

A=2
B=5
C=1