# factorising query

• Nov 12th 2005, 08:55 AM
x-disturbed-x
factorising query
factorise: 13x^2 + 14x - 27 = 0

(x-27)(13x+1) is this correct???
• Nov 12th 2005, 09:50 AM
CaptainBlack
Quote:

Originally Posted by x-disturbed-x
factorise: 13x^2 + 14x - 27 = 0

(x-27)(13x+1) is this correct???

$(x-27)(13x+1) \ \ \ = \ x(13x+1) -27(13x+1)$
$=\ 13x^2+x\ -351x-27$
$=\ 13x^2 -350x-27$

RonL
• Nov 12th 2005, 10:20 AM
Jameson
Quote:

Originally Posted by x-disturbed-x
factorise: 13x^2 + 14x - 27 = 0

(x-27)(13x+1) is this correct???

If you can't factor it just by spotting the zeroes, how else might you find the zeroes and put them in factored form?
• Nov 16th 2005, 09:35 PM
Fubini
Re: 13x^2 + 14x -27 = 0
The quadratic formula yields the roots of this degree 2 polynomial over the integers.

x= +/- square root [14^2 - 4 (-27)13] - 14 / 13x2

x= +/- 40 - 14 / 26

x= 1 or x = -27/13

Hence the polynomial factorises into (x-1)(x+ 27/13)=0
or (x-1)(13x + 27)=0.

QED (used for proofs but you get the picture).