factorise: 13x^2 + 14x - 27 = 0

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

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

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

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

$\displaystyle =\ 13x^2+x\ -351x-27$$\displaystyle =\ 13x^2 -350x-27$

So the answer is no.

RonL - Nov 12th 2005, 10:20 AMJamesonQuote:

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

- Nov 16th 2005, 09:35 PMFubiniRe: 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).