factorise: 13x^2 + 14x - 27 = 0
(x-27)(13x+1) is this correct???
$\displaystyle (x-27)(13x+1) \ \ \ = \ x(13x+1) -27(13x+1) $Originally Posted by x-disturbed-x
$\displaystyle =\ 13x^2+x\ -351x-27$$\displaystyle =\ 13x^2 -350x-27$
So the answer is no.
RonL
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).