Question 2598
by pure factorisation? 


1) cannot be factorised easily. You would have to do the "complete the square" method.


2) {{{4x^2 - 12x + 9 = 0 }}}
(2x - 3)(2x - 3) = 0
so 2x-3 = 0
2x = 3
x = 3/2


3) 5x(x+2) = 4(3x+1) --> + or - on the last term, seeing as you have "="?

{{{5x^2 + 10x = 12x + 4}}}
{{{5x^2 - 2x - 4 = 0}}}

this does not factorise easily. Again, do you want the "complete the square" method?


4) {{{2x+7 = x^2}}}
{{{x^2 - 2x - 7}}}


Same as 3).


5) {{{x^2 - 6x - 7}}}  Again you have 2 "=" signs..which is wrong?


6) {{{12x-3 = x^2}}}
{{{x^2 - 12x + 3 = 0}}}


Again this does not factorise easily.


jon.