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To tackle this question we start by thinking about how many brackets will we need to factorise this equation. As there are three parts to the equation it is most likely we are looking for two brackets. Knowing this we now position an x in the start of each bracket like so: (x+-a)(x+-b) where a and b are constants. This means when we multiply out the brackets we get x^{2} +/- (a+b)x +/- ab. THe next stage is to look for two factors of 6, so we will list all of the factors you can in this case as a process of elimination as it is a small number so there is 6 and 1 or 2 and 3. Having these as our options for a and b we look at the second part of the original equation which is a constant of x. In our equation the constant is minus 1. Knowing that our a and b must make -1 when combined in an addition; we can deduce that we want 3 and 2 as our a and b. The order we but these into the brackets at this point does not matter as we are yet to have chosen our subtraction or addition symbols. So far, our answer looks like this: (x +/- 3)(x +/- 2)To find our symbols for each bracket we turn to our original equation of which we are factorising, both constants are negative meaning only one of our brackets can be negative. Therefore we know through expansion that our factorisation must have -3 and +2 in order to obtain the original equation when expanded. our final answer is: (x - 3)(x + 2)This question could be asked on its own or as the start of a chain of questions based on this equation.