Q1) P(x) = (x-2)(x+2)(2x-3)(x+6) = 0

.: 4 roots are -6, -2, 2 and 1.5

How did I figure this out? Well I first tried the expected conventional approach of letting the 4 roots to be: -a, a, b and c and setting up the system of equations involving the coeffs of terms in P(x). I found, on first attempt, that the equations were relatively messy to solve; maybe I did not look at things the right way. So I changed tack.

I did the trial-&-error approach of seeking zeros of P(x), trying in succession x = 1, -1, 2 .... I found P(2) = 0. So I tried out x = -2; (to determine if 2 & -2 are the pair of roots) again I got 0. So P(x) has (x-2) and (x+2) , ( therefore x^{2}- 4) as factors. By inspection or by long division, I found P(x) = (x^{2}- 4)(2x^{2}+ 9x - 18). You can now solve for the 2nd quadratic eqn.

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