Anonymous

f(x) = 3x^2 +x -6. Solve f(x) = 0. 6x = -1(+/-)D, where D^2 = 1 + 4*3*6 = 73. Then D = rt73 and f(x) = (1/12)(6x+1-rt73)(6x+1+rt73).

Anonymous

3x^2 + x - 6 = x(3x + 1) - 6

Anonymous

3x^2+x-6= 3(x^2+x/3+(1/6)^2-(1/6)^2)-6= 3(x+1/6)^2-1/12-6= 3(x+1/6)^2-(sqr(73)/12)^2= [sqr(3)(x+1/6)+sqr(219)/6]* [sqr(3)(x+1/6)-sqr(219/6]= [sqr(3)x+sqr(3)(1+sqr(73))/6]* [sqr(3)x+sqr(3)(1-sqr(73))/6]= 3[x+(1+sqr(73))/6]* [x+(1-sqr(73))/6]

Anonymous

3x^2 +x -6 There are no rational factors Dear Argent (3x-2)(x+3).=3x²+7x -6 and not 3x²+x-6 as given

Anonymous

(3x-2)(x+3) is NOT correct. Try expanding it! 3x² + 7x - 6 Complete the square. The answer is not whole numbers as lenpol7 explains and is (approx according to my slide rule) 3(x + 1.59)(x - 1.26) Check: 3(x² + 0.333 - 2) 3x² + x - 6