It is not a quadratic, but rather a cubic equation
Trying for f(x) = 0 using ±1, ±2, ±3
I found f(3) = 27 - 63 + 30 + 6 = 0
so x-3 is a factor
Using synthetic division I found
f(x) = (x-3)(x^2 - 4x - 2)
I will leave you with solving the quadratic
x^2 - 4x - 2 = 0 to get the other two roots.
hint: use the formula, since it does not factor
you should get (2 ± √6)
f(x)=x^3-7x^2+10x+6 (find the zeros of this quadratic equation)
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