factor
b(x+2) + c(x+2)
(b+c)(x+2)
need more info for futher analysis.
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ax3 + bx2 + cx x(ax2 + bx + c) you get one answer as 0.
y= ax^3+bx^2+cx-42, assuming the point was (0, 42)
It is: CX = 110
x = cx = aa+1 If x = cx, then c = 1 aa = a2 if a2 + 1 = x, then a2 = x-1 short of parametrization, this is the answer for this equation, but if you doing advanced maths, then let x= t (teR) a2 = -1 +t (teR)
By inspection, the sum of the coefficients is 1 - 5 + 4 = 0Therefore x = 1 is a root. That is (x - 1) is a factor.Then suppose the other factor is ax^2 + bx + cSInce the coefficient of x^3 is 1, then a = 1 so that the second factor is x^2 + bx + cTherefore (x - 1)*(x^2 + bx + c) = x^3 - 5x + 4x^3 - x^2 + bx^2 - bx + cx - c = x^3 - 5x + 4Comparing coefficient of x^2 gives -1 + b = 0 so that b = 1Comparing coefficient of x gives -b + c = -5 so that c = b - 5 = -4Comparing constant terms gives -c = 4 or c = -4 which confirms above calculations.Thus, the second factor is (x^2 + x - 4) which does not have rational factors.