Re: Help with an equation please

thanks for your answers but is there as simpler way to do it, as
solving cubic equations is not part of my course?

ill give you the full question as this may help

A tangent at A(1,2) to the curve y = x^3 + x^2 meets the curve again
at B

Show that the equation of AB is y = 5x - 3

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dy/dx =  3x^2 + 2x
at A(1,2) , gradient = 3(1)^2 + 2(1) = 5

so y - 2 = 5 ( x - 1)

y = 5x - 3

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Show that the tangent and curve meet where x^3 + x^2 - 5x + 3 = 0

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(1^3) + (1^2) - 5(1) + 3

1 + 1 - 5 + 3 = 0

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Find the co-ordinates of B by solving the equation  x^3 + x^2 - 5x + 3
= 0 ( Note that A lies on the curve )

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so i already that x = 1 in one solution, is there an easier way to
find x = -3 ?

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