The cubic polynomial x^3-2x^2-x-6 is denoted by f(x). Show that (x-3) is a factor of f(x). Factorize F(x). Hence find the number of real roots of the equation f(x) = 0. Justifying your answer.

Hence write down the number of points of intersection of the graphs with equations
y = x^2-2x-1 and y = 6/x
justifying your answer.

Question

The cubic polynomial x^3-2x^2-x-6 is denoted by f(x). Show that (x-3) is a factor of f(x). Factorize F(x). Hence find the number of real roots of the equation f(x) = 0. Justifying your answer.

Hence write down the number of points of intersection of the graphs with equations
     y = x^2-2x-1         and        y = 6/x
justifying your answer.

javk · Answer

after factorization I got $$(x-3)(x ^{2}+x+2)$$
there is only one real root.

I am having trouble with the last part of the question where I find the number of points of intersection between the graphs.
how should I proceed when the question says `hence`

mww · Answer

The second part, hence indicates your result should lead into the first question.

Remember for two equations to intersect, you solve simultaneously and the solutions.

  y = x^2-2x-1         and        y = 6/x
x^2-2x-1 = 6/x
This eqn is readily convertible into a cubic. See that the original equation is restored. Just note the solutions to x^2-2x-1 = 6/x will obviously not include x = 0, but otherwise are the same solutions as that of x^3-2x^2-x-6