# Thread:

1. ## Need some help with the method

I've got an equation: 4x^3 - 24x^2 + 45 x - 23

A tangent is drawn at point P. (1,2)

dy/dx = 12x^2 - 48x +45

At f´(1) = 12 -48 + 45 = 9

2 = 9(1) +c

y = 9x - 7

Show that the x coordinate of point q is 4.

9x-7 = 4x^3 - 24x^2 + 45 x - 23

0 = 4x^3 - 24x^2 + 36 x - 16

I found x = 4 using polyrootfinder on my gdc and was wondering if there is any other way of coming to that answer?

2. I imagine $q$ is the point where the tangent cross the curve, doesn't it ?

Your method is good, but you can solve the equation if you notice that $x=1$ is solution, then put $(x-1)$ in factor..

3. That is correct. I see what you mean however I'm quite unfamiliar with factorisation with 3rd degree polynomials (I'm still a middle school student :S). Could you care to elaborate?

4. Of course !

$P(x)=2x^3-12x^2+18x-8$

You have to test some values such as 1, 2, 3, -1, -2, -3.

Indeed here : $P(1)=0$ so we can put $(x-1)$ in factor :

$P(x)=(x-1)(ax^2+bx+c)$

which we develop : $P(x)=ax^3+(b-a)x^2+(c-b)x-c$

And we identify each coefficient $\left\{\begin{array}{l}a=2\\b-a=-12\\c-b=18\\c=8\end{array}\right.$

After resolution we get : $\left\{\begin{array}{l}a=2\\b=-10\\c=8\end{array}\right.$

Hence $P(x)=(x-1)(2x^2-10x+8)$ and you know the factorisation 2nd degree polynomials.

Is my english correct ?

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