Maths homework

Okay, so we know (x-1) is a factor of  $$x^3+3x^2+x-5$$ .

Let's say we want to find the c in  $$x^3+3x^2+x-5 = (x-1)c$$

We want  $$x^3$$  so we know some part of c is  $$x^2$$

Let's write  $$x^3 + 3x^2 + x -5 = (x-1)(x^2+....)$$

We now have  $$x^3$$  but we also have  $$-x^2$$  because  $$(x-1)(x^2+...) = x^3 - x^2+...$$

We want  $$3x^2$$  so let's add  $$4x$$  which multiplies to  $$4x^2$$  and  $$4x^2-x^2 = 3x^2$$

Now we have  $$(x-1)(x^2+4x + ..) = x^3+3x^2-4x+..$$

we want  $$x$$  instead of  $$-4x$$  so we add  $$5$$  which multiplies to  $$5x$$

Now we have  $$(x-1)(x^2+4x+5) = x^3 -x^2 + 4x^2-4x + 5x - 5 = x^3 +3x^2+x-5$$

Which was what we were looking for.

$$x^2+4x+5$$  can't be further factorized so  $$(x-1)(x^2+4x+5)$$  is the final answer

Reinout 

I got the same result as you Reinout, and given part (b) of the question it was to be expected that the cubic had just one real root. (Differentiation of the quartic in (b) gets you 4 times the cubic).

For the second question

$$y = x^4 + 4x^3 + 2x^2 - 20x + 3$$

has as derivative

$$\frac{dy}{dx} = 4x^3+12x^2+4x-20 = 4(x^3+3x^2+x-5)$$

I just found

$$x^3 + 3x^2 + x - 5 = (x-1)(x^2+4x+5)$$

Which is  $$0$$  only for  $$x = 1$$

This is therefore the only stationairy point

$$y=x^4+4x^3+ 2x^2-20x+3 = 1^4+4*1^3+2*1^2-20*1+3 = 1 + 4 + 2 - 20 + 3 = -10$$

So the stationairy point is  $$(x,y) = (1,-10)$$

Reinout 

p.s. Thanks Bertie

Thank you Reinour-g and Bertie