y = (x+p)(x^2 - mx + n)

Let y = u*v such that:

u = (x+P) ==> u'=1

v = (x^2 - mx + n) ==> v' = 2x -m

dy = (u'v-uv')/v^2

= (x^2 -mx + n - (x+P)(2x-m)]/(x^2 - mx + n)^2

= (x^2 - mx + n -(2x^2 - mx + 2px - mp) /(x^2 - mx + n)^2

= (-x^2 - 2px + n + mp)/(x^2 - mx + n)^2

dy = 0

==> -x^2 - 2px + (n-mp) = 0

==> x^2 + 2px + (mp-n) = 0

y = (x+p)(x^2-mx+n)

Therefore .

We know y' = {u(x)v(x)}' = u'(x)v(x)+u(x)v'(x).............(1).

y = u(x)v(x), where,

u(x) =x+p implies u'(x) = (x+p)' = 1 and

v(x) = x^2-mx+n implies

v'(x) = (x^2-mx+n)' = (x^2)' +(-mx)' +(n)' = 2x-m.

Substitute in the formula (1):

y' = u'(x)v(x) +u(x)v'(x) = 1(x^2-mx+n) +(x+p)(2x-m)....(2)

If dy = 0, the dy/dx = y' = 0. So the expression at (2) is zero:

(x^2-mx+n)+(x+p)(2x-m) = 0

x^2-mx+n +(2x^2-mx+2px-mp) = 0

3x^2 +(-2mx+2px) +n-p

3x^2+2(p-m)x+n-p = 0. This is a quadratic equation in x. So we use the formula:

x1 = {-2(p-m) +sqrt[4(p-m)^2-4*3*(n-p)]}/(2*3)

x1 = {m-p +sqrt[(m-p)^2-3(n-p)]}/2

x2 = {m-p -sqrt[(m-p)^2 - 3(n-p)]}/3

For the beginning, let's differentiate the given function.

dy=(x+p)(x^2-mx+n)dx

Since the function is a product, we'll apply the product rule, when differentiating a product.

(f*g)' = f'*g + f*g'

We'll differentiate, to the right side, with respect to x:

[ (x+p)(x^2-mx+n) ]' = (x+p)' * (x^2-mx+n) + (x+p) * (x^2-mx+n)'

[ (x+p)(x^2-mx+n) ]' = x * (x^2-mx+n) + (x+p) * (2x-m)

We'll remove the brackets:

[ (x+p)(x^2-mx+n) ]' = x^3 - mx^2 + nx + 2x^2 - mx + 2px - mp

Now, we'll put dy = 0

We'll substitute the expression for dy:

x^3 - mx^2 + nx + 2x^2 - mx + 2px - mp = 0

We'll combine like terms:

x^3 + x^2*(-m+2) + x*(n-m+2p) - mp = 0

If we'll plug in values for the m, n, p, we'll calculate the values of x for dy = 0.