value of constant $C$ in cubic polynomial

Question

If $g(x)$ is a cubic polynomial with roots $p,q,r$ and $g'(x)$ divides $g(2x)$ then find value of $C$. where $Cpq=16r^2$

a solution I tried

$g(x)=A(x-p)(x-q)(x-r)$ and $g(2x)=A(2x-p)(2x-q)(2x-r)$

Help me how I solve that question

Answer 1

Let:

$$ g(x) = ax^3 + bx^2 + cx + d $$

with roots $r, p, q$.

$$ g'(x) = 3ax^2 + 2bx + c $$

$$ g(2x) = 8ax^3 + 4bx^2 + 2cx + d $$

If $g'(x)$ divides $g(2x)$ then:

$$ g(2x) = A(x-u)g'(x) $$

$$ 8ax^3 + 4bx^2 + 2cx + d = A[(x-u)(3ax^2 + 2bx + c)] $$

$$ 8ax^3 + 4bx^2 + 2cx + d = A3ax^3 + Ax^2(2b-3au) + Ax(c-2bu) - Acu $$

Comparing coefficients:

$$ 8a = 3Aa \implies A = \frac{8}{3} $$

$$ 4b = \frac{8}{3} (2b-3au) \implies b = \frac{4}{3}b - 2au \implies -6u = -\frac{b}{a} $$

$$ 2c = \frac{8}{3}(c-2bu) \implies \frac{3}{4}c = c-2bu \implies 2bu = \frac{c}{4} \implies -8u \frac{-b}{a} = \frac{c}{a}$$

$$ d = -\frac{8}{3}cu \implies -\frac{d}{a} = \frac{8u}{3} \frac{c}{a}$$

Note that:

$$\frac{-b}{a} = p+q+r$$ $$\frac{c}{a} = pq+pr+rq $$ $$\frac{-d}{a} = pqr $$

Thus:

$$ p +q +r = -6u $$ $$pq + pr + rq = 48u^2 $$ $$prq = 128u^3 $$

If you solve this, you will get $C$.