Solving simultaneous equations with multiple variables.

Question

If $(x+a)^2$ is a factor of $x^3 + 6px + k$, show that $k + 2a^3 = 0$

I've tried different ways to solve this, but is it even possible?

Answer 1

Original answer: using differentiation.

The fact that $p(x)=x^3+6px+k$ is divisible by $x+a$ at all means that $p(-a)=0$, i.e. $-a^3-6pa+k=0$.

The fact that $p(x)$ is divisible by $(x+a)^2$ also means that $p'(x)=3x^2+6p$ is divisible by $x+a$, i.e. $p'(-a)=3a^2+6p=0$.

Thus, from the second equation we get $6p=-3a^2$, and by substituting this in the first, we get $-a^3+3a^2\cdot a+k=0$, i.e. $2a^3+k=0$, q.e.d.

Update: Alternative solution which does not use differentiation.

Let us extract the factor $(x+a)^2=x^2+2ax+a^2$ from the polynomial $x^3+6px+k$. The remainder should be zero:

$$\begin{array}{rcl}x^3+6px+k&=&x(x^2+2ax+a^2)+(-2a)x^2+(-a^2+6p)x+k\\&=&x(x^2+2ax+a^2)+(-2a)(x^2+2ax+a^2)+4a^2x+2a^3+(-a^2+6p)x+k\\&=&(x-2a)(x^2+2ax+a^2)+(3a^2+6p)x+(2a^3+k)\end{array}$$

So, the quotient in division of $x^3+6px+k$ with $(x+a)^2$ is $x-2a$, and the remainder, which must be $0$, is $(3a^2+6p)x+(2a^3+k)$. The last polynomial (of degree $1$) is zero only if its coefficients are zero, which lets us conclude that $3a^2+6p=0$ and, most importantly, $2a^3+k=0$, as desired.

Answer 2

Let $p(x)$ denote the 3rd degree polynomial.

Your condition means that $-a$ is a double root of $p$, therefore it is also a root of the derivative $p(x)=3x^2+6p$.

As a consequence, we have :

$$\begin{cases}p(a)&=&-a^3-6pa+k&=&0\\p'(a)&=&3a^2+6p&=&0 \end{cases}$$

from which you easily derive your constraint.

Answer 3

Since $(x+a)^2$ divides $x^3 + 6px + k$ exactly, the cubic polynomial must be of the form $(x+a)^2(x - b)$. Since it has no $x^2$ term, the sum of the roots must be zero. So $(-a) + (-a) + b = 0$, hence $b = 2a$.

Therefore we have $x^3 + 6px + k = (x+a)^2(x-2a)$. The constant term $k$ must be equal to the product of the constant terms on the right. So $k = (a)(a)(-2a) = -2a^3$.