If $x^2 + kx + 1$ be a factor of $ax^4 + bx^3 + c,$ prove that $(a + c)(a - c)^2 = b^2c.$

Question

I don't know what method to approach to get the required proof.

Answer 1

If $x^2 + kx + 1$ is a factor, we can write:

$ax^4+bx^3+c = (x^2+kx+1)(\gamma_2x^2+\gamma_1x+\gamma_0)$ and solve for $\gamma_0, \gamma_1, \gamma_2$

Multiplying out the expressions:

$ax^4+bx^3+c = \gamma_2x^4 + (k\gamma_2+\gamma_1)x^3 + (\gamma_2+k\gamma_1+\gamma_0)x^2 + (\gamma_1+\gamma_0k)x+\gamma_0$.

So we must have $\gamma_2 = a$ (equate $x^4$ terms) and $\gamma_0 = c$ (equate constant terms)

From which it follows that: $\gamma_1 = -\frac{c+a}{k} = -ck$ (first equality from equating the $x^2$ terms (one is $0$), second from equating the the $x$ terms, one is $0$)

and that $k^2 = 1+\frac{a}{c}$ (using the two equivalent expresssions of $\gamma_1$ to solve for $k^2$)

and that $b = k(a-c)$ (equating the $x^3$ terms and substituting in $-ck$ for $\gamma_1$)

Then using all of this we have the following equivalences...

$$b^2c = k^2(a-c)^2c = (1+\frac{a}{c})(a-c)^2(c) = (c+a)(a-c)^2$$

as desired.