Two roots of $4x^3+8x^2+Kx-18=0$ are equal numerically but opposite in value. Find the value of K.

Question

Two roots of $4x^3+8x^2+Kx-18=0$ are equal numerically but opposite in value. Find the value of K.

I've tried plugging in two roots $x_0$ and $-x_0$, to get the relationship $k=-4x^2$. Plugging it back in doesn't achieve anything, I'm not sure how to proceed, I have seen people do this with the quadratic, but similarly trying to force this answer is too complicated, this is an SAT II Level 2 question and should be solvable in under a minute!

Answer 1

If $a$ is one of the two opposite roots, then $$ \begin{align} (x^2-a^2)(4x-r)&=4x^3+8x^2+Kx-18\\ 4x^3-rx^2-4a^2x+a^2r&=4x^3+8x^2+Kx-18 \end{align} $$ The coefficient of $x^2$ reveals the value of $r$. The constant term then reveals the value of $a^2$. And then the linear coefficient reveals the value of $K$.

Answer 2

Hint: by Vieta's relations the sum of all three roots is $-8/4=-2\,$. Now, if two of them sum to $0$ $\ldots$

Answer 3

Let the roots be $a, - a, b$. Then

$$4x^3+8x^2+Kx-18=4(x^2-a^2)(x-b)$$

So, $-4b=8$, $-4a^2=K$ and $4a^2b=-18$.

Answer 4

Reducing the polynomial to a monic one by extracting a factor of $ \ 4 \ $ produces $ \ 4·\left(x^3 + 2x^2 + \frac{K}{4}x - \frac92 \right) \ . \ $ The product of the roots $ \ -r^2·s \ $ is $ \ \frac92 \ $ and the sum of the three pairs of roots is $ \ -r^2 + rs - rs \ = \ -r^2 \ = \ \frac{K}{4} \ \ , \ $ so the sum of the reciprocals of the roots is $$ \ \frac{1}{r} \ + \ \frac{1}{-r} \ + \ \frac{1}{s} \ \ = \ \ \frac{1}{s} \ \ = \ \ \frac{-r^2}{-r^2·s} \ \ = \ \ \frac{K / 4}{9 / 2} \ = \ \frac{K}{18} \ \ \Rightarrow \ \ K·s \ = \ 18 \ \ . $$

As mentioned by the other posters, the sum of the roots is $ \ r - r + s \ = \ -2 \ \ , \ $ hence $ \ K \ = \ -9 \ \ . $

Synthetic or polynomial division, or even just "factoring by grouping", shows the factors to be $ \ (4x^3 + 8x^2) \ - \ (9x^2 + 18) \ = \ (x + 2)·(4x^2 - 9) \ = \ 4·(x + 2)·\left( x + \frac32 \right)·\left( x - \frac32 \right) \ \ . $