Cubic polynomial without $x^1$

Question

I factorized this equation

$4x^3+6x^2-1=0$

I think it's harder to factorize when the $x$ (degree one) doesn't exist here. So, this is what I did now:

$2x^2(2x+3)-1=0$

But the factorized equation looks like this: $(2x+1)(2x^2+2x-1)=0$

What happens in between these last steps?

I did synthetic division for the root $-1/2$, but how do I get the other two roots?

Answer 1

HINT

I think you are looking for the following result.

Based on it, we find out that $-1/2$ is a rational root.

Taking advantage of the knowledge of such solution, we can factor the given polynomial as next: \begin{align*} 4x^{3} + 6x^{2} - 1 = 0 & \Longleftrightarrow (4x^{3} + 2x^{2}) + (4x^{2} - 1) = 0\\\\ & \Longleftrightarrow 2x^{2}(2x + 1) + (2x + 1)(2x - 1) = 0\\\\ & \Longleftrightarrow (2x^{2} + 2x - 1)(2x + 1) = 0 \end{align*}

Can you take it from here?

Answer 2

as $0$ is not a root, we lose nothing by writing $x= \frac{1}{t},$ then multiplying the result by $ \; \; - t^3 \; . \; \; $ $$ t^3 - 6t -4 $$ If it factors, there is a rational root, and this $t$ value must be an integer that divides $4,$ so $\pm 1 , \; \pm 2 \; , \; \pm 4 $

Answer 3

To get the other two roots, it is possible to use the quadratic equation on $2x^{2} + 2x - 1$ by assigning the polynomial to $0$. Or, as stated by other users, it is possible to just use the Rational Root Theorem since the original polynomial does not have that many terms.