# Roots (1 Viewer)

#### NexusRich

##### Member
Is there a quick way of doing these without expanding everything out ? pls explain

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#### Lith_30

##### New Member
I don't think so, expanding everything out seems like the easiest way to solve it.

#### Trebla

This isn't a "shorter way" per se, but is an alternative if you're really that lazy to expand...

Let $\bg_white y=2\alpha+1$. Since $\bg_white \alpha$ is a root of the polynomial then

$\bg_white \dfrac{(y-1)^3}{8}-2\times\dfrac{(y-1)^2}{4}+4\times\dfrac{(y-1)}{2}+7=0$

$\bg_white (y-1)^3 - 4(y-1)^2 + 16(y-1)+56=0$

Since we get same result regardless of whether we define y in terms of $\bg_white \alpha,\beta,\gamma$ then this new equation has the roots $\bg_white 2\alpha+1,2\beta+1,2\gamma+1$.

Hence $\bg_white (2\alpha+1)(2\beta+1)(2\gamma+1)$ is just the product of the roots of that new equation.

By inspection of the expansion we can see that the constant term is -1- 4-16+56, or equivalently 35. The product of the roots is therefore -35.