# Finding a cubic equation with transformed roots using substitution method

So the question goes like this: The roots of the cubic equation $$2z^3+5z^2-3z-2$$ are $$\alpha, \beta, \gamma$$

Find the cubic equation with roots $$2\alpha + 1, 2\beta + 1, 2\gamma + 1$$

The original way I solved this is by first finding the coefficients - e.g $$\alpha + \beta + \gamma = \frac{b}{a}$$ and so on, but it turns out this way takes quite a long time. My textbook states that there is another method: the substitution method. The method involves a new variable $$w = 2z + 1$$. We write $$z$$ in terms of $$w$$ and substitute into the original equations, so since $$z = \frac{w-1}{2}$$, we do the following substitutions $$2(\frac{w-1}{2})^3 + 5(\frac{w-1}{2})^2 + 3(\frac{w-1}{2}) -2 = 0$$ The explanation in the book is as follows: This is a transformation of $$z$$ in the same way as the new roots are a transformation of the original $$z$$ roots. I don't get this part. How are these new roots (e.g $$2\alpha + 1$$) related in any way to the $$z$$ variable?

## 2 Answers

Define a new polynomial, given by $$Q(w) = P\left(\frac{z-1}{2}\right)$$. Then $$Q(w)$$ has roots $$2 \alpha+1, 2 \beta + 1, 2 \gamma + 1$$, because subbing these into $$Q$$ outputs $$P(\alpha), P(\beta), P(\gamma)$$, all of which are $$0$$.

Just write the polynomial explicitly

$$p(z) = 2(z-\alpha)(z-\beta)(z-\gamma)$$

Now, you see

$$p\left(\frac{w-1}{2}\right) = 2(\frac{w-1}{2}-\alpha)(\frac{w-1}{2}-\beta)(\frac{w-1}{2}-\gamma)$$ $$= \frac 14(w-(2\alpha + 1))(w-(2\beta + 1))(w-(2\gamma + 1))$$