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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?

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2 Answers 2

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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 $0$.

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Just write the polynomial explicitly

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

Now, you see

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

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