Q.Find the algebraic equation whose roots are 2 times the roots of $x^5 - 2x^4 + 3x^3 - 2x^2 + 4x + 3 = 0$.

To double every root of $P(x)=0$, substitute $x=y/2$ into $P(x)=0$ and clear denominators; the result in $x$ is the required equation.

Let the new roots be $y=2x$, i.e. $x=y/2$. Substitute into $x^5-2x^4+3x^3-2x^2+4x+3=0$:

$\frac{y^5}{32}-2\cdot\frac{y^4}{16}+3\cdot\frac{y^3}{8}-2\cdot\frac{y^2}{4}+4\cdot\frac{y}{2}+3=0$

$\frac{y^5}{32}-\frac{y^4}{8}+\frac{3y^3}{8}-\frac{y^2}{2}+2y+3=0.$

Multiply throughout by $32$:

$y^5-4y^4+12y^3-16y^2+64y+96=0.$

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