Why will any polynomial give the same zeroes as that polynomial multiplied by any number a? Let's say I have polynomial $x^3+x^2+5=0$. Even if I multiplied this by say 6, so $6(x^3+x^2+5=0)=0$, the roots will be the same. I know this seems blindingly obvious to many people, but it isn't to me. Can someone please explain.
I know it has something to do with it being equals to 0 but I don't know why. I mean x^3+x^2+5=0 is =0, but say I took $x^6+x^2=0$. They are both =0, but they won't yield the same roots. So why would $x^3+x^2+5=0$ and $6(x^3+x^2+5=0)=0$ yield the same roots, just because they are both equal to 0?
I also know that the ratio between all the terms in the polynomial will still be equal when you multiply them, but again, I don't get why that would make the multiplied polynomial yield the same zero as it's un-multiplied version.
Best Answer
Your notation is a bit strange: you have two equal signs: $$6(x^3+x^2+5=0)=0$$but this makes no sense. What you're considering is the polynomial $p(x):=x^3+x^2+5$, and you want to know its roots, which are the solutions to the equation $$(1)\qquad p(x)=0\iff x^3+x^2+5=0$$ Now you want to consider this polynomial multiplied/scaled by a factor of $6$. Thus, we want to consider the polynomial $q(x):=6p(x)=6(x^3+x^2+5)$. To find the roots of $q(x)$, we solve the equation $$(2)\qquad q(x)=0\iff 6(x^3+x^2+5)=0$$ Now, equations $(1)$ and $(2)$ have the same solution set because we can divide both sides of equation $(2)$ by $6$ to get $$6(x^3+x^2+5)=0\iff \frac{6}{6}(x^3+x^2+5)=\frac{0}{6}\iff x^3+x^2+5=0$$since $\frac{6}{6}=1$ and $\frac{0}{6}=0$. So performing this division gives us equation $(1)$, so equation $(1)$ and equation $(2)$ are equivalent. In general, you can do this with any $a\neq 0$, because dividing by any nonzero number preserves the equality.