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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.

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  • $\begingroup$ This is obviously not true if $a=0$. $\endgroup$ – Keen-ameteur Jul 17 '18 at 14:23
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    $\begingroup$ For the same reason why if $\,a \ne 0\,$ then $\,a\cdot b = 0 \implies b = 0\,$. $\endgroup$ – dxiv Jul 17 '18 at 20:40
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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.

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$f(x)=0$ if and only if $af(x)=0$ where $a \ne 0$.

To see this, let $g(x)=af(x)$ and prove that $f(x)=0$ if and only if $g(x)=0$.

If $f(x)=0$, then we can multiply $a$ to both sides and we have $af(x)=a\cdot 0=0$, that is $g(x)=0$.

Conversely, if $af(x)=0$, we can multiplied both sides by $a^{-1}$ since $a \ne 0$ and obtain $f(x)=0$.

Now, let's contrast this with suppose $f(x)=0$, and now let's consider the function $g(x)=f(x)+c$, where $c \ne 0$, the original position that is $0$, now take value $c\ne 0$. For multiplication, the initial value is $0$, multiplying with any non-zero value will preserve the value $0$. The essence is $0$ multiplied with any real number still give you $0$ but $0$ plus something non-zero gives you a non-zero number as it is the addition identity.

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If the roots of the polynomial are $$x_1,x_2,...x_k$$ the polynomial is factored as $$ P(x)= a(x-x_1)(x-x_2)...(x-x_k)$$ where a is the leading coefficient. If $a\ne 0$ then the zero's of P(x) and the zero's of $$Q(x)= b(x-x_1)(x-x_2)...(x-x_k)$$ are the same as long as $b\ne 0$

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It will help to think of a polynomial as a function $f(x)$ that takes as input a number $x$ and returns another number. For example, let $$f(x) = x^2 - 1.$$ Then $f(0)$ is $-1$, $f(0.5)$ is $-0.75$ and $f(1) = 0$. You can even imagine, and I recommend, plotting $y=f(x)$.

Now asking what are the roots of a polynomial is the same as asking for what values of $x$ does $f(x) = 0$. For our example, have already seen one root, $1$. And there is another one, which you can check: $-1$.

Now consider a new polynomial:

$$g(x) = 6 f(x) = 6x^2 - 6$$

What are it’s roots? Well, as you already know, the roots of $g$ are the same as those of $f$. But if you think of polynomials as functions, and the roots as those inputs for which the function is 0, you should be able to see why.

Thinking of $g$ as a function, you will see that, for any $x$, the number $g(x)$ is simply 6 times the number $f(x)$. What is six times 0?

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