What happens when we raise a polynomial with a power how many roots do we get?

Question

What happens if we square a quadratic equation?

For example $(x^2 +2x -4)^2$ how many roots will this question have?

Answer 1

Note that

$$(x-A)\cdot (x-B)=0 \iff [(x-A)\cdot (x-B)]^2 =0$$

therefore, more in general, $p(x)=0$ and $(p(x))^n=0$ with $n\in \mathbb N$ have the same number of distinct roots.

Answer 2

$[P(x)]^k = 0\iff P(x) = 0$.

So $P(x)^k$ will have the exact same roots as $P(x)$ but now that are "multiple roots".

.......

which probably needs a little clarification.

If $P(x) = a_mx^m + ...... + a_1x + a_0$ is a polynomial of degree $m$ then we say it has $m$ roots...

BUT some of the roots might not be real. (Example $x^2 + 1$ means $x^2 =-1$ doesn't have any real solutions. But it has two complex solutions $x = i$ and $x = -i$ where $i$ is an imaginary number with the property $i^2 =-1$. Depending on your course maybe you say $x^2 + 1$ doesn't have any real roots, or you say it has $2$ complex roots.)

AND some of the roots might be multiple roots. (Example $x^2 -4x + 4 = (x-2)^2$ has one "double root" $x = 2$. As to whether yous say $x^2 -4x + 4$ has two roots: $2$ and $2$, or whether you say $x^2 -4x +4$ has one double root depends on your class).

So what ever roots $P(x)$ has $P(x)^k$ will have the exact same roots but the multiplicity of each root is $k$ times larger.

Example if $P(x) = (x-2)^2(x+3) = (x^2 - 4x + 4)(x+3) = x^3 -x^2 -8x+12$ has a double root of $2$ and a single root of $-3$ then

$P(x)^5 = [(x-2)^2(x+3)]^5= (x-2)^{10}(x+3)^5$ will have a 10times root of $2$ and a $5$ times root of $-3$.

Technically that is $15$ roots.