My professor told me this for factoring polynomials:
Add up the coefficients and if they equal 0 then the polynomial has root of 1.
Add up, but switch the signs of the coefficients with odd number powers. If sum = 0 then there is a root of -1.
I just want to make sure when he means this is the $c$ term added up in this neat little trick.
I didn't think so, but I want to make sure.
Do you guys know?
On
Not sure what you mean by "this is the $c$ term added up", but here are some examples:
$P(x) = 3 x^4 - 2 x^3 + x^2 - 2$: $3 - 2 + 1 + 0 - 2 = 0$ so $1$ is a root. $3 + 2 + 1 - 0 - 2 = 4$ so $-1$ is not a root.
On
As requested, here is my solution in answer form.
if p(x) = $\sum {a_i}{x^i}$ is a polynomial then p(1) = $\sum {a_i}$ is the sum of the coefficients. similarly, p(-1) is the sum of the even numbered coefficients minus the sum of the odd numbered coefficients.
To be clear: the constant term is the degree 0 term, an even numbered coefficient, so its sign does not change
This 'trick' your professor taught you is actually what is called the remainder theorem. Basically what it says is that if you have a polynomial: $p(x)$ and you calculate the value of $p(d)$ then $p(d)$ is the remainder when $p(x)$ is divided by $(x-d)$. For example $p(x) = x^2 -2x+1$. Let $d=1$ and we get:$$p(d) = p(1) = 1^2-2(1)+1 = 1-2+1 = 0$$ Thus $p(x)$ is divisible by $(x-1)$ so it has root $d = 1$. Notice that in the last step ($1-2+1$) all we are doing is adding the coefficients and because the sum is equal to $0$ it is indeed divisible by $1$. The same applies for $d=-1$. When we put $-1$ in the polynomial we are switching the signs of all the coefficients with odd powers just like your professor taught you to.