How to find multiplicity for my zeros?
I have the polynomial $P(x) = -2x^3 - x^2+x$
Factored form: $-x(x+1)(2x-1)$
The zeros are: $x=-1, x=-\frac12$
Multiplicity: $3,2,1$
Y-Int: $0$
Leading Term: $-2x^3$
I'm trying to graph this equation and can't figure out which multiplicity belongs to which zeros. Thanks for the help!
$\endgroup$ 23 Answers
$\begingroup$Each zero has multiplicity 1 in fact. Looking at your factored polynomial: $$-2x^3-x^2+1=(-x)(x+1)(2x-1)$$ The multiplicity of each zero is the exponent of the corresponding linear factor. If we re-write the factorization in the suggestive form: $$-2x^3-x^2+1=-(x)^1(x+1)^1(2x-1)^1$$ The multiplicity of the root -1 is the exponent of the factor $(x+1)$; so it has multiplicity 1. The same applies for the other two roots.
$\endgroup$ 1 $\begingroup$The values $3,2,1$ are not multiplicities of zeroes, they are the degrees of the nonzero terms in the polynomial. Not the same thing.
The polynomial has degree $3$ (the highest degree among nonzero terms), which is the sum of the multiplicities of all zeroes. Each linear factor has degree $1$, and contributes a zero.
$\endgroup$ $\begingroup$If $P(x)$ is a polynomial such that $P(\alpha)=0$, then we call $\alpha$ a zero of the polynomial $P(x)$. If $\alpha$ is a zero of $P(x)$, then $(x-\alpha)\mid P(x)$.
We can then write $P(x)=(x-\alpha)R(x)$ with $\deg(P)>\deg(R)$.
Repeating this process, we can continue factoring out $(x-\alpha)$ until we reach the point: $P(x)=(x-\alpha)^n Q(x)$ where $Q(\alpha)\neq 0$.
This value of $n$ is what we call the multiplicity of the zero $\alpha$.
In simpler terms, once you write your polynomial in factored form and combining like factors, the exponents are the multiplicities for the corresponding roots.
Here, we have $P(x)=-2x^3-x^2+x=-2(x-0)^1(x-(-1))^1(x-\frac{1}{2})^1$ according to your work.
We see then that $0$ is a zero of multiplicity one, $-1$ is a zero of multiplicity one, and $\frac{1}{2}$ is a zero of multiplicity one.
$\endgroup$
