# Does the degree of a polynomial give the number of roots?

I am aware of the fundamental theorem of algebra, i.e., the degree of a polynomial is the number of roots of the polynomial. For example, $$x^2 - 9 = 0$$ would have two solutions: $$x=3$$ and $$x=-3$$. However, sometimes I come across quadratic polynomials that only have one root, e.g.,

$$t^2 - 2 t + 1 = (t-1)(t-1) = 0$$

which only has the solution $$1$$, or so I think. Is there some underlying concept that I am overlooking?

• Welcome to Math Stack Exchange. A degree n polynomial with complex coefficients has n complex roots when they are counted with multiplicity. In your example, root $1$ has multiplicity $2$ – J. W. Tanner Apr 7 at 4:40
• Can you go further in depth , I still don't quite understand – RonRon Scores Apr 7 at 4:49
• It appears someone else did ;-) – J. W. Tanner Apr 7 at 4:55
• You can see it as "1 appears twice, so we double count it." This is usually written "the solution 1 has multiplicity 2". A particular solution showing up more than once has an effect, so we count each time it shows up as one solution. Think of the graphs of (x-1)(x-1)=0 and x-1=0. These are different as one is a parabola and one is a straight line. Hope this helps. – Edcookie274 Apr 7 at 5:38

This concept is called multiplicity. The Fundamental Theorem of Algebra tells us that if we have a polynomial $$p$$ of degree $$n$$ with complex coefficients, then we can express $$p$$ as $$p(x)=(x-r_1)(x-r_2)(x-r_3)\cdots(x-r_{n-1})(x-r_{n})$$, where $$r_1,r_2,r_3,...,r_{n-1},r_n$$ are complex numbers (and the roots).
The Fundamental Theorem of Algebra is not violated in your example because $$t^2-2t+1=(t-1)(t-1),$$ and our root $$r=1$$ is definitely a complex number (you can also think of it think of it like $$r_1=r_2=1)$$. Therefore, we say that the multiplicity of our root $$r=1$$ is $$2.$$
Also, by the quadratic formula, if $$b^2-4ac=0,$$ then $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt0}{2a}=\frac{-b\pm0}{2a}=\frac{-b}{2a}$$, which implies that we will only have one root $$r=\frac{-b}{2a}$$ with multiplicity $$2$$.
2. Complex roots, $$x^2+1=0$$ has no real roots, but two complex roots i, and -i.