Factoring a Polynomial with Real Roots

GUIDE: Mathematics of the Discrete Fourier Transform (DFT). Factoring a Polynomial with Real Roots

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Factoring a Polynomial with Real Roots

Remember “factoring polynomials”? Consider the second-order polynomial

\

It is second-order because the highest power of $x$ is $2$ (only non-negative integer powers are allowed). The polynomial is also monic because its leading coefficient, the coefficient of $x^2$, is $1$. Since it is second order, there are at most two real roots (or zeros) of the polynomial. Suppose they are denoted $x_1$ and $x_2$. Then we have $p(x_1)=0$ and $p(x_2)=0$, and we can write

\

This is the factored form of the monic polynomial $p(x)$. (For a non-monic polynomial, we may simply divide all coefficients by the first to make it monic, and this doesn’t affect the zeros.) Multiplying out the symbolic factored form gives

\

Comparing with the original polynomial, we find we must have

\



This is a system of two equations in two unknowns. Unfortunately, it is anonlinear system of two equations in two unknowns.2.1Nevertheless, because it is so small, the equations are easily solved. In beginning algebra, we did them by hand. However, nowadays we can use a computer program such as Mathematica:
In[]:=
        Solve[{x1+x2==5, x1 x2 == 6}, {x1,x2}]
Out[]:
        {{x1 -> 2, x2 -> 3}, {x1 -> 3, x2 -> 2}}
Note that the two lists of substitutions points out that it doesn’t matter which root is 2 and which is 3. In summary, the factored form of this simple example is

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Note that polynomial factorization rewrites a monic $n$th-order polynomial as the product of $n$ first-order monic polynomials, each of which contributes one zero (root) to the product. This factoring business is often used when working with digital filters.

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