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Real Polynomials#

Definition: Real Polynomial

A real polynomial is a polynomials over the real numbers.

Binomial Theorem#

The Binomial Theorem

The expansion of the expression

\[ (x + y)^n, \]

where \(n \in \mathbb{N}\) is given by the real polynomials

\[ (x + y)^n = \sum_{k = 0}^n \binom{n}{k}x^{n-k}y^k = \sum_{k = 0}^n \binom{n}{k}x^{k}y^{n-k} \]

Tip

The expansion has \(n+1\) terms. The \(k\)-th term (\(k \in \{1, \dotsc, n + 1\}\)) is given by the number of combinations \(C_n^{k-1} x^{n-(k-1)} y^{k-1}\).

Proof

TODO

Polynomial Division#

Theorem: Polynomial Division

Let \(A(x)\) and \(B(x)\) be two real polynomials such that the degree of \(A\) is greater than or equal to the degree of \(B\).

If \(B(x)\) is nonzero, then there exist unique polynomials \(Q(x)\) and \(R(x)\) such that

\[ A(x) = Q(x)B(x) + R(x), \]

where

  • \(\deg(Q) = \deg(A) - \deg(B)\)
  • \(\deg(R) \lt \deg(B)\) or \(R(x)\) is the zero polynomial.

We call \(A\) the dividend, \(B\) the divisor, \(Q\) the quotient and \(R\) the remainder.

Proof

TODO

Definition: Divisibility

If \(R(x) = 0\), then we say that \(A\) is divisible by \(B\).

Theorem: Factorization Theorem

Every real polynomial \(A(x) = \sum_{k=0}^n a_k x^k\) can be factorized into a product of real polynomials of degree \(\le 2\). More specifically,

\[ A(x) = a_n(x-r_1)^{p_1}\cdots(x-r_l)^{p_l} (x^2 + c_1 x + d_1)^{q_1}\cdots(x^2+c_s x + d_s)^{q_s}, \]

where:

  • \(r_i\) are the distinct real roots of \(A(x) = 0\);
  • \(p_i\) is the multiplicity of \(r_i\);
  • \(c_j, d_j \in \mathbb{R}\);
  • \(q_j \in \mathbb{N}_0\).
Proof

TODO

Polynomial Remainder Theorem (Little Bézout's Theorem)

The remainder \(R\) upon the division of a real polynomial \(A(x)\) with a real polynomial \(B(x) = x - p\) is the value of \(A\) at \(p\).

\[ R = A(p) \]
Proof

The polynomial division theorem tells us that

\[ A(x) = (x-p)Q(x) + R(x) \]

Since the degree of \(B(x)\) is \(1\), the degree of \(R\) must either be \(1\), i.e. \(R\) is a constant and contains no variables or \(R\) must be [[Zero Polynomial|zero]]. We can therefore denote \(R(x)\) as just \(R\). For \(x = p\) we obtain

\[ A(p) = (p-p)Q(p) + R = 0 + R = R \]
Algorithm: Horner's Method

Horner's method is a way to easily divide a polynomial \(A(x)\) by a polynomial \(B(x)\) of degree one.

  1. Write \(B(x)\) as \(B(x) = x - p\). Since \(\deg(B) = 1\), the remainder \(R\) is just a real number because \(\deg (R) \lt \deg(B) \implies \deg (R) = 0\). This means that we have
\[ A(x) = (x-p)Q(x) + R \]
  1. To determine \(Q(x)\) and \(R\), construct the following table:
\(a_n\) \(a_{n-1}\) \(\cdots\) \(a_1\) \(a_0\)
\(p\) \(q_{n-1} = a_n\) \(q_{n-2} = pq_{n-1} + a_{n-1}\) \(\cdots\) \(q_0 = pq_1 + a_1\) \(R = pq_0 + a_0\)
  • We calculate the table from left to right.
  • The first coefficient \(q_{n-1}\) of \(Q(x)\) is equal to the first coefficient of \(A(x)\), i.e. \(q_{n-1} = a_n\).
  • For all other coefficients of \(Q(x)\) we have \(q_i = pq_{i+1} + a_{i-1}\).
  • At the end of the calculation, the right-most cell will contain \(R\).
Example

Let \(A(x) = 2x^5+3x^3-11x^2+6\) and \(B(x) = x-3\). Create the table.

\(2\) \(0\) \(3\) \(-11\) \(0\) \(6\)
\(p = 3\)

Perform the algorithm step by step.

\(2\) \(0\) \(3\) \(-11\) \(0\) \(6\)
\(p = 3\) 2
\(2\) \(0\) \(3\) \(-11\) \(0\) \(6\)
\(p = 3\) 2 6
\(2\) \(0\) \(3\) \(-11\) \(0\) \(6\)
Algorithm: Change of Variables

We are given a real polynomial \(A(x)\) and want to transform it into another real polynomial \(B(y)\) where \(y = x - p\) for some \(p \in \mathbb{R}\).

  1. Divide \(A\) by \((x-p)\).
\[ A(x) = (x-p) Q_1(x) + R_0 \]
  1. Divide \(Q_1\) by \((x-p)\).
\[ Q_1(x) = (x-p) Q_2(x) + R_1 \]
  1. Repeat step 2, dividing \(Q_i\) by \((x-p)\) in order to obtain \(Q_{i + 1}\) and \(R_1\) until \(Q_{i+1}\) is just a number.
  2. Finally,
\[ B(y) = R_0 + R_1y + R_2y^2 + R_3y^3 + \cdots \]

Tip

You can use Horner's method to speed up the process.

Example

Let \(A(x) = x^4 + 8x^3 + 24x^2 + 50x + 90\) and let \(y = x + 2\). We can write \(y\) as \(y = x - p\), where \(p = -2\).

Apply Horner's method:

\(1\) \(8\) \(24\) \(50\) \(90\)
\(-2\) \(1\) \(6\) \(12\) \(26\) \(38 = R_0\)
\(-2\) \(1\) \(4\) \(4\) \(18 = R_1\)
\(-2\) \(1\) \(2\) \(0 = R_2\)
\(-2\) \(1\) \(0 = R_3\)
\(-2\) \(1 = R_4\)

We have

\[ \begin{aligned} B(y) &= R_0 + R_1 y + R_2 y^2 + R_3 y^3 + R_4 y^4 \\ &= 38 + 18y + y^4 \\ &= y^4 + 18y + 38 \\ &= (x+2)^4 + 18(x + 2) + 38 \end{aligned} \]