Polynomials - Factor & Remainder Theorem

Factor & Remainder Theorems go "hand in hand"

The factor theorem tells provides with two tools:

• if we're told that \(f(c) = 0\) then we can state (without further proof) that \(\begin{pmatrix}x - c\end{pmatrix}\) is a factor of \(f(x)\)
• if we're told that \(\begin{pmatrix}x - c\end{pmatrix}\) is a factor of \(f(x)\), then we can state (without further proof) that \(f(c) = 0\).

In essence, the factor theorem is "just" a special case of the remainder theorem.

Indeed, with the remainder theorem in mind, when the remainder \(R\) of \(\frac{f(x)}{x-c}\) equals to zero, \(f(c)=R=0\), then \(\begin{pmatrix}x-c\end{pmatrix}\) is, by very definition, a factor of \(f(x)\).

Furthermore since the remainder theorem tells us:

It's Just Like with Whole Numbers

Note: in the same way that \(4\) isn't a factor of \(13\) because: \[13 \div 4 = 3 \ R \ 1\] but \(4\) is a factor of \(12\) because: \[12 \div 4 = 3 \ R \ 0\] if the remainder upon division by \(\begin{pmatrix} x - c \end{pmatrix}\) isn't equal to \(0\) then \(\begin{pmatrix} x - c \end{pmatrix}\) isn't a factor and \(f(c) \neq 0\): \[\frac{f(x)}{x-c} = Q(x) + \frac{R}{x-c} \quad , \ R \neq 0\] if the remainder upon division by \(\begin{pmatrix} x - c \end{pmatrix}\) equals \(0\) then \(\begin{pmatrix} x - c \end{pmatrix}\) is a factor and \(f(c) = 0\): \[\frac{f(x)}{x-c} = Q(x)\]

Factoring Polynomials

The factor theorem provides us with a method for factoring polynomials. Indeed, if we know that a number \(c\) is a zero (or root) of a polynomial \(f(x)\), that is if: \[f(c) = 0\] then the factor theorem tells us that \(\begin{pmatrix}x - c \end{pmatrix}\) is factor of \(f(x)\). This means that there must be a quotient function, \(Q(x)\), such that: \[f(x) = \begin{pmatrix}x - c \end{pmatrix}.Q(x)\] The quotient function, \(Q(x)\), is a polynomial function whose degree is \(1\) less than the degree of \(f(x)\): \[\text{Deg}\begin{bmatrix} Q(x)\end{bmatrix} = \text{Deg}\begin{bmatrix} f(x)\end{bmatrix} - 1\] to find what the quotient polynomial is we use synthetic division of polynomials.

Step-by-Step Worked Example

Multiplying both sides of: \[\frac{f(x)}{x+1} = x^2 - 7x + 10 + \frac{0}{x+1}\] by \(\begin{pmatrix} x+1 \end{pmatrix}\) leads to: \[f(x) = \begin{pmatrix}x+1\end{pmatrix}\begin{pmatrix}x^2 - 7x + 10 \end{pmatrix}\] which highlights the fact that \(\begin{pmatrix}x+1\end{pmatrix}\) is a factor of \(f(x)\).

Multiplying both sides of: \[\frac{f(x)}{x+1} = x^2 - 7x + 10 + \frac{0}{x+1}\] by \(\begin{pmatrix} x+1 \end{pmatrix}\) leads to: \[f(x) = \begin{pmatrix}x+1\end{pmatrix}\begin{pmatrix}x^2 - 7x + 10 \end{pmatrix}\] which highlights the fact that \(\begin{pmatrix}x+1\end{pmatrix}\) is a factor of \(f(x)\).

Since \(Q(x)\) is also a polynomial function, provided we can find its zeros, we can factor it as well. Repeatedly factoring the quotient polynomial we get we can potentially write a polynomial of degree \(n\): \[f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0\] in factored form, as a product of upto \(n\) linear factors: \[f(x) = a_n \begin{pmatrix}x - c_1 \end{pmatrix}\begin{pmatrix}x - c_2 \end{pmatrix} \dots \begin{pmatrix}x - c_n \end{pmatrix}\] The following couple of tutorials show examples of how the factor theorem, as well as synthetic division can be used to write a cubic polynomial in factored form.

Factoring Polynomials

Since the quotient function \(Q(x)\) is also a polynomial function we can look for its zeros and therefore its factors. Say \(x=c_2\) is one of the zeros of \(Q(x)\) then we can write: \[Q(x) = \begin{pmatrix}x-x_2\end{pmatrix}Q_2(x)\] Going back to \(f(x)\), and letting \(c = c_1\) and \(Q(x) = Q_1(x)\) we can therefore write: \[f(x) = \begin{pmatrix}x-c_2 \end{pmatrix}\underbrace{\begin{pmatrix}x - c_1 \end{pmatrix} Q_2(x)}_{Q_1(x)}\] By repeatedly doing this, a polynomial of degree \(n\), \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x +a_0\), can be written as a product of upto \(n\) linear factors, \(x-c_1\), \(x-c_1\), ... , \(x-c_n\) in \(\mathbb{R}\).

Note: when working within the set of complex numbers \(\mathbb{C}\), a polynomial of degree \(n\) will always have \(n\) zeros and therefore \(n\) linear factors \(x-c_1\), \(x-c_1\), ... , \(x-c_n\): \[f(x)=a_nx^n+a_{n-1}x^{n-1} + \dots + a_1x + a_0 = a_n \begin{pmatrix}x-c_n\end{pmatrix}\begin{pmatrix}x-c_{n-1}\end{pmatrix} \dots \begin{pmatrix}x-c_2\end{pmatrix}\begin{pmatrix}x-c_1\end{pmatrix}\]

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