Remainder Theorem

In algebra, the polynomial remainder theorem or little Bézout's theorem (named after Étienne Bézout) is an application of Euclidean division of polynomials. It states that the remainder of the division of a polynomial $f(x)$ by a linear polynomial $x-r$ is equal to $f(r).$ In particular, $x-r$ is a divisor of $f(x)$ if and only if $f(r)=0,$ a property known as the factor theorem.

Examples

Example 1

Let $f(x)=x^{3}-12x^{2}-42$ . Polynomial division of $f(x)$ by $(x-3)$ gives the quotient $x^{2}-9x-27$ and the remainder $-123$ . Therefore, $f(3)=-123$ .

Example 2

Show that the polynomial remainder theorem holds for an arbitrary second degree polynomial $f(x)=ax^{2}+bx+c$ by using algebraic manipulation:

{\begin{aligned}{\frac {f(x)}{x-r}}&={\frac {a{x^{2}}+bx+c}{x-r}}\\&={\frac {a{x^{2}}-arx+arx+bx+c}{x-r}}\\&={\frac {ax(x-r)+(b+ar)x+c}{x-r}}\\&=ax+{\frac {(b+ar)(x-r)+c+r(b+ar)}{x-r}}\\&=ax+b+ar+{\frac {c+r(b+ar)}{x-r}}\\&=ax+b+ar+{\frac {a{r^{2}}+br+c}{x-r}}\end{aligned}}

Multiplying both sides by (x − r) gives

f(x)=ax^{2}+bx+c=(ax+b+ar)(x-r)+{a{r^{2}}+br+c}

.

Since $R=ar^{2}+br+c$ is the remainder, we have indeed shown that $f(r)=R$ .

Proof

The polynomial remainder theorem follows from the theorem of Euclidean division, which, given two polynomials $f (x)$ (the dividend) and $g (x)$ (the divisor), asserts the existence (and the uniqueness) of a quotient $Q (x)$ and a remainder $R (x)$ such that

f(x)=Q(x)g(x)+R(x)\quad {\text{and}}\quad R(x)=0\ {\text{ or }}\deg(R)<\deg(g).

If the divisor is $g(x)=x-r,$ where r is a constant, then either R(x) = 0 or its degree is zero; in both cases, $R (x)$ is a constant that is independent of $x$ ; that is

f(x)=Q(x)(x-r)+R.

Setting $x=r$ in this formula, we obtain:

f(r)=R.

A slightly different proof, which may appear to some people as more elementary, starts with an observation that $f(x)-f(r)$ is a linear combination of terms of the form $x^{k}-r^{k},$ each of which is divisible by $x-r$ since $x^{k}-r^{k}=(x-r)(x^{k-1}+x^{k-2}r+\dots +xr^{k-2}+r^{k-1}).$

Applications

The polynomial remainder theorem may be used to evaluate $f(r)$ by calculating the remainder, $R$ . Although polynomial long division is more difficult than evaluating the function itself, synthetic division is computationally easier. Thus, the function may be more "cheaply" evaluated using synthetic division and the polynomial remainder theorem.

The factor theorem is another application of the remainder theorem: if the remainder is zero, then the linear divisor is a factor. Repeated application of the factor theorem may be used to factorize the polynomial.

Factor Theorem

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.

The factor theorem states that a polynomial $f(x)$ has a factor $(x-k)$ if and only if $f(k)=0$ (i.e. $k$ is a root).

Factorization of polynomials

Two problems where the factor theorem is commonly applied are those of factoring a polynomial and finding the roots of a polynomial equation; it is a direct consequence of the theorem that these problems are essentially equivalent.

The factor theorem is also used to remove known zeros from a polynomial while leaving all unknown zeros intact, thus producing a lower degree polynomial whose zeros may be easier to find. Abstractly, the method is as follows:

"Guess" a zero $a$ of the polynomial $f$ . (In general, this can be very hard, but math textbook problems that involve solving a polynomial equation are often designed so that some roots are easy to discover.)
Use the factor theorem to conclude that $(x-a)$ is a factor of $f(x)$ .
Compute the polynomial ${\textstyle g(x)={\frac {f(x)}{(x-a)}}}$ , for example using polynomial long division or synthetic division.
Conclude that any root $x\neq a$ of $f(x)=0$ is a root of $g(x)=0$ . Since the polynomial degree of $g$ is one less than that of $f$ , it is "simpler" to find the remaining zeros by studying $g$ .

Example

Find the factors of

x^{3}+7x^{2}+8x+2.

To do this one would use trial and error (or the rational root theorem) to find the first x value that causes the expression to equal zero. To find out if $(x-1)$ is a factor, substitute $x=1$ into the polynomial above:

{\begin{aligned}x^{3}+7x^{2}+8x+2&=(1)^{3}+7(1)^{2}+8(1)+2\\&=1+7+8+2\\&=18\end{aligned}}

As this is equal to 18 and not 0, this means $(x-1)$ is not a factor of $x^{3}+7x^{2}+8x+2$ . So, we next try $(x+1)$ (substituting $x=-1$ into the polynomial):

(-1)^{3}+7(-1)^{2}+8(-1)+2.

This is equal to $0$ . Therefore $x-(-1)$ , which is to say $x+1$ , is a factor, and $-1$ is a root of $x^{3}+7x^{2}+8x+2.$

The next two roots can be found by algebraically dividing $x^{3}+7x^{2}+8x+2$ by $(x+1)$ to get a quadratic:

{x^{3}+7x^{2}+8x+2 \over x+1}=x^{2}+6x+2,

and therefore $(x+1)$ and $x^{2}+6x+2$ are factors of $x^{3}+7x^{2}+8x+2.$ Of these, the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic $-3\pm {\sqrt {7}}.$ Thus the three irreducible factors of the original polynomial are $x+1,$ $x-(-3+{\sqrt {7}}),$ and $x-(-3-{\sqrt {7}}).$