Polynomial Division

Division with polynomials (done with either long division or synthetic division) is analogous to long division in arithmetic: we take a dividend divided by a divisor to get a quotient and a remainder (which will be zero if the divisor is a factor of the dividend).

More formally, given a dividend $f(x) ,$ divisor $g(x) ,$ quotient $q(x)$ and remainder $r(x),$ we know $\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)} .$

With long division, the division continues step by step until a remainder $r(x)$ is reached whose degree is less than the degree of the divisor $g(x)$.

Synthetic division is long division that is written in a format convenient to do by hand. It is designed only for dividing with divisors in the format $(x - k),$ but this is a very common circumstance when looking for factors of a polynomial.

Main page: Division Algorithm

Let $f(x)$ and $g(x)$ be two polynomial functions and suppose that $g(x)$ is a non-zero polynomial. Then there exists unique polynomial functions $q(x)$ and $r(x)$ such that

$$f(x)=g(x)\cdot q(x)+r(x)$$

holds for all values of $x$, and $r(x)$ is either zero constant or a polynomial of degree lower than the degree of $g(x)$. $_\square$

Divide $6x^{4}+5x^{3}+4x-4$ by $2x^{2}+x-1.$

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Using long division, we have the following:

Now, since $4x-3$ has a lower degree than $2x^{2}+x-1$, we can stop there. Thus, we can conclude that $3x^{2}+x+1$ is the quotient polynomial and $4x-3$ is the remainder polynomial, implying

$$6x^4+5x^{3}+4x-4=(2x^{2}+x-1)(3x^{2}+x+1)+4x-3. \ _\square$$

Now we can see that in the division algorithm, $f(x)$ is the dividend polynomial, $g(x)$ is the divisor polynomial, $q(x)$ is the quotient polynomial and $r(x)$ is the remainder polynomial. Therefore, the identity $f(x)=g(x)\cdot q(x)+r(x)$ expresses the fact that

$$\text{(dividend) = (divisor)(quotient) + (remainder)}.$$

We often refer to the value of $\frac{f(x)}{g(x)}$ as the quotient, but the word used in this way must not be confused with the quotient polynomial $q(x)$ in the division algorithm.

Main Article: Synthetic Division

Synthetic division is a process to find the quotient and remainder when dividing a polynomial by a monic linear binomial (a polynomial of the form $x-k$). Consider dividing $x^2+2x+6$ by $x-1.$ First, by the long division algorithm:

This is what the same division looks like with synthetic division:

Main page: Remainder factor theorem

Remainder Theorem

For a polynomial $f(x)$, the remainder of $f(x)$ upon division by $x-c$ is $f(c)$. $_\square$

Dividing $f(x)$ by $x-c$, we obtain $$f(x)=(x-c)q(x)+r(x),$$ where $r(x)$ is the remainder. Since $x-c$ has degree 1, it follows that the remainder $r(x)$ has degree 0, and thus is a constant. Let $r(x)=R$. Substituting $x=c$, we obtain $f(c)=(c-c)q(c)+R=R$. Thus, $R=f(c)$ as claimed. $_\square$

Factor Theorem

Let $f(x)$ be a polynomial such that $f(c) =0$ for some constant $c$. Then $x-c$ is a factor of $f(x)$. Conversely, if $x-c$ is a factor of $f(x)$, then $f(c)=0$. $_\square$

Forward direction:
Since $f(c)=0$, by the remainder theorem, we have $f(x) = (x-c)h(x) + R = (x-c)h(x)$. Hence, $x-c$ is a factor of $f(x)$.

Backward direction:
If $x-c$ is a factor of $f(x)$, then (by definition) the remainder of $f(x)$ upon division by $x-c$ would be 0. By the remainder theorem, this is equal to $f(c)$. Hence, $f(c)=0$. $_\square$

Note: There are no restrictions on the constant $c$. It could be a real number, a complex number, or even a matrix!

The remainder-factor theorem is often used to help factorize polynomials without the use of long division. When combined with the rational roots theorem, this gives us a powerful factorization tool.

Find the remainder of the polynomial division $\displaystyle \frac{f(x)}{g(x)}$ using the remainder theorem, where $$$$ $$\begin{array}{c}&f(x)=x^{3}-2x^{2}+8x, &g(x)=x+2.\end{array}$$

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Since $x+2=x-(-2)$,

$$\begin{aligned} f(-2) &=(-2)^{3}-2(-2)^{2}+8(-2)\\ &=-8-8-16\\ &=-32. \end{aligned}$$

Therefore, $-32$ is the remainder when $x^{3}-2x^{2}+8x$ is divided by $x+2$. $_\square$

Find the value of $k$ such that $f(x)=x^{3}+2x^{2}-3kx-10$ upon division by $x+3$ has a remainder of $8$.

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From the remainder theorem, we have

$$\begin{aligned} f(-3)=8 \Leftrightarrow (-3)^{3}+2(-3)^{2}-3k(-3)-10&=8\\ 9k&=27\\ k&=3. \ _\square \end{aligned}$$

When the polynomial $f(x)$ is divided by by $x+2$, the remainder is $3$, and when $f(x)$ is divided by $x-1$, the remainder is $2$. Find the remainder when $f(x)$ is divided by $(x+2)(x-1)$.

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Let the remainder upon division by $(x+2)(x-1)$ be written in the form $ax+b$. This is so because the degree of the divisor is $2$. Then from division algorithm we have

$$f(x)=(x+2)(x-1)q(x)+ax+b.$$

By the remainder theorem we have

$$\begin{aligned} f(-2) &=(-2+2)(-2-1)q(-2)+a(-2)+b\\ &=-2a+b\\ &=3. \qquad \qquad (1) \\ \\ f(1) &=(1+2)(1-1)q(1)+a(1)+b\\ &=a+b \\ &=2. \qquad \qquad (2) \end{aligned}$$

Solving the simultaneous equations $(1)$ and $(2)$ gives

$$a=-\frac{1}{3}, b=\frac{7}{3}.$$

Therefore the remainder is

$$r(x)=-\frac{1}{3}x+\frac{7}{3}. \ _\square$$

When the polynomial $p(x)=ax^{4}+bx^{3}+cx-8$ is divided by $x-1$, the remainder is $2$. If $x+1$ and $x-2$ are factors of $p(x)$, what are the values of $a, b$ and $c$?

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By the remainder theorem, we have

$$\begin{aligned} f(1)=2 \Leftrightarrow a(1)^{4}+b(1)^{3}+c(1)-8&=2\\ a+b+c&=10 &\qquad (1)\\ \\ f(-1)=0 \Leftrightarrow a(-1)^{4}+b(-1)^{3}+c(-1)-8&=0\\ a-b-c&=10 &\qquad (2)\\ \\ f(2)=0 \Leftrightarrow a(2)^{4}+b(2)^{3}+c(2)-8&=0\\ 16a+8b+2c&=8. &\qquad (3) \end{aligned}$$

Solving the simultaneous equations $(1), (2)$ and $(3)$ gives

$$a=9, b=-23, c=24. \ _\square$$