# Polynomial Division

When we divide one polynomial by another, we continue the **long division** procedure until the remainder is either the zero polynomial or polynomial of a lower degree than the divisor. Suppose that we have two polynomials: \(f(x)\) of degree \(n\) and \(g(x)\) of degree \(r\), where \(r\leq n\), and that we carry out the division

to obtain quotient \(q(x)\) and remainder \(r(x)\). Then

- The division continues step by step until a remainder \((r(x))\) is reached whose degree is less than the degree of the divisor \((g(x))\).
- When the remainder is zero, it is said to be exact (the divisor is a factor of the dividend).
- By reviewing the steps, we have \[f(x)=g(x)\cdot q(x)+r(x).\]

## Division algorithm

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 \(2x^{2}+x-1.\)

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.

## Remainder factor theorem

Main page: Remainder factor theorem

Remainder TheoremFor 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 TheoremLet \( f(x)\) 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 \[\] \[\begin{array}&f(x)=x^{3}-2x^{2}+8x, &g(x)=x+2.\end{array}\]

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

\[\begin{align} f(-2) &=(-2)^{3}-2(-2)^{2}+8(-2)-1\\ &=-8-8-16-1\\ &=-33. \end{align}\]

Therefore, \(-33\) 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 \(8\).

From the remainder theorem, we have

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

## 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 \((x+2)(x-1)\).

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

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

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 \(c\)?

By the remainder theorem, we have

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

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

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

**Cite as:**Polynomial Division.

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