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

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

\[\begin{align} f(-2) &=(-2)^{3}-2(-2)^{2}+8(-2)\\ &=-8-8-16\\ &=-32. \end{align}\]

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

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

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

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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

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

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 \(c\)?

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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

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