Section 3.2

Real Number Division Algorithm

Definition 35 (Division Algorithm (Real Numbers))

Let \(a\) and \(b\) be any real number where \(b\ne0\). Then there exists a unique real number \(q\) and \(r\) such that

\[a=b\cdot q+r\]

where \(0<r<b\) or \(r=0\).

Example 45

If we look at \(12\) and \(3\) we know that

\[12=3\cdot 4 + 0\]

This can be shown by doing long division:

\[\begin{alignat*}{10} & \phantom{11}4\\ 3 & )\overline{12}\\ - & \underline{\phantom{)}12} & \leftarrow4(3)\phantom{--}\\ & \phantom{)1}0 & \leftarrow12-12 \end{alignat*}\]

If we look at \(13\) and \(3\) we know that

\[13 = 3\cdot 4 + 1\]

This can be shown by doing long division:

\[\begin{alignat*}{10} & \phantom{11}4\\ 3 & )\overline{13}\\ - & \underline{\phantom{)}12} & \leftarrow4(3)\phantom{--}\\ & \phantom{)1}1 & \leftarrow13-12 \end{alignat*}\]

Since the division algorithm for real numbers is \(a=bq+r\) we know that if we divide both sides by \(b\) we have

\[\frac{a}{b} = q+\frac{r}{b}\]

We will use this fact to rewrite polynomial over polynomial expressions later on. For example, we know that \(13=3(4)+1\). If we divide both sides by \(3\) we have: \(\frac{13}{3} = 4+\frac{1}{3}\).

Division Algorithm for Polynomials

Definition 36 (Division Algorithm for Polynomials)

Let \(f(x)\) and \(g(x)\) be polynomials with \(g(x)\) of lesser degree than \(f(x)\) and \(g(x)\) is of degree one or higher. Then there exists a unique \(q(x)\) and \(r(x)\) such that

\[f(x)=g(x)\cdot q(x)+r(x)\]

where \(0<\text{deg}(r(x))<\text{deg}(g(x))\) or \(\text{deg}(r(x))=0\).

Example 46

If we look at \(x^2+3x+2\) and \(x+1\) we know that

\[x^2+3x+2=(x+1)(x+2)+0\]

by factoring \(x2+3x+2\). This can also be done by polynomial long division.

\[\begin{alignat*}{10} & \phantom{)x^{2}+)}x+2\\ x+1 & )\overline{x^{2}+3x+2}\\ - & \underline{\phantom{)}x^{2}+x} & \leftarrow x(x+1)\phantom{)-(x^{2}+x)}\\ & \phantom{)x^{2}+)}2x+2 & \leftarrow(x^{2}+3x)-(x^{2}+x)\\ & \underline{-\phantom{x^{2}))}2x+2} & \leftarrow2(x+1)\phantom{))))(2x+2}\\ & \phantom{-x^{2}))2x+)}0 & \leftarrow(2x+2)-(2x+2) \end{alignat*}\]

From the previous example, we also know that

\[\frac{x^2+3x+2}{x+1}=(x+2)\]

Let’s do another polynomial long-division problem.

Example 47

Use polynomial long division to evaluate \(\frac{3x^3-2x^2-150}{x^2-4}\).

Solution:

We should first notice that the numerator and denominator are missing an \(x\) term. This would be like saying \(12\) is the same as \(102\). Therefore, we must first rewrite the quotient.

\[\frac{3x^3-2x^2+0x-150}{x^2+0x-4}\]

Next, we have the following setup

\[\begin{align*} x^{2}+0x-4 & )\overline{3x^{3}-2x^{2}+0x-150} \end{align*}\]

Then we start the polynomial long division

\[\begin{alignat*}{10} & \phantom{)3x^{3}-2x^{2}+)}3x-2\\ x^{2}+0x-4 & )\overline{3x^{3}-2x^{2}+0x-150}\\ - & \underline{\phantom{)}3x^{3}-0x^{2}-12x} & \leftarrow3x(x^{2}+0x-4)\phantom{x)-(3x^{3}-2x^{2}-12)}\\ & \phantom{)3^{3}}-2x^{2}+12x-150 & \leftarrow(3x^{3}-2x^{2}+0x)-(3x^{3}-2x^{2}-12x)\\ - & \underline{\phantom{xx}-2x^{2}+0x\phantom{x}+8} & \leftarrow-2(x^{2}+0x-4)\phantom{150)-(-2x^{2}+0x+))}\\ & \phantom{xxxxxxxx}12x-158 & \leftarrow(-2x^{2}+12x-150)-(-2x^{2}+0x+8)\\ \end{alignat*}\]

Therefore, we have

\[\frac{3x^3-2x^2-150}{x^2-4}=3x-2+\frac{12x-158}{x^2-4}\]

Synthetic Division

In special cases, we can use a method called synthetic division instead of polynomial long division. We can use synthetic division when a polynomial is divided by a binomial \(x-k\). This means the previous example could not be done by synthetic division.

Consider the following polynomial long division of \(\frac{3x^3-2x^2+0x-150}{x-4}\). Notice that the denominator satisfied the \(x-k\) requirement. We then have

\[\begin{alignat*}{10} & \phantom{)3x^{3}-}3x^{2}+10x+40\\ x-4 & )\overline{3x^{3}-2x^{2}\phantom{xx}+0x-150}\\ - & \underline{\phantom{x}3x^{3}-12x^{2}} & \leftarrow3x^{2}(x-4)\phantom{x)-(3x^{3}-2x^{2}-12xxxx)}\\ & \phantom{)3^{3}xxxx}10x^{2}+0x & \leftarrow(3x^{3}-2x^{2}+0x)-(3x^{3}-2x^{2}-12x)\\ - & \underline{\phantom{xxxxxx}10x^{2}-40x} & \leftarrow-2(x^{2}+0x-4)\phantom{150)-(-2x^{2}+0x+))}\\ & \phantom{xxxxxxxxxxx}40x-150 & \leftarrow(-2x^{2}+12x-150)-(-2x^{2}+0x+8)\\ - & \underline{\phantom{xxxxxxxxxxx}40x-160} & \leftarrow40(x-4)\phantom{0)-(4x-160xxxxxxxxxxx)}\\ & \phantom{xxxxxxxxxxxxxxxx}10 & \leftarrow(40x-150)-(4x-160)\phantom{xxxxxxxxxx)} \end{alignat*}\]

Next, we want to do this again but remove all the \(x^n\) factors.

\[\begin{alignat*}{10} & \phantom{)3x^{3}-}3\phantom{xx}10\phantom{xxxx}40\\ -4 & )\overline{3\phantom{xx}-2\phantom{xx}0\phantom{x}-150}\\ - & \underline{\phantom{)}3\phantom{xx}-4}\\ & \phantom{)3^{3}}10\phantom{xxxx}0\\ - & \underline{\phantom{xxx}10\phantom{xxx}40}\\ & \phantom{xxxxxxxx}40\phantom{x}-150\\ - & \underline{\phantom{xxxxxxxx}40\phantom{x}-160}\\ & \phantom{xxxxxxxxxxxxx}10 \end{alignat*}\]

where \(q(x)=3x^2+10x+40\) and \(r(x)=10\) still. We can still remove some items from the work:

\[\begin{alignat*}{10} & \phantom{)3x^{3}-}3\phantom{xx}10\phantom{xxxx}40\\ -4 & )\overline{3\phantom{xx}-2\phantom{xx}0\phantom{x}-150}\\ - & \underline{\phantom{)}\phantom{xxx}-4}\\ & \phantom{)3^{3}xx}10\phantom{xxxx}\\ - & \underline{\phantom{xxxxi}10\phantom{xx}40}\\ & \phantom{xxxxxxxx}40\phantom{x}\\ - & \underline{\phantom{xxxxxxxx}40\phantom{x}-160}\\ & \phantom{xxxxxxxxxxxxx}10 \end{alignat*}\]

Next, we can smash everything together to get

\[\begin{align*} & \phantom{)3x^{3}-}3\phantom{xx}10\phantom{xxxx}40\\ -4 & )\overline{3\phantom{xx}-2\phantom{xxxxx}0\phantom{x}-150}\\ - & \underline{\phantom{xxx}-12\phantom{xx}-40\phantom{x}-160}\\ & \phantom{xxxxx}10\phantom{xxxx}40\phantom{xxx}10 \end{align*}\]

Finally, if we use \(4\) instead of \(-4\) and add down instead of subtract down we have synthetic division:

\[\begin{align*} & \phantom{)3x^{3}-}3\phantom{xxxxx}10\phantom{xxxx}40\\ 4 & )\overline{3\phantom{xx}-2\phantom{xxxxx}0\phantom{x}-150}\\ + & \underline{\phantom{xxxxx}12\phantom{xxxx}40\phantom{xxx}160}\\ & \phantom{xxxxx}10\phantom{xxxx}40\phantom{xxxx}10 \end{align*}\]

Again, doing this shows that \(q(x)=3x^2+10x+40\) and \(r(x)=10\). We also have

\[\frac{3x^3-2x^2+0x-150}{x-4}=3x^2+10x+40+\frac{10}{x-4}\]

Synthetic Division Examples

Example 48

Rewrite \(\frac{5x^3-6x^2-28x-2}{x+2}\) as \(q(x)+\frac{r(x)}{x+2}\).

Solution:

In this case \(x-k\) we have \(x+2=x-(-2)\). This means \(k=-2\).

\[\begin{align*} & \phantom{xxxxxx}5\phantom{xx}-16\phantom{xxxx}4\\ -2 & )\overline{5\phantom{xx}-6\phantom{xx}-28\phantom{xx}-2}\\ + & \underline{\phantom{xxx}-10\phantom{xxxx}32\phantom{xx}-8}\\ & \phantom{xxx}-16\phantom{xxxxx}4\phantom{xx}-10 \end{align*}\]

Therefore,

\[\frac{5x^3-6x^2-28x-2}{x+2} = 5x^2 -16x +4 + \frac{-10}{x+2}\]

Example 49

Evaluate \(\frac{5x^3-6x^2+3x}{x-1}\)

Solution:

Here \(k=1\).

\[\begin{align*} & \phantom{xxxxxx}5\phantom{xx}-1\phantom{xxxx}2\\ 1 & )\overline{5\phantom{xx}-6\phantom{xxx}3\phantom{xxxxx}0}\\ + & \underline{\phantom{xxxxxx}5\phantom{x}-1\phantom{xxxxx}2}\\ & \phantom{xxx}-1\phantom{xxxxx}2\phantom{xxxx}2 \end{align*}\]

Therefore, we have \(\frac{5x^3-6x^2+3x}{x-1}=5x^2-x+2+\frac{2}{x-1}\).

---

Quick reminder: \(i=\sqrt{-1}\), \(i^2=-1\), \(i^3=-i\), and \(i^4=1\).

The following are examples of how to do arithmetic involving \(i\).

\[\begin{align*} (2-i)(-8+4i) & = -16 + 8i + 8i - 4(-1)\\ & = -16+16i+4\\ & = -12+16i \end{align*}\]

\[30+(4+i) = 34 + i\]

\[(14+5i)+(-4-8i)=10-3i\]

---

Example 50

Let \(f(x)=x^3+9x^2+28x+30%\). Use synthetic division with \(k=-3+i\).

Solution:

Here is the setup

\[\begin{align*} & \phantom{xxxxxx)}1\\ -3+i & )\overline{\phantom{x}1\phantom{xxxx}9\phantom{xxxx}28\phantom{xxxx}30}\\ + & \underline{\phantom{xxxxxxxxxxxxxxxxxxxx}}\\ & \phantom{xxxxxxxxxxxxxxxxxxxx} \end{align*}\]

Then we will compute \(1(-3+i)=-3+i\) and then compute \(9+(-3+i)=6+i\). This then gives:

\[\begin{align*} & \phantom{xxxxxx)}1\phantom{xxx}6+i\\ -3+i & )\overline{\phantom{x}1\phantom{xxxx}9\phantom{xxxx}28\phantom{xxxx}30}\\ + & \underline{\phantom{xxx}-3+i\phantom{xxxxxxxxxxxx}}\\ & \phantom{xxxxx}6+i \end{align*}\]

Then we will compute (\(i^2=-1\)) \((6+i)(-3+i)=-19+3i\) and then compute \(28-(-19+3i)\). This then gives:

\[\begin{align*} & \phantom{xxxxxx)}1\phantom{xxxxx}6+i\phantom{xxxxx}9+3i\\ -3+i & )\overline{\phantom{x}1\phantom{xxxx}9\phantom{xxxxxx}28\phantom{xxxxxxx}30}\\ + & \underline{\phantom{xxx}-3+i\phantom{x}-19+3i\phantom{xxxxxxxx}}\\ & \phantom{xxxxx}6+i\phantom{xxxx}9+3i \end{align*}\]

Then we will compute \((9+3i)(-3+i)=30\) and then compute \(30+(-30)=0\). This then gives:

\[\begin{align*} & \phantom{xxxxxx)}1\phantom{xxxxx}6+i\phantom{xxxxx}9+3i\\ -3+i & )\overline{\phantom{x}1\phantom{xxxx}9\phantom{xxxxxx}28\phantom{xxxxxxx}30}\\ + & \underline{\phantom{xxx}-3+i\phantom{x}-19+3i\phantom{xxxxx}30}\\ & \phantom{xxxxx}6+i\phantom{xxxx}9+3i\phantom{xxxxxx}0 \end{align*}\]

Therefore, we have \(\frac{x^3+9x^2+28x+30}{x+3-i}=x^2+(6+i)x+(9+3i)\)

Remainder Theorem

We know that \(f(x)=g(x)q(x)+r(x)\) is from the division algorithm. If we force \(g(x)=x-k\) then we have the following identity

\[f(x)=(x-k)q(x)+r(x)\]

where \(r(x)\) is going to be a constant value. In fact, this leads to an important theorem.

Definition 37 (Remainder Theorem)

If a polynomial \(f(x)\) is divided by \(x-k\), then the remainder is equal to \(f(k)\).

Example 51

Let \(f(x)=-x^4+3x^2-4x-5\). Evaluate \(f(-3)\).

Solution:

Instead of evaluating \(f(-3)\) as the following:

\[\begin{align*} f(-3) & = -(-3)^4+3(-3)^2-4(-3)-5\\ & = -81 + 27 + 12 - 5\\ & = -47 \end{align*}\]

We can do the following:

\[\begin{align*} & \phantom{xxxxx}-1\phantom{xxx}3\phantom{xxx}-6\phantom{xxxxx}14\\ -3 & )\overline{\phantom{x}-1\phantom{xx}0\phantom{xxxx}3\phantom{xx}-4\phantom{xxxx}-5}\\ + & \underline{\phantom{xxxxxxx}3\phantom{xx}-9\phantom{xxx}18\phantom{xxx}-42}\\ & \phantom{xxxxxxx}3\phantom{xx}-6\phantom{xxx}14\phantom{xxx}-47 \end{align*}\]

Would also say, \(f(-3)=-47\).

Root (or Zero) of a Function

Definition 38 (Root (or Zero) or a Function)

A root (or zero) of a polynomial function \(f(x)\) is a number \(k\) such that \(f(k)=0\).

Currently, finding roots of a polynomial function is difficult when considering all possible real numbers as potential roots. However, in the following section, we can narrow down the search. For now, we will consider testing \(x=0\), \(x=1\), or \(x=-1\).

Example 52

Let \(f(x)=x^4-6x^3+8x^2+6x-9\). Find all the roots for \(f(x)\).

Solution:

It should be obvious that \(x=0\) is not a root. Therefore, we will move on to \(x=1\). Using the remainder theorem we will see if \(f(1)=0\).

\[\begin{align*} & \phantom{xxxxxxx}1\phantom{xx}-5\phantom{xxx}3\phantom{xxx}9\\ 1 & )\overline{\phantom{x}1\phantom{xx}-6\phantom{xxxx}8\phantom{xxx}6\phantom{x}-9}\\ + & \underline{\phantom{xxxxxxx}1\phantom{xx}-5\phantom{xxx}3\phantom{xxx}9}\\ & \phantom{xxxxx}-5\phantom{xxxx}3\phantom{xxx}9\phantom{xxx}0 \end{align*}\]

This means, \(f(1)=0\) and

\[f(x)=(x-1)(x^3-5x^2+3x+9)\]

We then want to see if \(f(-1)=0\). Using the remainder theorem we will evaluate \(x^3-5x^2+3x+9\) at \(x=-1\).

\[\begin{align*} & \phantom{xxxxxxx}1\phantom{xx}-6\phantom{xxxx}9\\ 1 & )\overline{\phantom{x}1\phantom{xx}-5\phantom{xxxx}3\phantom{xxxx}9}\\ + & \underline{\phantom{xxxxxxx}1\phantom{xxxx}6\phantom{xx}-9}\\ & \phantom{xxxxx}-6\phantom{xxxx}3\phantom{xxxx}0 \end{align*}\]

This means, \(f(-1)=0\) and

\[f(x)=(x-1)(x+1)(x^2-6x+9)\]

Looking at \(x^2-6x+9\) we know that it factors to \((x-3)(x-3)\). Therefore,

\[f(x)=(x-1)(x+1)(x-3)^2\]

and all the roots of \(f\) is \(\{1,-1,3\}\).

In the next section, we will discover how to find all possible rational roots. This will help find zero of polynomials more efficiently.