4.9 Synthetic Division and the Remainder Theorem

Learning Outcomes

Divide polynomials using synthetic division.
Evaluate a polynomial function using the Remainder Theorem.

Synthetic Division

As we have seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a degree $$\require{color}\require{enclose}1$$ polynomial whose leading coefficient is [latex]1[/latex].

Synthetic Division

Synthetic division is a shortcut that can be used when the divisor is a binomial in the form [latex]x–k[/latex], for a real number [latex]k[/latex]. In synthetic division, only the coefficients are used in the division process.

To illustrate the process, let’s review an example from the previous section of dividing [latex]x^2-4x-12[/latex] by [latex]x+2[/latex] using long division.

[latex]\begin{array}{l} \color{Gray}{\hspace{7pt}\textsf{deg.}\hspace{10pt}2\hspace{20pt}1\hspace{17pt}\textsf{c.t.}}\\[-10pt] \phantom{x+2 -x^2}\hspace{.5pt} {\color{Gray}{\Bigl|}} \phantom{{}-4}x {\color{Gray}{\Bigl|}} -6\\[-10pt] x+2 \enclose{longdiv}{\phantom{-}x^2 {\color{Gray}{\Bigl|}} -4x {\color{Gray}{\Bigl|}} -12}\\[-10pt] \phantom{x+2\hspace{3pt}} \enclose{bottom}{- x^2 {\color{Gray}{\Bigl|}} - 2x {\color{Gray}{\Bigl|}} \phantom{{} -12}}\\[-10pt] \phantom{x+2 -x^2 {\color{Gray}{\Bigl|}} \hspace{-9pt}} -6x {\color{Gray}{\Bigl|}} -12\\[-10pt] \phantom{x+2 -x^2 {\color{Gray}{\Bigl|}} \hspace{-10.5pt}} \enclose{bottom}{{}+6x {\color{Gray}{\Bigl|}} +12}\\[-10pt] \phantom{x+2 -x^2 {\color{Gray}{\Bigl|}} \hspace{-8.5pt} {\color{red}{{}+6x}} {\color{Gray}{\Bigl|}} {\color{red}{{}+{}}}}0 \end{array}[/latex]

Since the degree of the divisor is $$1$$ and the degree of the quotient is always the difference of the degrees of the dividend and the divisor, the degree of the quotient will be exactly $$1$$ less than the degree of the divisor. In our example, the degree of the quotient is $$2-1=1$$. Since the leading term of the divisor is $$x$$, the coefficients of the quotient will be the leading coefficients of subsequent current dividends. In particular, the leading coefficient of the quotient will always match the leading coefficient of the (original) dividend. The leading coefficient of each subsequent (current) dividend is obtained by subtracting the product of the coefficient of the quotient term of the degree matching the degree of that current dividend and the constant term of the divisor from the coefficient of the term of the same degree in the previous dividend. In our example, $$-6$$, the leading coefficient of the “second” dividend of degree $$1$$, is obtained by subtracting the product of the coefficient of the quotient term of degree $$1$$, $$1$$, and the constant term of the divisor, $$2$$, from the coefficient of the term of degree $$1$$ in the previous (original) dividend, $$-4$$: $$-4 -1\cdot 2=-6$$. That number, $$-6$$, becomes the constant term of the quotient. Recall that to avoid mistakes in subtraction of polynomials, in long division diagrams, we decided to write the opposite of the product of the given term of the quotient and the divisor to add instead of subtracting. In our example, the highlighted term, $${\color{red}{\enclose{box}{-\ 2x}}}$$ (see below) has been obtained by multiplying $${\color{Orange}{\enclose{box}{x}}}$$, the first term of the quotient, by $${\color{blue}{\enclose{box}{+\ 2}}}$$, the constant term of the divisor, and writing the opposite of the product, $${\color{red}{\enclose{box}{-\ 2x}}}$$.

[latex]\begin{array}{l} \color{Gray}{\hspace{7pt}\textsf{deg.}\hspace{11pt}2\hspace{21pt}1\hspace{18pt}\textsf{c.t.}}\\[-10pt] \phantom{x\enclose{box}{+\ 2} -x^2}\hspace{.5pt} {\color{Gray}{\Bigl|}} \phantom{-\ 4}{\color{Orange}{\enclose{box}{x}}} {\color{Gray}{\Bigl|}} -6\\[-10pt] x{\color{blue}{\enclose{box}{+\ 2}}} \enclose{longdiv}{\phantom{-}x^2 {\color{Gray}{\Bigl|}}\hspace{2pt} -4x {\color{Gray}{\Bigl|}} -12}\\[-10pt] \phantom{x\enclose{box}{+\ 2}\hspace{3pt}} \enclose{bottom}{- x^2 {\color{Gray}{\Bigl|}} {\color{red}{\enclose{box}{-\ 2x}}}{\color{Gray}{\Bigl|}} \phantom{{} -12}}\\[-10pt] \phantom{x+2 -x^2 {\color{Gray}{\Bigl|}} \hspace{-5pt}} -6x {\color{Gray}{\Bigl|}} -12\\[-10pt] \phantom{x+2 -x^2 {\color{Gray}{\Bigl|}} \hspace{-6.5pt}} \enclose{bottom}{{}+6x {\color{Gray}{\Bigl|}} +12}\\[-10pt] \phantom{x+2 -x^2 {\color{Gray}{\Bigl|}} \hspace{-4.5pt} {\color{red}{{}+6x}} {\color{Gray}{\Bigl|}} {\color{red}{{}+{}}}}0 \end{array}[/latex]

Note that $$x+2 = x-(-2)$$. To expedite the process of multiplication and addition in synthetic division, we will use the opposite of the constant term of the divisor. In our example, multiplying $${\color{Orange}{x}}$$ by $${\color{red}{-}}{\color{blue}{2}}$$ would immediately result in $${\color{red}{-2x}}$$. Let’s rewrite our long division diagram (so far without the quotient) and note which numbers actually matter.

[latex]\begin{array}{l} {\color{lightgray}{x-(}}{-2{\color{lightgray}{)}}} \enclose{longdiv}{\phantom{-}1{\color{lightgray}{x^2}} {\color{Gray}{\Bigl|}} {-4}{\color{lightgray}{x}} {\color{Gray}{\Bigl|}} {-12}}\\[-10pt] \phantom{{x-}{(-2)}\hspace{3pt}} \enclose{bottom}{{\color{lightgray}{-1 x^2}} {\color{Gray}{\Bigl|}} {-2}{\color{lightgray}{x}} {\color{Gray}{\Bigl|}} \phantom{-12}}\\[-10pt] \phantom{{x-}{(-2)} -1x^2 {\color{Gray}{\Bigl|}}}\hspace{.5pt} {-6}{\color{lightgray}{x}} {\color{Gray}{\Bigl|}} {\color{lightgray}{-12}}\\[-10pt] \phantom{{x-}{(-2)} -1x^2 {\color{Gray}{\Bigl|}} \hspace{-1.5pt}} \enclose{bottom}{{\color{lightgray}{+6x}} {\color{Gray}{\Bigl|}} {+12}}\\[-10pt] \phantom{{x-}{(-2)} -1x^2 {\color{Gray}{\Bigl|}} \hspace{8pt} {\color{red}{+6x}} {\color{Gray}{\Bigl|}}}0 \end{array}[/latex]

We collapse the diagram above into a table by moving each of the rows up to fill any vacant spots:

[latex]\begin{array}{r|rrr} -2 & 1 & -4 &-12\\ & & -2 & +12\\ \hline & & -6 & 0 \end{array}[/latex]

Note that $$-2$$ in the upper left corner is the opposite of the constant term of the divisor. We will refer to it as the number associated with the divisor. The remaining numbers in the first row are the coefficients of the terms of the (original) dividend in descending order of degrees of the terms of the dividend. The number in the lower right corner, $$0$$, is the remainder, and $$-6$$ to the left of it is the constant term of the quotient. We’re only missing the leading coefficient of the quotient, $$1$$, which simply matches the leading coefficient of the dividend. We will just “drop down” that leading coefficient of the dividend to the third row.

[latex]\begin{array}{rrrr} {\color{Gray}{\textsf{deg.}}} & {\color{Gray}{2}} & \phantom{-}{\color{Gray}{1}} & \hspace{4pt} {\color{Gray}{\textsf{c.t.}}} \end{array}[/latex]
[latex]\begin{array}{c|rrr} {\color{blue}{-2}} & {\color{Green}{1}} & {\color{Green}{-4}} & {\color{Green}{-12}}\\ \phantom{\textsf{deg.}} & \downarrow & {\color{red}{-2}} & {\color{red}{+12}}\\ \hline & {\color{Orange}{1}} & {\color{Orange}{-6}} & {\color{Plum}{0}} \end{array}[/latex]
[latex]\begin{array}{rrrr} {\color{Gray}{\ \textsf{deg.}}} & {\color{Gray}{1}} & \hspace{1pt} {\color{Gray}{\textsf{c.t.}}} & {\color{Gray}{\textsf{rem.}}} \end{array}[/latex]

The degree indicators are included only for convenience and clarity. They are not a part of the diagram. The diagram above is equivalent to the long division diagram below.

[latex]\begin{array}{l} \color{Gray}{\hspace{8pt}\textsf{deg.}\hspace{12pt}2\hspace{21pt}1\hspace{17pt}\textsf{c.t.}}\\[-10pt] \phantom{x+2 -1x^2}\hspace{.5pt} {\color{Gray}{\Bigl|}} \phantom{{}-{}}{\color{Orange}{1x}} {\color{Gray}{\Bigl|}} {\color{Orange}{{}-6}}\\[-10pt] {\color{blue}{x+2}} \enclose{longdiv}{\phantom{-}{\color{Green}{1x^2}} {\color{Gray}{\Bigl|}} {\color{Green}{{}-4x}} {\color{Gray}{\Bigl|}} {\color{Green}{{}-12}}}\\[-10pt] \phantom{x+2\hspace{3pt}} \enclose{bottom}{{\color{red}{-1x^2}} {\color{Gray}{\Bigl|}} {\color{red}{{}-2x}} {\color{Gray}{\Bigl|}} \phantom{{} -12}}\\[-10pt] \phantom{x+2 -1x^2 {\color{Gray}{\Bigl|}} \hspace{-9pt}} {\color{Green}{{}-6x}} {\color{Gray}{\Bigl|}} {\color{Green}{{}-12}}\\[-10pt] \phantom{x+2 -1x^2 {\color{Gray}{\Bigl|}} \hspace{-10.5pt}} \enclose{bottom}{{\color{red}{{}+6x}} {\color{Gray}{\Bigl|}} {\color{red}{{}+12}}}\\[-10pt] \phantom{x+2 -1x^2 {\color{Gray}{\Bigl|}} \hspace{-8.5pt} {\color{red}{{}+6x}} {\color{Gray}{\Bigl|}} {\color{red}{{}+{}}}}{\color{Plum}{0}} \end{array}[/latex] or [latex]\begin{array}{l} \color{Gray}{\hspace{17pt}\textsf{deg.}\hspace{14pt}2\hspace{24pt}1\hspace{13pt}\textsf{c.t.}}\\[-10pt] \phantom{{x-}{(-2)} -\enclose{box}{1}x^2}\hspace{.5pt} {\color{Gray}{\Bigl|}} \phantom{-}{\color{Orange}{\hspace{3pt}1{\color{lightgray}{x}}}} {\color{Gray}{\Bigl|}} {\color{Orange}{-6}}\\[-10pt] {\color{lightgray}{x-(}}{{\color{blue}{-2}}{\color{lightgray}{)}}} \enclose{longdiv}{\phantom{-}{\color{Orange}{\enclose{box}{\color{Green}{1}}}}{\color{lightgray}{x^2}} {\color{Gray}{\Bigl|}} \hspace{3pt}{\color{Green}{-4{\color{lightgray}{x}}}} {\color{Gray}{\Bigl|}} {\color{Green}{-12}}}\\[-10pt] \phantom{{x-}{(-2)}\hspace{3pt}} \enclose{bottom}{{\color{lightgray}{\hspace{4pt} -1x^2}} {\color{Gray}{\Bigl|}} \hspace{3pt}{\color{red}{-2{\color{lightgray}{x}}}} {\color{Gray}{\Bigl|}} \phantom{ -12}}\\[-10pt] \phantom{{x-}{(-2)} -1x^2 {\color{Gray}{\Bigl|}} \hspace{4pt}} {\color{Orange}{\enclose{box}{\color{Green}{-6}}}}{\color{lightgray}{x}} {\color{Gray}{\Bigl|}} {\color{lightgray}{-12}}\\[-10pt] \phantom{{x-}{(-2)} -1x^2 {\color{Gray}{\Bigl|}} \hspace{4pt}} \enclose{bottom}{{\color{lightgray}{\hspace{2pt}+6x}} {\color{Gray}{\Bigl|}} {\color{red}{+12}}}\\[-10pt] \phantom{{x-}{(-2)} -1x^2 {\color{Gray}{\Bigl|}} \hspace{16pt} {\color{red}{+6x}} {\color{Gray}{\Bigl|}}}{\color{Plum}{0}} \end{array}[/latex]

emphasizing only the numbers that matter.

Recall that the $${\color{Orange}{\enclose{box}{\color{Green}{1}}}}$$ from the original dividend that matched the $${\color{Orange}{1}}$$ in the quotient has been “dropped down” in the synthetic division diagram, and the $${\color{Orange}{\enclose{box}{\color{Green}{-6}}}}$$ from the “second” dividend that matched the $${\color{Orange}{-6}}$$ in the quotient has been obtained by adding the $${\color{Green}{-4}}$$ from the original dividend to the $${\color{red}{-2}}$$, the result of multiplying the $${\color{Orange}{1}}$$ by the $${\color{blue}{-2}}$$, the number associated with the divisor, etc.

Okay, but how do we recreate the complete synthetic division diagram without going through long division first?

In setting up the synthetic division diagram for our division problem, we start with writing $${\color{blue}{-2}}$$, the number associated with the divisor, in the upper left corner, and the coefficients of the divisor in columns of proper degree (on top) to complete the first row. Then we “drop down” the leading coefficient, $$1$$, to the third row indicating it by an arrow, $$\downarrow$$, in the second row.

Then we multiply that “dropped down” $${\color{Orange}{1}}$$ from the last row of the diagram by the number associated with the divisor, $${\color{blue}{-2}}$$, place the resulting product, $${\color{red}{-2}}$$, in the second row of the diagram in the first vacant spot, add to the number above it, $${\color{Green}{-4}}$$, and place the sum, $${\color{Orange}{-6}}$$, below it in the first vacant spot of the last row.

[latex]\begin{array}{c|rrr} {\color{blue}{-2}} & {\color{Green}{1}} & {\color{Green}{-4}} & {\color{Green}{-12}}\\ \phantom{\textsf{deg.}} & \downarrow & {\color{red}{-2}} & {\phantom{+12}}\\ \hline & {\color{Orange}{1}} & {\color{Orange}{-6}} & {\phantom{0}} \end{array}[/latex]

In similar way, we multiply that $${\color{Orange}{-6}}$$ by the number associated with the divisor, $${\color{blue}{-2}}$$, place the resulting product, $${\color{red}{+12}}$$, in the second row of the diagram in the first vacant spot, add to the number above it, $${\color{Green}{-12}}$$, and place the sum, $${\color{Plum}{0}}$$, below it in the first vacant spot of the last row.

[latex]\begin{array}{c|rrr} {\color{blue}{-2}} & {\color{Green}{1}} & {\color{Green}{-4}} & {\color{Green}{-12}}\\ \phantom{\textsf{deg.}} & \downarrow & {\color{red}{-2}} & {\color{red}{+12}}\\ \hline & {\color{Orange}{1}} & {\color{Orange}{-6}} & {\color{Plum}{0}} \end{array}[/latex]

Since it was the lower right corner of the diagram, the $${\color{Plum}{0}}$$ “completed” the diagram and is the remainder of the division. We hope you’ve already noted the pattern in creating the synthetic division diagram for our division problem.

That pattern of properly placing the number associated with the divisor, the coefficients of the dividend, “dropping down” the leading coefficient of the dividend, and “generating” the remaining numbers in the diagram, generalizes to division of any polynomial by a degree $$1$$ binomial with the leading coefficient of $$1$$. Let’s review it making sure we understand all steps properly.

How To: DIVIDE A polynomial, $$P(x)$$, BY A BINOMIAL, $$x-k$$, using synthetic division

Make sure you view the dividend, $$P(x)$$, in descending powers of the variable and note the degree of $$P(x)$$. Include terms of all degrees less than or equal to the degree of $$P(x)$$ including all coefficients with a coefficient of zero for all missing degrees of terms.
Set up the diagram:

[latex]\begin{array}{r|l} {\color{blue}k} & {\color{Green}\textsf{dividend coefficients}}\\ & \\ \hline & \end{array}[/latex]

Remember that $${\color{blue}k}$$, the number associated with the divisor, is the number “behind” the “minus” in the divisor. In other words, $${\color{blue}k}$$ is the opposite of the constant term of the divisor. The dividend coefficients must be written in the same order as they appear in the form of $$P(x)$$ described above (including the zeros).

Drop down the leading coefficient of $$P(x)$$ to the last row of the diagram.
Repeat the following “division step” until you “complete” the diagram:

Multiply the last number from the last row by $${\color{blue}k}$$, place the resulting product in the second row of the diagram in the first vacant spot, add to the number above it, and place the sum below it in the first vacant spot of the last row.

The completed diagram will have the following form:

[latex]\begin{array}{r|l} {\color{blue}k} & \enclose{box}{\color{Green}\mathstrut\quad \textsf{dividend}\hspace{8.5pt} \textsf{coefficients}\quad}\\ & \downarrow\ \enclose{box}{\mathstrut\color{red}\textsf{“division step products”}}\\ \hline & \enclose{box}{\color{Orange}\mathstrut\textsf{quotient coefficients}}\hspace{3.5pt} \enclose{box}{\color{Plum}\mathstrut\textsf{rem.}} \end{array}[/latex]

Remember that the degree of the quotient is $$1$$ less than the degree of $$P(x)$$.

ExAMPLE

Use synthetic division to divide $$2x^3-3x^2-x+3$$ by $$x+3$$.

Show Solution

We view $$P(x)=2x^3-3x^2-x+3$$ as $${\color{Green}{2}}x^3+({\color{Green}{-3}})x^2+({\color{Green}{-1}})x+{\color{Green}{3}}$$. Note that the degree of $$P(x)$$ is $$3$$ and no terms of any degree less than $$3$$ are missing. Note that $$k={\color{blue}{-3}}$$ since $$x+3=x-({\color{blue}{-3}})$$. We set up the synthetic division diagram including degrees of the dividend for clarity, and drop down the leading coefficient of $$P(x)$$, $$2$$.

[latex]\begin{array}{rrrrr} {\color{Gray}{\textsf{deg.}}} & {\color{Gray}{3}} & \phantom{-}{\color{Gray}{2}} & \hspace{8pt} {\color{Gray}{1}} & \hspace{1pt} {\color{Gray}{\textsf{c.t.}}} \end{array}[/latex]
[latex]\begin{array}{c|rrrr} {\color{blue}{-3}} & {\color{Green}{2}} & {\color{Green}{-3}} & {\color{Green}{-1}} & {\color{Green}{+3}}\\ \phantom{\textsf{deg.}} & \downarrow & & & \\ \hline & {\color{Orange}{2}} & & & \end{array}[/latex]

The first division step is to multiply the last number from the last row, $${\color{Orange}{2}}$$, by $${\color{blue}-3}$$, place the resulting product, $${\color{red}{-6}}$$, in the second row of the diagram in the first vacant spot, add to the number above it, $${\color{Green}{-3}}$$, and place the sum, $${\color{Orange}{-9}}$$, below it in the first vacant spot of the last row:

[latex]\begin{array}{c|rrrr} {\color{blue}{-3}} & {\color{Green}{2}} & {\color{Green}{-3}} & {\color{Green}{-1}} & {\color{Green}{+3}}\\ \phantom{\textsf{deg.}} & \downarrow & {\color{red}{-6}} & & \\ \hline & {\color{Orange}{2}} & {\color{Orange}{-9}} & & \end{array}[/latex]

We continue with the remaining division steps (the second one being to multiply the last number from the last row, $${\color{Orange}{-9}}$$, by $${\color{blue}-3}$$, place the resulting product, $${\color{red}{+27}}$$, in the second row of the diagram in the first vacant spot, add to the number above it, $${\color{Green}{-1}}$$, and place the sum, $${\color{Orange}{26}}$$, below it in the first vacant spot of the last row, etc.) until we complete the diagram remembering that the number in the lower right corner is the remainder. It might be more convenient to include the variable factors in proper powers instead of degrees below the third row to easier identify the quotient.

[latex]\begin{array}{c|rrrr} {\color{blue}{-3}} & {\color{Green}{2}} & {\color{Green}{-3}} & {\color{Green}{-1}} & {\color{Green}{3}}\\ \phantom{\textsf{deg.}} & \downarrow & {\color{red}{-6}} & {\color{red}{+27}} & {\color{red}{-84}}\\ \hline & {\color{Orange}{2}} & {\color{Orange}{-9}} & {\color{Orange}{26}} & {\color{Plum}{-81}} \end{array}[/latex]
[latex]\begin{array}{ccccc} \phantom{\textsf{deg.}\ } & {\color{Gray}{x^2}} & \; {\color{Gray}{x}}\; & \phantom{+}{\color{Gray}{1}}\phantom{-} & {\color{Gray}{\textsf{rem.}}} \end{array}[/latex]

The quotient is $${\color{Orange}{2x^2-9x+26}}$$, the remainder is $${\color{Plum}{-81}}$$, and we can write the division statement in the form

[latex]\displaystyle\frac{{\color{Green}{2x^3-3x^2-x+3}}}{{\color{blue}{x+3}}} = {\color{Orange}{2x^2-9x+26}} + \frac{{\color{Plum}{-81}}}{{\color{blue}{x+3}}}[/latex]

or, in equivalent form, as

[latex]\displaystyle\frac{{\color{Green}{2x^3-3x^2-x+3}}}{{\color{blue}{x+3}}} = {\color{Orange}{2x^2-9x+26}} {\color{Plum}{{}-{}}} \frac{{\color{Plum}{81}}}{{\color{blue}{x+3}}}[/latex]

We verify the correctness of this statement by simplifying $$({\color{blue}{x+3}})\cdot \left({\color{Orange}{2x^2-9x+26}}\right) + ({\color{Plum}{-81}})$$ to verify if it matches $${\color{Green}{2x^3-3x^2-x+3}}$$. We have

[latex]\begin{align} (x+3)\cdot \left(2x^2-9x+26\right) + (-81) &= 2x^3-9x^2+26x\\[-10pt] &\quad\; \phantom{2x^3}+6x^2-27x+84-81\\ &= 2x^3-3x^2-x+3\ \large\checkmark \end{align}[/latex]

The next example will include a polynomial with a missing term of a specific degree.

Example

Use synthetic division to divide [latex]-9x^4+10x^3+7x^2-6[/latex] by [latex]x - 1[/latex].

Show Solution

Notice there is no degree $$1$$ term in $$P(x)=-9x^4+10x^3+7x^2-6$$, so we view it as $${\color{Green}{-9}}x^4+{\color{Green}{10}}x^3+{\color{Green}{7}}x^2+{\color{Green}{0}}x+({\color{Green}{-6}})$$. Also, note that $$k={\color{blue}{1}}$$. We set up the synthetic division diagram including degrees of the dividend for clarity, and drop down the leading coefficient of $$P(x)$$, $$-9$$.

[latex]\begin{array}{rrrrrr} {\color{Gray}{\textsf{deg.}}} & \phantom{-}{\color{Gray}{4}} & \phantom{+1}{\color{Gray}{3}} & \phantom{+}{\color{Gray}{2}} & \hspace{8pt}{\color{Gray}{1}} & \hspace{1pt} {\color{Gray}{\textsf{c.t.}}} \end{array}[/latex]
[latex]\begin{array}{c|rrrrr} {\color{blue}{1}} & {\color{Green}{-9}} & {\color{Green}{+10}} & {\color{Green}{+7}} & {\color{Green}{+0}} & {\color{Green}{-6}}\\ \phantom{\textsf{deg.}} & \downarrow & & & &\\ \hline & {\color{Orange}{-9}} & & & & \end{array}[/latex]

We continue with the division steps (the first one being to multiply the last number from the last row, $${\color{Orange}{-9}}$$, by $${\color{blue}1}$$, place the resulting product, $${\color{red}{-9}}$$, in the second row of the diagram in the first vacant spot, add to the number above it, $${\color{Green}{10}}$$, and place the sum, $${\color{Orange}{1}}$$, below it in the first vacant spot of the last row, etc.) until we complete the diagram remembering that the number in the lower right corner is the remainder. We include the variable factors in proper powers instead of degrees to easier identify the quotient.

[latex]\begin{array}{c|rrrrr} {\color{blue}{1}} & {\color{Green}{-9}} & {\color{Green}{+10}} & {\color{Green}{+7}} & {\color{Green}{+0}} & {\color{Green}{-6}}\\ \phantom{\textsf{deg.}} & \downarrow & {\color{red}{-9}} & {\color{red}{+1}} & {\color{red}{+8}} & {\color{red}{+8}}\\ \hline & {\color{Orange}{-9}} & {\color{Orange}{1}} & {\color{Orange}{8}} & {\color{Orange}{8}} & {\color{Plum}{2}} \end{array}[/latex]
[latex]\begin{array}{cccccc} \phantom{\textsf{deg.}\ } &\ \ {\color{Gray}{x^3}} & {\ \ \ \ \color{Gray}{x^2}} & \ \ \ {\color{Gray}{x}}& \ \ \ {\color{Gray}{1}} & {\color{Gray}{\textsf{rem.}}} \end{array}[/latex]

The quotient is $${\color{Orange}{-9x^3+x^2+8x+8}}$$, the remainder is $${\color{Plum}{2}}$$, and we can write the division statement in the form

[latex]\displaystyle\frac{{\color{Green}{-9x^4+10x^3+7x^2-6}}}{{\color{blue}{x-1}}} = {\color{Orange}{-9x^3+x^2+8x+8}} + \frac{{\color{Plum}{2}}}{{\color{blue}{x-1}}}[/latex]

We verify the correctness of this statement by simplifying $$({\color{blue}{x-1}})\cdot \left({\color{Orange}{-9x^3+x^2+8x+8}}\right) + {\color{Plum}{2}}$$ to verify if it matches $${\color{Green}{-9x^4+10x^3+7x^2-6}}$$. We have

[latex]\begin{align} (x-1)\cdot \left(-9x^3+x^2+8x+8\right) + 2 &= -9x^4+x^3+8x^2+8x\\[-10pt] &\quad\; \phantom{-9x^4}+9x^3-x^2-8x-8+2\\ &= -9x^4+10x^3+7x^2-6\ \large\checkmark \end{align}[/latex]

These next videos contain further examples of using synthetic division.

Synthetic division is a powerful tool in examining polynomial functions. You will learn more about it in a College Algebra class. We will present one application of synthetic division in evaluating polynomial functions.

The Remainder Theorem and Evaluation of Polynomial Functions

Let’s revisit the example above where the dividend, $$P(x)=-9x^4+10x^3+7x^2-6$$, was divided by the divisor, $$x-1$$. The quotient was $$-9x^3+x^2+8x+8$$ and the remainder was $$2$$. Note that

[latex]\begin{align} P({\color{Green}{1}})&=-9\cdot{\color{Green}{1}}^4+10\cdot{\color{Green}{1}}^3+7\cdot{\color{Green}{1}}^2-6\\ &=-9+10+7-6\\ &=2 \end{align}[/latex]

It’s not a coincidence that the value, $$P(1)$$, matches the remainder of the division of $$P(x)$$ by $$x-1$$. Recall that an equivalent form of a general polynomial division statement is

$$\textsf{dividend} = \textsf{divisor}\cdot\textsf{quotient}+\textsf{remainder}$$

If we rewrite it using function notation, denoting the divisor $$d(x)$$, the quotient $$q(x)$$, and the reminder $$r(x)$$, we get

$$P(x)=d(x)\cdot q(x)+r(x)$$.

This functional equation is true for every real number $$x$$. Let $$k$$ be a real number. If we let $$d(x)=x-k$$, then the remainder must be of degree les than $$1$$, meaning a constant. Let’s denote that constant $$R$$. We get

$$P(x)=(x-k)\cdot q(x)+R$$

and

[latex]\begin{align} P({\color{Green}{k}})&=({\color{Green}{k}}-k)\cdot q({\color{Green}{k}})+R\\ &=0\cdot q({\color{Green}{k}})+R\\ &=0+R\\ &=R \end{align}[/latex]

We have just proven the following Remainder Theorem.

THE REMAINDER THEOREM

For any polynomial [latex]P(x)[/latex] of degree $$1$$ or higher and any real number $$k$$, $$P(k)$$ is equal to the remainder of the division of $$P(x)$$ by $$x-k$$. Algebraically:

If [latex]\dfrac{P(x)}{x-k}=q(x)+\dfrac{R}{x-k}[/latex], then [latex]P(k)=R[/latex].

As always, an equality is true “both ways”. In the Remainder Theorem, we can say that

the value of $$P(x)$$ at $$x=k$$, $$P(k)$$, is equal to the remainder of the division of $$P(x)$$ by $$x-k$$ (and we can identify that remainder using synthetic division with $$k$$ as the number associated with the divisor), or
the remainder of the division of $$P(x)$$ by $$x-k$$ is equal to the value of $$P(x)$$ at $$x=k$$, $$P(k)$$.

ExAMPLE

Given the function $$f(x)=x^4+x^3-5x^2+x-6$$, calculate $$f(-2)$$, $$f(1)$$, $$f(2)$$, and $$f(3)$$ using synthetic division and the Remainder Theorem.

Show Solution

Of course, $$f(x)$$ takes a role of $$P(x)$$ in the Remainder Theorem. We view $$f(x)$$ as $${\color{Green}{1}}x^4+{\color{Green}{1}}x^3+({\color{Green}{-5}})x^2+{\color{Green}{1}}x+({\color{Green}{-6}})$$. Note that no terms of any degree less than the degree of $$f(x)$$ are missing. We need four synthetic division diagrams, one for each value of $$k$$:

[latex]\begin{array}{c|rrrrr} {\color{blue}{-2}} & {\color{Green}{1}} & {\color{Green}{+1}} & {\color{Green}{-5}} & {\color{Green}{+1}} & {\color{Green}{-6}}\\ & \downarrow & {\color{red}{-2}} & {\color{red}{+2}} & {\color{red}{+6}} & {\color{red}{-14}}\\ \hline & {\color{Orange}{1}} & {\color{Orange}{-1}} & {\color{Orange}{-3}} & {\color{Orange}{7}} & {\color{Plum}{-20}} \end{array}[/latex]

So, $$f({\color{blue}{-2}})={\color{Plum}{-20}}$$. (We changed the usual green color indicating the input of evaluation in this whole example to emphasize that it corresponds to the number associated with the divisor when we use the Remainder Theorem.)

[latex]\begin{array}{c|rrrrr} {\color{blue}{1}} & {\color{Green}{1}} & {\color{Green}{+1}} & {\color{Green}{-5}} & {\color{Green}{+1}} & {\color{Green}{-6}}\\ & \downarrow & {\color{red}{+1}} & {\color{red}{+2}} & {\color{red}{-3}} & {\color{red}{-2}}\\ \hline & {\color{Orange}{1}} & {\color{Orange}{2}} & {\color{Orange}{-3}} & {\color{Orange}{-2}} & {\color{Plum}{-8}} \end{array}[/latex]

So, $$f({\color{blue}{1}})={\color{Plum}{-8}}$$.

[latex]\begin{array}{c|rrrrr} {\color{blue}{2}} & {\color{Green}{1}} & {\color{Green}{+1}} & {\color{Green}{-5}} & {\color{Green}{+1}} & {\color{Green}{-6}}\\ & \downarrow & {\color{red}{+2}} & {\color{red}{+6}} & {\color{red}{+2}} & {\color{red}{+6}}\\ \hline & {\color{Orange}{1}} & {\color{Orange}{3}} & {\color{Orange}{+1}} & {\color{Orange}{3}} & {\color{Plum}{0}} \end{array}[/latex]

So, $$f({\color{blue}{2}})={\color{Plum}{0}}$$.

[latex]\begin{array}{c|rrrrr} {\color{blue}{3}} & {\color{Green}{1}} & {\color{Green}{+1}} & {\color{Green}{-5}} & {\color{Green}{+1}} & {\color{Green}{-6}}\\ & \downarrow & {\color{red}{+3}} & {\color{red}{+12}} & {\color{red}{+21}} & {\color{red}{+66}}\\ \hline & {\color{Orange}{1}} & {\color{Orange}{4}} & {\color{Orange}{7}} & {\color{Orange}{22}} & {\color{Plum}{60}} \end{array}[/latex]

So, $$f({\color{blue}{3}})={\color{Plum}{60}}$$.

We encourage you to check our answers in the example above by plugging the numbers into $$f(x)$$ and simplifying to compute the indicated values.

Note that although synthetic division is very useful, the numbers in the diagram can become large in magnitude making the process a bit more tedious. Fractions or any real numbers can be used as coefficients possibly making the computations quite cumbersome.

We finish with an example in which we determine the remainder of the division using the Reminder Theorem “backward”.

ExAMPLE

Determine the remainder of the division of $$-3x^3+2x^2-3x+4$$ by $$x+2$$ using the Remainder Theorem.

Show Solution

Module 4: Rational Expressions, Functions, and Equations