Synthetic Division

How synthetic division works

Synthetic division divides a polynomial $P(x)$ by a binomial of the form $x-a$. Some problems call this Ruffini's rule; it is exactly the same procedure.

Write the coefficients of $P(x)$ in a row, using $0$ for any missing power. Put $a$ to the left. Then follow these steps:

1. Bring down the first coefficient.

2. Multiply the number you just wrote on the bottom row by $a$, and write the product under the next coefficient.

3. Add that column and write the sum on the bottom row.

4. Repeat steps 2 and 3 until you reach the last column.

The last number on the bottom row is the remainder. The other bottom numbers are the coefficients of the quotient, which has degree one less than $P(x)$. Each column is one step of long division, with the subtraction built into the addition.

Evaluating a polynomial: the Remainder Theorem

Look again at the first example: dividing by $x-3$ left remainder $4$. Is that a coincidence? Plug $x=a$ into $P(x)=(x-a)Q(x)+R$. The first term vanishes and you get $P(a)=R$.

So synthetic division is a fast way to evaluate a polynomial: the last number in the table is $P(a)$. For high-degree polynomials this beats substituting term by term, since you only multiply and add.

Testing zeros and factors: the Factor Theorem

If the remainder is $0$, then $P(a)=0$. In that case $a$ is a zero (or root) of $P$ and the division is exact.

Factoring when you know one zero

Once you know a zero $a$, synthetic division hands you the quotient: a polynomial of one degree lower. Keep factoring the quotient, using synthetic division again or ordinary factoring.

Finding all the zeros

What if nobody tells you a zero? You can narrow down the candidates.

Why? If $\frac{p}{q}$ is a zero, multiplying $P\!\left(\frac pq\right)=0$ by $q^n$ shows $p$ must divide the constant term and $q$ must divide the leading coefficient. List the candidates, test the easy ones ($\pm1,\pm2,\dots$) with a quick synthetic division, and when one works, continue with the quotient.

When you reach a quadratic quotient, factor it or use the quadratic formula. If its discriminant $b^2-4ac$ is negative, the quadratic has no real zeros and is irreducible over the real numbers. Keep it as one factor.

Unknown coefficients from a division condition

Sometimes a coefficient is a letter. A condition such as "divisible by $x-a$" or "remainder $c$" turns into an equation: run the synthetic division with the letter in it, and set the last number equal to what the condition says.

Building a polynomial from its zeros

The Factor Theorem also works backward. If $r_1$ and $r_2$ are zeros, then $(x-r_1)$ and $(x-r_2)$ are factors, and the missing information fills in the rest.

When the divisor is $cx-a$

Synthetic division needs a divisor of the form $x-a$. A fractional $a$ is fine. To divide by $x-\tfrac12$, use $a=\tfrac12$.

For a divisor like $2x-3$, the coefficient of $x$ is not $1$. Factor it out: $2x-3=2\!\left(x-\tfrac32\right)$. Run the table with $a=\tfrac32$, then divide the quotient by $2$. The remainder stays the same.

Finding dimensions: area and volume

Area is length times width, and volume is length times width times height. So if you know one dimension, dividing finds the product of the others. A real dimension must divide exactly, so the remainder must be $0$.

Now back to Jayden's box, with $V(x)=x^3+6x^2+11x+6$ and height $x+1$.

Zeros in real-world models

Many models ask "when is the output zero?": profit zero (break-even), height zero, position zero. Set the polynomial equal to $0$, find the zeros, and then keep only the ones that make sense in the situation.

A pattern: dividing $x^n-a^n$ by $x-a$

Divide $x^4-81$ by $x-3$. The row is $1,\,0,\,0,\,0,\,-81$ and $a=3$.

The bottom row is $1,\,3,\,9,\,27$, the powers of $a=3$. Each product just multiplies the previous entry by $a$, and the zeros in the middle never disturb it. The remainder is $a^n-a^n=0$.

Polynomial inequalities

To solve $P(x)\ge0$ (or $<$, $>$, $\le$), the sign of $P$ can only change at its zeros. So:

1. Move everything to one side so the other side is $0$.

2. Factor $P(x)$ completely, using synthetic division to find the zeros.

3. Mark the zeros on a number line. They cut it into intervals.

4. Test one value in each interval to get its sign.

5. Pick the intervals that match the inequality. Use brackets for $\ge,\le$ (the zeros count) and parentheses for $>,<$.

Summary

Back to Jayden's box: dividing $V(x)$ by $x+1$ gave a base area $x^2+5x+6=(x+2)(x+3)$, so the box is $(x+1)$ by $(x+2)$ by $(x+3)$ inches.