Lesson 51: Using Synthetic Division

New Concept

If a polynomial $f(x)$ is divided by $x - k$, the remainder is $r = f(k)$.

Why it matters

This theorem reveals a profound link between division and function evaluation, turning a complex calculation into a simple substitution. Mastering this is key to efficiently finding the roots of polynomial equations, a central challenge in higher algebra.

What’s next

Next, you’ll use synthetic division as a powerful shortcut to apply this theorem and evaluate complex polynomials in just a few steps.

To divide a polynomial $f(x)$ by $x-k$, you can use synthetic division. Write $k$ and the coefficients of $f(x)$ on the first line. Bring down the first coefficient, then repeatedly multiply the newest quotient coefficient by $k$ and add it to the next coefficient of $f(x)$. The final result is the remainder.

Example 1: To divide $2x^3 - 3x^2 - 5x + 7$ by $x + 1$, use $k=-1$. The quotient is $2x^2 - 5x$ and the remainder is $7$.

Example 2: To divide $x^3 + 4x^2 - 9$ by $x - 2$, use $k=2$ and a placeholder for the missing $x$ term. The quotient is $x^2 + 6x + 12$ and the remainder is $15$.

Think of synthetic division as a cheat code for polynomial division, but only when you're dividing by a simple term like $x - k$. It strips away all the variables, letting you work just with the numbers in a neat little grid. It's a fast-paced rhythm of multiplying and adding that makes complex division problems much cleaner and quicker to solve.

If a polynomial $f(x)$ is divided by $x - k$, the remainder is $r = f(k)$.

Example 1: To find the remainder of $f(x) = x^3 - 2x^2 - 5x + 6$ when divided by $x-3$, we know the remainder is $f(3)$.
$f(3) = (3)^3 - 2(3)^2 - 5(3) + 6 = 27 - 18 - 15 + 6 = 0$. The remainder is $0$.
Example 2: The remainder when $P(x) = 2x^3 - 7x + 1$ is divided by $x+2$ is $P(-2)$.
$P(-2) = 2(-2)^3 - 7(-2) + 1 = -16 + 14 + 1 = -1$. The remainder is $-1$.

The Remainder Theorem is a fantastic mathematical shortcut that connects division to evaluating functions. It tells us that the leftover 'remainder' from a synthetic division problem is the exact same number you'd get if you plugged the divisor's value into the polynomial. So, dividing $f(x)$ by $x-3$ gives a remainder that equals $f(3)$. Cool, right?

By the Remainder Theorem, you can divide $f(x)$ by $x - k$ to find $f(k)$. If synthetic division is used to divide, the process is called synthetic substitution.

Example 1: Find $f(4)$ for $f(x) = -x^4 + 3x^3 + 10x - 5$ using synthetic substitution.

The remainder is $-29$, so $f(4) = -29$.
Example 2: For $P(x) = 12x^3 - 16x^2 + 150x + 154$, find $P(6)$.

The remainder shows that $P(6) = 3070$.

Synthetic substitution is the practical use of the Remainder Theorem. Instead of painstakingly plugging a number into a huge polynomial, you just perform a quick synthetic division. The final remainder you get is the answer you were looking for! It's a powerful and efficient method for evaluating polynomials, especially when the numbers get big and messy.