Synthetic substitution
Use synthetic substitution to evaluate polynomials at a value: list coefficients, bring down the first, multiply and add repeatedly to reach the remainder that equals p(a).
Key Concepts
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. \n $$ \begin{array}{c|ccccc} 4 & 1 & 3 & 0 & 10 & 5 \\ & & 4 & 4 & 16 & 24 \\ \hline & 1 & 1 & 4 & 6 & 29 \end{array} $$ \n The remainder is $ 29$, so $f(4) = 29$. \n Example 2: For $P(x) = 12x^3 16x^2 + 150x + 154$, find $P(6)$. \n $$ \begin{array}{c|cccc} 6 & 12 & 16 & 150 & 154 \\ & & 72 & 336 & 2916 \\ \hline & 12 & 56 & 486 & 3070 \end{array} $$ \n 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.
Common Questions
What is synthetic substitution and how does it differ from direct substitution?
Synthetic substitution is a streamlined numeric process that evaluates a polynomial using only addition and multiplication, avoiding higher-power arithmetic. Direct substitution plugs the value into each term separately. Synthetic substitution is faster and less error-prone for polynomials of degree 3 or higher.
What are the steps for synthetic substitution?
Write all coefficients in order, including zero for any missing degree. Bring down the first coefficient. Multiply it by the evaluation value a, write the product under the next coefficient, and add. Repeat across all coefficients. The final sum is p(a).
How is synthetic substitution connected to the Remainder Theorem?
By the Remainder Theorem, dividing p(x) by (x-a) gives remainder p(a). Synthetic substitution performs exactly this division. The bottom-row numbers are the coefficients of the quotient, and the last number is the remainder equal to p(a).