Synthetic Substitution
The first way you were taught to evaluate a po lynomial for a given value of a variable was direct
substitu tion. Simply put , that means you p lugged that value into the expression and found the result.
EXAMPLE: If P(x) =
2x2
+ 3x – 10, find P when x = 5 . By substituting, P = 2(5)² + 3(5) – 10
P(5) = 55
The process of direct substitu tion can become a real pain for higher degree polynomial expressions . T here
is another method called SYNTHE TIC SUB STITUTION that w ill make evaluating a polynomial a very
simple process.
Given some polynom ial Q = 3x² + 10x² – 5x – 4 in one variable. You can evaluate Q when x = 2 by
plugging in that value as we did before.
Q(x) = 3x³ + 10x² – 5x – 4
= 3(2)³ + 10(2)² – 5(2) – 4
= 3(8) +10(4) – 10 – 4
= 24 + 40 + – 10 – 4
= 50
So the value of Q is 50 when x is 2 .
Or by us ing S YNTHE TIC SUBS TITUTION , we would write the coefficients of Q;
Q = 3 x³ + 10x² – 5x – 4
3 1 0 –5 –4
Now, we’ll leave a space under those coefficients and draw a line. We will also write down the value of the
variable to be pl ugged in .
2
3 10 –5 –4
Once we do that, we are set up to evaluate Q when x = 2.
To accomplish that, we bring down t he first number, 3, and mu ltiply by 2, then add. Keep repeating this
process. The last value will be the value of Q when x is 2 .
2
3 10 –5 –4
6 32 54
3 16 27 50
Notice, we did get 50 as we did before.
EXAMPLE: Evaluate 2x4 – x³ + 5x + 3 when x = 3
In this example, notice there is no quadratic term, no x². W hen we write the coefficients, we’ll need to
write zero for the coefficient of that missi ng term.
3
2 –1 0 5 3
6 15 45 150
2 5 15 50 153
The value of that polynomial expression when x = 3 is 153.
