𝑓(𝑥) = 5𝑥4 + 32𝑥3 − 24𝑥2 − 20𝑥 + 7
Try 𝑥 = 1
1 | 5 32 -24 -20 7
| 5 37 13 -7
5 37 13 -7 0
Hurray, the synthetic division has a 0 on the end of the bottom
row, and that tells us that 𝑥 = 1 is a zero of 𝑓(𝑥) and the other
numbers across the bottom of the synthetic division tells us that
we have factored 𝑓(𝑥) as
𝑓(𝑥) = (𝑥 - 1)(5𝑥3 + 37𝑥2 - 13𝑥 - 7)
We think that what's in the second parentheses might be factorable,
too, so we try 𝑥 = -7 on it:
-7 | 5 37 13 -7
| -35 -14 7
5 2 -1 0
Hurray, as before, the synthetic division has a 0 on the end of the
bottom row, and that tells us that 𝑥 = -7 is a zero of both
5𝑥3 + 37𝑥2 - 13𝑥 - 7 as well as 𝑓(𝑥) and the other
numbers across the bottom of the synthetic division tells us that
we have factored 𝑓(𝑥) further as:
𝑓(𝑥) = (𝑥 - 1)(𝑥 + 7)(5𝑥2 + 2𝑥 - 1)
What's in the last parentheses cannot be factored, so we find the
zeros by setting 𝑓(𝑥) = 0 and use the zero-factor principle:
(𝑥 - 1)(𝑥 + 7)(5𝑥2 + 2𝑥 - 1)
𝑥 - 1 = 0; 𝑥 + 7 = 0, 5𝑥2 + 2𝑥 - 1 = 0
𝑥 = 1; 𝑥 = -7;
To find the other two zeros, we must use the quadratic formula:
5𝑥2 + 2𝑥 - 1 = 0
Edwin
