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Answer:
yes
Step-by-step explanation:
There are several ways to go at this.
My first choice is to use a graphing calculator. It shows the function has a zero at x=5, so x-5 is a factor.
Another good choice is to use synthetic division (2nd attachment). If the remainder is zero, then x-5 is a factor. It is and it is.
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You can also evaluate the function at x=5. The remainder theorem tells you that if the value is zero, then x-5 is a factor. Evaluating the polynomial written in Horner form is a lot like synthetic division.
(((x -4)x -15)x +58)x -40 for x=5 is ... (-10·5 +58)5 -40 = 40-40 = 0
The value of h(5) is zero, so x-5 is a factor of h(x).
