Will choose brainliest (:
is the given linear expression a factor of the polynomial? show your work.
h(x)=x^4-4x^3-15x^2+58x-40; x-5
! (:

  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).

to rewrite a literal equation you may need to divide by a number, or by a variable.

hope i .

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.