How do you find all the zeros of #F (x) = 5x^3 (x+ 3)^ 4 (x-7)# with all its multiplicities?

1 Answer
Aug 20, 2017

Answer:

The zeros are:

#x=0" "# with multiplicity #3#

#x=-3" "# with multiplicity #4#

#x=7" "# with multiplicity #1#

Explanation:

  • Each linear factor corresponds to a zero. That is: #a# is a zero if and only if #(x-a)# is a factor.

  • The multiplicity of each factor is the multiplicity of the corresponding zero.

We can rewrite the given #F(x)# slightly to bring it into a form with factors with the form #(x-a)# as follows:

#5x^3(x+3)^4(x-7) = 5(x-0)^3(x-(-3))^4(x-7)#

So the zeros are:

#x=0" "# with multiplicity #3#

#x=-3" "# with multiplicity #4#

#x=7" "# with multiplicity #1#