How do you determine p(c) given #p(x)=x^3+2x^2+3x+4# and c=-1?

Question

2072 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-08T15:31:18

#2#

We have: #p(x) = x^(3) + 2 x^(2) + 3 x + 4#; #c = - 1#

First, let's substitute #x# with #c#:

#=> p(c) = (c)^(3) + 2 (c)^(2) + 3 (c) + 4#

Then, the value of #c# is given as #- 1#.

Let's substitute #c# with #- 1#:

#=> p(- 1) = (- 1)^(3) + 2 (- 1)^(2) + 3 (- 1) + 4#

#=> p(- 1) = - 1 + 2 - 3 + 4#

#=> p(- 1) = 1 + 1#

#=> p(- 1) = 2#