Question

if q(x)=x^3+3x^2+3x+1,find q(-2)+q(-3)+q(1/2)

Answer 1

hope it helps u...........

Answer 2

The answer is $q(-2)+q(-3)+q(\frac{1}{2}) =\frac{-45}{8}$

Step-by-step explanation:

Given: $q(x)=x^3+3x^2+3x+1$

To find: q(-2)+q(-3)+q($\frac{1}{2}$)

Solution:

First finding the values of q(-2), q(-3), and q($\frac{1}{2}$)

• $q(-2) =(-2)^3+3(-2)^2+3(-2)+1$

$q(-2) =(-8)+3(4)-6+1$

$q(-2) =-8+12-5$

$q(-2) =-13+12$

$q(-2) =-1$

• $q(-3) =(-3)^3+3(-3)^2+3(-3)+1$

$q(-3) =(-27)+3(9)-9+1$

$q(-3) =-27+27-8$

$q(-3) =-8$

• $q(\frac{1}{2}) =(\frac{1}{2})^3+3(\frac{1}{2})^2+3(\frac{1}{2})+1$

$q(\frac{1}{2}) =(\frac{1}{8})+3(\frac{1}{4})+(\frac{3}{2})+1$

$q(\frac{1}{2}) =\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$

$q(\frac{1}{2}) =\frac{1+6+12+8}{8}$

$q(\frac{1}{2}) =\frac{27}{8}$

• Now adding the values

$q(-2)+q(-3)+q(\frac{1}{2}) =-1 -8 +\frac{27}{8}$

$=-9 +\frac{27}{8}$

$=\frac{-72 +27}{8}$

$=\frac{-45}{8}$

• Thus $q(-2)+q(-3)+q(\frac{1}{2}) =\frac{-45}{8}$