Answer :
x^4+ 3x^3+4x^2-8 | x+2 = x^3+x^2+2x-4
-x^4-2x^3
-----------------------
x^3+4x^2-8
-x^3-2x^2
-------------------------
2x^2-8
-2x^2-4x
--------------
-4x-8
4x+8
----------------------------
/ /
is a factor because (x+2)(x^3+x^2+2x-4)=x^4+3x^3+4x^2-8
-x^4-2x^3
-----------------------
x^3+4x^2-8
-x^3-2x^2
-------------------------
2x^2-8
-2x^2-4x
--------------
-4x-8
4x+8
----------------------------
/ /
is a factor because (x+2)(x^3+x^2+2x-4)=x^4+3x^3+4x^2-8
Answer:
(x+2) is a factor of P(x)
Step-by-step explanation:
Determine if x + 2 is a factor of p(x) = x ^4 + 3x^ 3 + 4x ^2 - 8
[tex]p(x) = x ^4 + 3x^ 3 + 4x ^2 - 8[/tex]
To determine (x+2) is a factor , we set x+2=0 and solve for x
[tex]x+2=0 , x=-2[/tex]
then we replace x with -2 in p(x)
[tex]p(x) = x^4 + 3x^ 3 + 4x ^2 - 8[/tex]
[tex]p(-2) = (-2)^4 + 3(-2)^ 3 + 4(-2)^2 - 8[/tex]
[tex]p(-2) = 16-24+16- 8=0[/tex]
WE got p(-2)=0 , so (x+2) is a factor of P(x)