Answer :

dannielle
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

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)

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