Examine whether x + 2 is a factor of x³ + 3x² + 5x + 6 and of 2x + 4.

Step-by-step explanation:

x+2=0 x=-2

x^3+3x^2+5x+6

(-2)^3+3(-2)2+5(-2)+6

-8+12-10+6

0

hence x+2 is a factor of x^3+3x^2+5x+6

x+2=0

x=-2

2x+4

2(-2)+4

-4+4

0

also x+2 is a factor of 2x+4

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

two? xd

step-by-step explanation: