Question

evaluate 2x⁴+5x³+7x²-x+41 when x=-2-√3i

Answer 1

hope this helps you☺️

Answer 2

Answer:

$2x^4+5x^3+7x^2-x+41=6$ when $x=-2-\sqrt3 i$.

Step-by-step explanation:

Given : $x=-2-\sqrt3 i$

To find : Evaluate $2x^4+5x^3+7x^2-x+41$

Solution :

$x=-2-\sqrt3 i=-(2+\sqrt3 i)$ ....(1)

Squaring both side,

$x^2=(-(2+\sqrt3 i))^2$

$x^2=4+3i^2+4\sqrt3 i$

$x^2=4-3+4\sqrt3 i$

$x^2=1+4\sqrt3 i$ .....(2)

Multiply (1) and (2),

$x\times x^2=(-(2+\sqrt3 i))(1+4\sqrt3 i)$

$x^3=-(2+8\sqrt3 i+\sqrt3 i+12i^2)$

$x^3=-(2+9\sqrt3 i-12)$

$x^3=-(-10+9\sqrt3 i)$

$x^3=10-9\sqrt3 i$ .....(3)

Squaring (2) both side,

$(x^2)^2=(1+4\sqrt3 i)^2$

$x^4=1+48i^2+8\sqrt3 i$

$x^4=1-48+8\sqrt3 i$

$x^4=-47+8\sqrt3 i$ .....(4)

Substitute (1),(2),(3) and (4) in the expression,

$=2(-47+8\sqrt3 i)+5(10-9\sqrt3 i)+7(1+4\sqrt3 i)-(-2-\sqrt3 i)+41$

$=-94+16\sqrt3 i+50-45\sqrt3 i+7+28\sqrt3 i+2+\sqrt3 i+41$

$=6+0\sqrt3 i$

$=6$

Therefore, $2x^4+5x^3+7x^2-x+41=6$ when $x=-2-\sqrt3 i$.