Question

Solve the function.

Answer 1

Answer: x=-4,  x=1±√2i

Step-by-step explanation:

$x^3+2x^2-5x+12=0\\x^3+2x^2+2x^2-2x^2-5x+12=0\\(x^3+4x^2)-(2x^2+5x-12)=0\\(x^2*(x+4))-(2x^2+5x+3x-3x-12)=0\\x^2*(x+4)-((2x^2+8x)-(3x+12))=0\\x^2*(x+4)-(2x*(x+4)-3*(x+4))=0\\x^2*(x+4)-(x+4)*(2x-3)=0\\(x+4)*(x^2-(2x-3))=0\\(x+4)*(x^2-2x+3)=0\\x+4=0\\x=-4$

$\displaystyle\\x^2-2x+3=0\\a=1\ \ \ \ b=-2\ \ \ \ c=3\\D=b^2-4ac\\D=(-2)^2-4*1*3\\D=4-12\\D=-8\\\sqrt{D}=\sqrt{-8} \\\sqrt{D}=\sqrt{-1*4*2} \\\sqrt{D}=\sqrt{-1}*\sqrt{4}*\sqrt{2} \\ \sqrt{D}=2\sqrt{2}i$

$\displaystyle\\x=\frac{-bб\sqrt{D} }{2a} \\\\x=\frac{-(-2)б2\sqrt{2}i }{2*1} \\\\x=\frac{2б2\sqrt{2}i }{2} \\\\x=\frac{2*(1б\sqrt{2}i) }{2}\\ x=1б\sqrt{2}i$