Respuesta :
Answer:
The answer is "non-real solution: "i-3 and -i-3" and real solution is -2".
Step-by-step explanation:
In the given equation there is some mistype error so, its correct equation can be defined as follows:
Given value:
[tex]\bold{ f(x) = x^3+ 8x^2 +22x +20}[/tex]
use hit and trial method we put x value are -2:
[tex]f(-2) = (-2)^3+ 8(-2)^2 +22(-2) +20\\\\[/tex]
[tex]= -8 + 8 \times 4 +22 \times -2 +20\\\\= -8 + 32 -44 +20\\\\= -52 + 52\\\\=0[/tex]
So, the real solution to the equation is "-2".
In the above solution, we calculate the f(x) =-2 is its real solution. so, the calculation for non-real solution:
divide the value [tex]x^3+8x^2+ 22x +20[/tex] by ([tex]x+2[/tex]):
Quotient: [tex]x^2+6x+10[/tex]
Remainder: 0
[tex]\to x^2+6x+10 \\\\compare \ with \ ax^2+bx+c: \\\\a= 1\\b= 6\\c=10\\\\\bold{Formula:}\\\\\to D= b^2-4ac\\\\[/tex]
[tex]= 6^2- 4 \times 1 \times 10\\\\ = 36- 40\\\\ = -4[/tex]
Formula for calculate non-real solution:
[tex]\bold{\frac{-b\pm \sqrt{D}}{2}}[/tex]
non-real solution: "i-3 and -i-3"
