Find all possible roots: 1. X^4-2x^2-16x-15=0 2. 2x^3+7x^2+6x-5=0 3. 2x^3-11x^2+16x-6=0 4. 3x^4+4x^3-x^2+4x-4=0

possible-roots

#1

Find all possible roots:

  1. X^4-2x^2-16x-15=0
  2. 2x^3+7x^2+6x-5=0
  3. 2x^3-11x^2+16x-6=0
  4. 3x^4+4x^3-x^2+4x-4=0

Answer:

1) We must transform the equation in simple factors:
x^4-2x^2-16x-15=x^4+2x^3-2x^3+5x^2-4x^2-3x^2-10x-6x-15 = x^4+2x^3+5x^2-2x^3-4x^2-10x-3x^2-6x-15 = x^2(x^2+2x+5)-2x(x^2+2x+5)-3(x^2+2x+5) = (x^2+2x+5)(x^2-2x-3) = (x^2+2x+5)(x^2-3x+x-3) = (x^2+2x+5)[x(x-3)+(x-3)] = (x-3)(x+1)(x^2+2x+5)
Now this result is equal with 0 => (x-3)(x+1)(x^2+2x+5)=0
From this equation results that at least one factor is equal with 0 =>
a) x-3=0 => x=3
b) x+1=0 => x=-1
c) x^2+2x+5=0
Now we calculate the discriminant of the quadratic polynomial:
\Delta=b^2-4ac => \Delta=4-20=-16 which means that the roots aren’t real, they are complex and we use the formulas:
x1=(-b-i*\sqrt{-\Delta})/2a => x1=(-2-4i)/2=-1-2i
x2=(-b+i*\sqrt{-\Delta})/2a => x1=(-2+4i)/2=-1+2i
The final results are:
-real roots: -1 and 3
-complex roots: -1-2i and -1+2i
2) In order to resolve this equation we do the same steps
2x^3+7x^2+6x-5 = 2x^3+8x^2-x^2+10x-4x-5 = 2x^3+8x^2+10x-x^2-4x-5 = 2x(x^2+4x+5)-(x^2+4x+5) = (x^2+4x+5)(2x-1)
(x^2+4x+5)(2x-1)=0 =>
a) 2x-1=0 => x=1/2
b) x^2+4x+5=0
\Delta=16-20=-4 =>
x1=(-4-2i)/2=-2-i
x2=(-4+2i)/2=-2+i
The final results are:
-real root: 1/2
-complex roots: -2-i and -2+i
3) Again I’ll do the same steps
2x^3-11x^2+16x-6 = 2x^3-8x^2-3x^2+4x+16x-6 = 2x^3-8x^2+4x-3x^2+12x-6 = 2x(x^2-4x+2)-3(x^2-4x+2) = (2x-3)(x^2-4x+2)
(2x-3)(x^2-4x+2)=0 =>
a) 2x-3=0 => x=3/2
b) x^2-4x+2=0
\Delta=16-8=8 this time the discriminant is positive and the formulas for roots are:
x1=(-b-\sqrt{-\Delta})/2a => x1=(4-2*\sqrt{2})/2=2-\sqrt{2}
x1=(-b+\sqrt{-\Delta})/2a => x1=(4+2*\sqrt{2})/2=2+\sqrt{2}
The final results are:
-real roots: 3/2, 2-\sqrt{2} and 2+\sqrt{2}
**4)**Again I’ll do the same steps
3x^4+4x^3-x^2+4x-4 = 3x^4+4x^3+3x^2-4x^2+4x-4 = 3x^3+3x^2+4x^3+4x-4x^2-4 = 3x^2(x^2+1)+4x(x^2+1)-4(x^2+1) = (x^2+1)(3x^2+4x-4) = (x^2+1)(3x^2+6x-2x-4) = (x^2+1)[3x(x+2)-2(x+2)] = (x+2)(3x-2)(x^2+1)
(x+2)(3x-2)(x^2+1)=0 =>
a) x+2=0 => x=-2
b) 3x-2=0 => x=2/3
c) x^2+1=0
\Delta=0-4=-4 =>
x1=(-2i)/2=-i
x2=(2i)/2=i
The final results are:
-real roots: -2 and 2/3
-complex roots: -i and i