Find three roots of the equation x^3-8x^2+11x+2=0 , giving the two non-integer roots in the exact form p+/- sqrt(q)
Any help will be appreciated
Note that (x-a)(x-b)(x-c) = x^3 -...x^2 +...x...-abc
So the product of the root a,b,c is -2. This means the integer root can only be -1,1,-2,2.
Substitute x=-1 to check and this is not a root.
Then x=1, this is not a root either.
x=2 works, so we know one root is x=2 with factor is (x-2).
Divide the cubic expression by (x-2), you should get x^2 - 6x -1 and solving the equation
x^2 - 6x - 1=0 gives the other 2 roots. 3+/-sqrt(10).
+2
nice
factors are +/-1 and +/-2
plug in +1
1-8+11+2= not zero
plug in -1
[only the odd exponents change signs]
-1-8-11+2=not zero
plug in +2
8-32+22+2=0
x=2, which means the factor is (x-2)
synthetic division
start with the 1 of x^3
multiply by 2
add to the next coefficient
repeat
gives you
1
-6
-1
0(remainder)
which translates into
1x^2-6x-1
quadratic formula
b^2-4ac
36+4
40
+/-sqrt40
+/-2sqrt10
rest of q.f.
(-b+/-2sqrt10)/2a
(6+/-2sqrt10)/2
(3+/-sqrt10)
your three roots:
x=2, (3+/-sqrt10)
Comments
Note that (x-a)(x-b)(x-c) = x^3 -...x^2 +...x...-abc
So the product of the root a,b,c is -2. This means the integer root can only be -1,1,-2,2.
Substitute x=-1 to check and this is not a root.
Then x=1, this is not a root either.
x=2 works, so we know one root is x=2 with factor is (x-2).
Divide the cubic expression by (x-2), you should get x^2 - 6x -1 and solving the equation
x^2 - 6x - 1=0 gives the other 2 roots. 3+/-sqrt(10).
+2
nice
factors are +/-1 and +/-2
plug in +1
1-8+11+2= not zero
plug in -1
[only the odd exponents change signs]
-1-8-11+2=not zero
plug in +2
8-32+22+2=0
x=2, which means the factor is (x-2)
synthetic division
start with the 1 of x^3
multiply by 2
add to the next coefficient
repeat
gives you
1
-6
-1
0(remainder)
which translates into
1x^2-6x-1
quadratic formula
b^2-4ac
36+4
40
+/-sqrt40
+/-2sqrt10
rest of q.f.
(-b+/-2sqrt10)/2a
(6+/-2sqrt10)/2
(3+/-sqrt10)
your three roots:
x=2, (3+/-sqrt10)