Algebra: Polynomials?

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

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)

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