Предмет: Алгебра, автор: bekautembaev74

помогите решить ету проблему
1) x^4-8x^3-2x^2+16x-3=0

(z^2+3z+6)^2+2=(z2+3z+6)-3z^2=0

Ответы

Автор ответа: waflee
0

Ответ:
Let’s solve the equation step by step:

x^4 - 8x^3 - 2x^2 + 16x - 3 = 0

We can try to factor the polynomial. One possible way is to use synthetic division to test for roots. We can start by testing for x = 2:

2 | 1 -8 -2 16 -3

 |    2 -12 -28 -24

 -----------------

  1 -6 -14 -12 -27

The remainder is not zero, so x = 2 is not a root.

We can then try to test for x = -1:

-1 | 1 -8 -2 16 -3

  |    -1   9 -7  -9

  -----------------

   1 -9   7   9 -12

We get a remainder of -12, so x = -1 is a root.

We can then use polynomial long division to divide (x + 1) into the polynomial:

      x^3 - 9x^2 + 16x -19

(x+1) | x^4 -8x^3 -2x^2 +16x-3

      x^4 + x^3

      ----------

        -9x^3-2x^2+16x

        -9x^3-9x^2

        ----------

                7x^2+16x

                7x^2+7x

                --------

                     9x-3

We get a quotient of q(x) = x^3 -9x^2 +16x-19 and a remainder of -3.

We can then solve for the remaining roots by factoring q(x) using the rational root theorem or other methods. However, since this is a cubic polynomial, we may need to use numerical methods or approximations to find the roots.

Therefore, the solutions to the equation are x = 2, -1, and two other complex roots that we need to approximate.
Объяснение:
Let’s solve the equation step by step:

(z^2+3z+6)^2+2=(z^2+3z+6)-3z^2=0

We can simplify the left-hand side of the equation by expanding the square of the trinomial:

z^4 + 6z^3 + 17z^2 + 36z + 40 = z^2 + 3z + 6 - 3z^2

We can then move all the terms to one side of the equation:

z^4 + 6z^3 + 20z^2 - 33z - 34 = 0

We can then try to factor the polynomial. One possible way is to use synthetic division to test for roots. We can start by testing for z = -1:

-1 | 1   6   20   -33   -34

  |    -1  -5    -15   -5

  -----------------------

   1   5   15   -48   -39

The remainder is not zero, so z = -1 is not a root.

We can then try to test for z = -2:

-2 | 1   6   20   -33   -34

  |    -2  -8    16    34

  -----------------------

   1   4   12   -17     0

We get a remainder of zero, so z = -2 is a root.

We can then use polynomial long division to divide (z + 2) into the polynomial:

      z^3 + 4z^2 + 4z -17

(z+2) | z^4 + 6z^3 +20z^2-33z-34

      z^4 +2z^3

      --------

        4z^3+20z^2

        4z^3+8z^2

        ----------

               12z^2-33z

               12z^2-24z

               --------

                     -9z-34

                     -9z-18

                     -----

                      -16

We get a quotient of q(z) = z^3 + 4z^2 +4z-17 and a remainder of -16.

We can then solve for the remaining roots by factoring q(z) using the rational root theorem or other methods. However, since this is a cubic polynomial, we may need to use numerical methods or approximations to find the roots.

Therefore, the solutions to the equation are z = -1, -2, and two other complex roots that we need to approximate.      

i'm so sorry for the English ♥

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