Solve this equation 2x^5+3x^4-30x^3-57x^2-2x+24 = 0 for all x

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation 2x^5+3x^4-30x^3-57x^2-2x+24 = 0 has to be solved for x.

2x^5+3x^4-30x^3-57x^2-2x+24 = 0

=> 2x^5 + 2x^4 + x^4 + x^3 - 31x^3 - 31x^2 - 26x^2 - 26x + 24x + 24 = 0

=> 2x^4(x + 1) + x^3(x + 1) - 31x^2(x + 1) - 26x(x + 1) + 24(x + 1) = 0

=> (x + 1)(2x^4 + x^3 - 31x^2 - 26x + 24) = 0

=> (x + 1)(2x^4 - 8x^3 + 9x^3 - 36x^2 + 5x^2 - 20x - 6x + 24) = 0

=> (x + 1)(2x^3(x - 4) + 9x^2(x - 4) + 5x(x - 4) - 6(x - 4)) = 0

=> (x + 1)(x - 4)(2x^3 + 9x^2 + 5x - 6) = 0

=> (x + 1)(x - 4)(2x^3 + 3x^2 + 6x^2 + 9x - 4x - 6) = 0

=> (x + 1)(x - 4)(x^2(2x + 3) + 3x(2x + 3) - 2x(2x + 3)) = 0

=> (x + 1)(x - 4)(2x + 3)(x^2 + 3x - 2x) = 0

=> x = -1, x = 4, x = -3/2

The roots of x^2 + 3x - 2x = 0 are

x4 = `(-3 + sqrt(9 + 8))/2 = -3/2 + sqrt17/2`

x5 = `-3/2 - sqrt 17/2`

The roots of the equation 2x^5+3x^4-30x^3-57x^2-2x+24 = 0 are `{-3/2, -1, -3/2 - sqrt 17/2, -3/2 + sqrt17/2, 4}`

We’ve answered 318,989 questions. We can answer yours, too.

Ask a question