What is the solution of 2*x^3-19*x^2+25*x-8 =0

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The equation 2*x^3-19*x^2+25*x-8 =0 has to be solved.

2*x^3-19*x^2+25*x-8 =0

=> 2x^3 - 2x^2 - 17x^2 + 17x + 8x - 8 = 0

=> 2x^2(x - 1) - 17x(x - 1) + 8(x-1) = 0

=> (x - 1)(2x^2 - 17x + 8) = 0

=> (x - 1)(2x^2 - 16x - x + 8) = 0

=> (x - 1)(2x(x - 8) - 1(x - 8)) = 0

=> (x - 1)(2x - 1)(x - 8)= 0

=> x = 1, x = 1/2 and x = 8

**The solution of the equation 2*x^3-19*x^2+25*x-8 =0 is {1/2, 1, 8}**

Let's solve for `x ` of the given equation

Multiply `2` by `x^3` to get `2x^3`

`2x^3 - 19 * x^2 + 25 * x - 8 = 0`

Multiply `-19` by `x^2` to get `-19x^2`

`2x^3 - 19x^2 + 25 * x - 8 = 0`

Multiply `25` by `x` to get `25x` .

`2x^3 - 19x^2 + 25x - 8 = 0`

Find every combination of `+-p/q` These are the possible roots of the polynomial function.

`+- 1, +- 1/2, +- 2, +- 4, +- 8`

Since `1` is a known root, divide the polynomial by `(x-1)` to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

`(2x^3 - 19x^2 + 25x - 8)/(x-1)`

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by `1` .

`2x^2 - 17x + 8`

The polynomial can be written as a set of linear factors.

`(x-1)(x - 1/2)(x - 8)`

These are the roots (i.e. zeros) of the polynomial `2x^3 - 19x^2 + 25x - 8`

Therefore, the solution of the equation is `x = 1, 1/2, 8`

**The graph shows that the solution has a corresponding `0` y-value**

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