x(x-4)(3x+2)= 240 can you find the value of X
You need to open the brackets such that:
`3x^3 + 2x^2 - 12x^2 - 8x -240 = 0`
Collecting like terms yields:
`3x^3- 10x^2 - 8x - 240 = 0`
You need to use the rational root test to find the roots, hence you should form the set of possible rational roots. The numerator of these possible rational roots needs to be one factor of constant coefficient and the denominator needs to be one factor of leading coefficient.
The sketch of the graph of function may help you to eliminate some rational roots from the set.
Notice that the graph of function intercepts x axis at x=6, hence you need to substitute 6 for x to check if this value cancel the polynomial such that:
`3*6^3 -360 -48 - 240 = 0=gt 240 - 240 = 0`
This proves that x = 6 is a root for `f(x) = 3x^3 - 10x^2 - 8x - 240.`
Since you know one root, you may find the next two roots using the factored form of polynomial such that:
`3x^3 - 10x^2 - 8x - 240 = (x-6)(ax^2 + bx + c)`
You need to open the brackets to the right side:
`3x^3 - 10x^2 - 8x - 240 = ax^3 + bx^2 + cx - 6ax^2 - 6bx - 6c`
You need to group like powers to the right side such that:
`3x^3 - 10x^2 - 8x - 240 = ax^3 + x^2(b- 6a) + x(c - 6b) - 6c`
Equating coefficients of like powers yields:
`b - 6a = -10 =gt b - 18 = -10 =gt b = 18 - 10 =gt b = 8`
`-6c = -240 =gt c = 40`
Hence, you may find the next two roots if you solve the quadratic `3x^2 + 8x + 40 = 0` .
`x_(1,2) = (-8+-sqrt(64 - 480))/6 =gt x_(1,2) = (-8+-sqrt(-416))/6`
`x_(1,2) = (-8+-4isqrt26)/6 =gt x_(1,2) = (-4+-2isqrt26)/3`
Hence, evaluating the solutions to equation yields `x_1 = 6, x_(2,3) = (-4+-2isqrt26)/3.`