Factor x^3+3x-40?

Answer 1

If this expressiion has some rational roots, then they will be

+/- 1, +/- 2, +/- 4, +/- 5, +/- 8,+/- 10,+/- 20 and/or +/- 40

If none of them are, then there are no rational roots. Neither 1 nor -1 are roots, anyway. So, try synthetic division with the other ones. Here an example

.........1.......... 0.......... 3....... - 40
2.................... 2........... 4 ......... 14
......... 1.......... 2........... 7......... -26


.........1.......... 0.......... 3....... - 40
- 2.................- 2........... 4 ........- 14
......... 1.........- 2........... 7.........- 56

And a similar thing will happen for any negative value, let's take -k (where k is a positive real number) and observe that the remainder is less than -40 so it never will be 0

.........1.......... 0.......... 3.......... - 40
- k.................- k........... k^2 ........-3k - k^3
......... 1.........- k........... 3+k^2.....- 40 - 3k - k^3 <0 forall k >0

This only leaves these possible rational roots:
4, 5, 8,10, 20 and/or 40. So, just try them.

.........1.......... 0.......... 3....... - 40
4.................... 4...........16 .........76
......... 1.......... 4...........19......... 36

And, if you plug greater values, the remainder will be even greater, so, it will never be 0. Let's prove this:


.........1......... 0.......... 3.......... - 40
k................... k........... k^2 ........3k + k^3
......... 1.........k........... 3+k^2.....-40+37k + k^3

If k > 4, then k^3 > 64 and 37k > 148. So k^3 + 37 k - 40 > 0

So, the roots are irrational.

You can take the expression's derivative to see where the extrems are, but they are surelly between 2 and 4.

So, now let's study if there are some other roots.

The quotient of the expression, divided by x-k, and since we know that the remainder must 0 for at least 1 real value, is x^2 + kx+3+k^2. Remember that a 3rd grad polynomial *must* have at least one real root.

Let's now calculate the sign of the discriminant Delta

Delta = b^2 - 4ac = k^2 - 12 - 4k^2 = - 3k^2 - 12 < 0. So, there are no other real roots. The only root belongs belongs to the interval (2,4). After you have found it in an approximated way, plug it in Q(x) and you can find an approximated factoring of your equation.

P(x) = (x-k)(x^2 + kx+3+k^2)

Ilusion

Answer 2

x(x² + 3) -40 is about as far as it will go over the real numbers.
If it's part of a cubic equation (like x^3+3x-40 = 0) then you'll have to go to the Tartaglia-Cardano solution (look it up on the web) to get the roots.

Doug

Answer 3

x^3+8x-5x-40 8 + -5= 3
8 x -5= -40

x (x+8)- 5(x + 8)

(x + 8) (x -5)

yo

Answer 4

OK, Now what??

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