Question

Find the exact zeros.
G(x)=x^4-3x^3-5x^2+29x-30

Answer 1

\[x^4-3 x^3-5 x^2+29 x-30=(x-2) (x+3) \left(x^2-4 x+5\right) \]x = 2 and -3
The other two zeros are the solutions to the following equation:\[x^2-4 x+5=0 \]

Answer 2

I believe it, but how did you get the x-2 and x+3 factors?

Answer 3

(x - 2) is zero when x = 2
(x + 3) is zero when x = -3

With a little more experience, you will be able to solve these types of expressions by inspection.

Answer 4

You could tell that (x-2) and (x+3) would factor out of that polynomial by inspection?! How would a non-robot do it?

Answer 5

When I have used the Matematica "Factor" function for this problem expression and the others in the past, I have often wondered myself how they do it.

Answer 6

I thought they'd found solutions for up to quintic polynomials, so they could be using that. They could also be doing some numerical analysis (e.g. Newton-Raphson) to find a couple real zeros to get it down to the more easily-factorable cubic/quadradic polynomials. Or magic.

Answer 7

use the possible rational zeroes method
in this case, the factors of 30 is 1x30, 2x15, 3x10, 5x6
and so, the possible rational zeroes are 1, 2, 3, 5, 6, 10, 15, and 30.
try to sub everyone of this into x, and if y is 0
then you got the factor is 2, 3 !!! :)

Answer 8

@ koo1992
Thanks for the insight.