find all the real zeros of the polynomial

Question

P(x)=x³-2x²-9x+4

its real zeros are

x1=

x2=

x3=

with x1≤x2≤x3

If the polynomial has only two real zeros, input them at x1 and x2 ; if the polynomial has only one real zero, input it at x1

Answer 1

P(x) = x^3 - 2x^2 - 9x + 4

 

Possible numerators of rational zeros are factors of 4 and

possible denominators of rational zeros are factors of 1.

 

Possible rational zeros are: ±1, ±2, ±4.

 

±1 don't work by mental arithmetic.

 

2 | 1 -2 -9 4

. . . . 2 . 0 -18

. . .1 .0.-9 . not zero

 

-2 | 1 -2 -9 4
. . . . .-2 .8 2
. . . 1 -4.-1 . not zero

 

4 | 1 -2 -9 4
. . . . 4 . 8 -4
. . .1 .2.-1 . 0

So P(x) = (x-4)(x^2+2x-1)

 

If x^2+2x-1 = 0, then

 

x = (-2 ± √(2^2-4(1)(-1)))/(2(1))

x = (-2 ± √(2(4))/2

x = -1 ± √(2)

 

So the real zero of P(x) is:

 

x1 = 4

 

and the other two are complex conjugates.

Answer 2

Graph x³-2x²-9x+4 on your graphing calculator. The zeros are where the graph crosses the x-axis.