What are the zeros of this polynomial?

Question

P(x) = x^3 + 2x^2 + 3x - 6.

Thank you!!

Answer 1

x^3 + 2x^2 + 3x -6 =0
x^3 - x^2 + 3x^2 - 3x + 6x - 6=0
x^2(x-1)+3x(x-1)+6(x-1)=0
(x-1)(x^2+3x+6)=0
Therefore, zeros are at points below :
x=1
x= [-3+sqrt(15).i]/2
x= [-3-sqrt(15).i]/2

So, it has 1 real-root & 2 imaginary-roots(those include 'i').

Answer 2

Try graphing it

x=1 will cause P(1)=1+2+3-6=0

factoring also could be helpful to factor to

(x-1)*(x^2+3x+6)

set both sides to equal zero and solve
x-1=0
x=1
and
(x^2+3x+6)=0
x^2+3x=-6
which can never happen
the lowest value of (x^2+3x+6) is actually 3.75 (not that this is particularly important once you have proved it can't be zero)

so 1 is the only zero

the easiest way to do these programs is to graph the equation and see where it crosses the x axis (aka y=0)

Answer 3

x^3 + 2x^2 + 3x -6 = x^3 - x^2 + 3x^2 - 3x + 6x -6 =
(x-1)(x^2 + 3x +6)
Only one real root x=1.

Answer 4

for the polynomial

x³ + 2x² + 3x - 6 = 0
x =
-1.5 - 1.9364916731037085 i ....or
-1.5 - 1.9364916731037085 i .....or
1
where i = √-1

Answer 5

1 is an evident root, since the sum of the coefficients are 0

Use the synthetic division and solve the quadratic equation that you get

Ana

Answer 6

Underneath the mushroom.