Question

can you help me solve solve -x^3-3x^2+4=0 with steps

Answer 1

Have you considered the Rational Root Theorem?

Answer 2

well x = 1 looks like a solution to me

you can use the factor theorem to check...

and then a little division... so you can get the quadratic factor...

then see if it can be factored

Answer 3

how do i divide with an x^1 missing?

Answer 4

well if x = 1 is a solution then x - 1 is a factor...

so if you don't know polynomial division or synthetic division...

then use the rational root theorem

Answer 5

i do know how but there is a remainder of 8. what do i do with that?

Answer 6

Put in a zero (0) as a place-holder.

Answer 7

factor out -1

then you have

\[-1(x^3 + 3x^2 - 4) = 0\]

it may be easier to see that x = 1 is a solution..

so

(x -1) is a factor

Answer 8

If x - 1 is a factor of -x^3 - 3x^2 + 4 then

-x^3 + x^2 - 4x^2 + 4x - 4x+ 4 = 0
-x^2(x - 1) - 4x(x - 1) - 4(x - 1) = 0
(x - 1)(-x^2 - 4x - 4) = 0
-(x - 1)(x^2 + 4x + 4) = 0
-(x - 1)((x + 2)^2 = 0

Answer 9

@chyu

Answer 10

So now applying zero product property:

x - 1 = 0
x + 2 = 0

And the values of x should be obvious at this point.