Approximate the real zeros of f(x)=2x^3+3x^2-1 to the nearest tenth.?

Question

what are the steps for this.

Answer 1

set f(x) = 0 and then solve the equation 2x^3 + 3x^2 - 1 = 0

The answers are -1 and 0.5

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

Answer 2

Let's look for rational zeros first.
If a/b is a rational zero then b must divide 2
and a must divide -1.
The possible rational zeros are 1, -1, ½ and -½.
Since f(-1) = 2(-1) + 3 -1 = 0, -1 is a root, so
x+1 is a factor of f(x). By synthetic division or
long division, the quotient is
2x² + x -1 = (2x-1)(x+1).
So your equation has roots x = -1, -1 and x = 1/2 = 0.5.
No need to approximate here!
BTW: Since x = -1 is a double root of f(x), the
graph of f(x) will be tangent to the x-axis at x = -1.

Answer 3

The first thing I would do would be to plot this function to get a sense of where its zeroes might occur. Once you do that, you see there is a zero in the neighborhood of minus one, which leads you to notice that if you set x=-1, f(-1)=-2+3-1=0.

Knowing that x=-1 is a root of the equation, divide your original function by (x+1) to obtain the quadratic:

2x^2+x-1

This factors to: (2x-1)(x+1), so the zeroes of f are at: x=1/2 and a double root at x = -1

Answer 4

I guess you mean 'real roots', and not 'real zeros'

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

By rearranging get:
f(x) = 2x^3 + 4x^2 + 2x - x^2 - 2x - 1
= 2x^3 + 4x^2 + 2x - (x^2 + 2x + 1)
= 2x(x^2 + 2x + 1) - (x^2 + 2x + 1)

Look very well, see that (x^2 + 2x + 1) is a factor

f(x) = (2x - 1)(x^2 + 2x + 1)

And again (x^2 + 2x + 1) can be factored as (x + 1)^2, or
as (x + 1)(x + 1)
Therefore:
----------------------------------
f(x) = (2x - 1)(x + 1)(x + 1)
-----------------------------------

If f(x) = 0, then (2x - 1)(x + 1)(x + 1) = 0, from which we can get the three values of x as:

x = 1/2, x = -1 (twice)

Answer 5

f(x)=2x^3+3x^2-1=0
=> f(-1)=2(-1)^3+3(-1)^2-1=0
(x+1) is a factor


(2x^3+3x^2-1)/ (x+1)
=> 2x^2 +x - 1=0
=> (2x -1)(x+1)=0
=> x = 1/2 OR x= -1

QED

Answer 6

what language r u using ???? ^=???????