please help me solve, 3x^3+5x^2-3x-5=0?

the ratio of coefficient of x^3 and x^2 is same as that off x^1 and constant so we group it that way

x^2(3x+5) - 1(3x+5)
= 3x+5)(x^2-1)
= (3x + 5)(x+1)(x-1)

so roots are -5/3, -1 and 1

3x^3+5x^2-3x-5=0
x^2(3x + 5) -(3x + 5) = 0

(x^2 - 1)(3x + 5) = 0
(x + 1)(x - 1)(3x+ 5) = 0

the values are -1, 1, and -5/3

3x³ + 5x² - 3x - 5 = 0
by inspection, you can see x=1 is a solution (3 + 5 - 3 -5).
(x - 1) (3x² + 8x + 5) = 0 .... factor out (x - 1) by synthetic division
(x - 1) (3x + 5) (x + 1) = 0 .. then factor 3x² + 8x + 5

Answer: x = -5/3, 1, -1

3x^3+5x^2-3x-5=0

3 5 -3 -5
-1
-3 -2 5
3 2 -5 0

(x+1)(3x^2+2x-5)= 0
(x+1) (3x^2-3x+5x-5)=0
(x+1)[(3x(x-1)+5(x-1)]=0
(x+1)(x-1)(3x+5)=0
is the

10 points answer huray.......

f (x) = 3x³ + 5x² - 3x - 5
f (1) = 3 + 5 - 3 - 5 = 0
Thus x - 1 is a factor of f (x)
Use synthetic division to find other factor.

1 | 3_____5_____-3_____-5
_ |______ 3_____ 8_____ 5
_ | 3_____8_____ 5_____ 0

(x - 1)(3x² + 8x + 5) = 0

(x - 1)(3x + 5)(x + 1) = 0

x = 1 , x = - 5/3 , x = - 1

this is simple. factor out (3x +5) in each part. it's obvious.

3x^3+5x^2-3x-5=0
x^2(3x + 5) -(3x + 5) = 0

(x^2 - 1)(3x + 5) = 0
(x + 1)(x - 1)(3x+ 5) = 0

the values are -1, 1, and -5/3

3x^3 + 5x^2 - 3x - 5 = 0
(3x^3 + 5x^2) - (3x + 5) = 0
x^2(3x + 5) - 1(3x + 5) = 0
(3x + 5)(x^2 - 1) = 0
(3x + 5)(x + 1)(x - 1) = 0

3x + 5 = 0
3x = -5
x = -5/3

x + 1 = 0
x = -1

x - 1 = 0
x = 1

∴ x = -5/3 , -1 , 1

3x^3+5x^2-3x-5=0
There may be other ways, but the simplest way would be to change the equation to:
3x^3+5x^2=3x+5
Now factor out the x^2 on the left and you get:
x^2(3x+5)=3x+5
then divide out the 3x+5 and you get x^2=1 so you have to check to see if either -1 or +1 is correct. Usually a polynomial will have as many solutions as the highest power. I think, but you can check your textbook, or work out a bunch that you think up to see if this is true.

use polynomial division like find the factor that can divide the equation eg X-1 and do long division on the equation using this factor so that the equation becomes a simple quadratic then factorise it you will have 3 answers of X fairly simple stuff

this is simple. factor out (3x +5) in each part. it's obvious.

3x^3+5x^2-3x-5=0
x^2(3x + 5) -(3x + 5) = 0

(x^2 - 1)(3x + 5) = 0
(x + 1)(x - 1)(3x+ 5) = 0

the values are -1, 1, and -5/3