[SIZE=6]y=f(x) is a polynomial let f(x)=x³+4x-6 find all the solutions of x making f(x)=0[/SIZE]

first graph x^3+4x-6=0 to see where zeros are approximately so, there will be one real zero, between 1 and 1.5 use Newton’s Method to find zero f(x)=x^3+4x-6 first derivative is f'(x)=3x^2+4 let initial value x₀=0 checking all above, we see that -> closest to zero so, solution is x[0]=1.135

Thanks a lot for taking time to solve it Like you I also use Newton's method to solve ,I also write a program (excell) to solve ,and I find 3 solutions two negative and one positive positive solution:x=1.6739 negative solutions: x=-0.917 and x=-3.257

i write a program using excel to solve the question I must put into a seed first when i keey in 1 I get x=1.673 i keey in -1 I get x=-0.917 I keey in -3 I get x=- 3.257

Using Newton's method I write a program using excel to solve the question I must put into a seed first when I keey in 1 I get x=1.673 I keey in -1 I get x=-0.917 I keey in -3 I get x=- 3.257 You may take a reference of the graph shown below

sorry You are right I mistype the fuctionn in my program I try again with the right function and put in the seed 1 and I get the solition x=1.1347