# How do you find the zeros of #f(x) = 3x^4 +4x^3 +6x^2 -4#?

##### 1 Answer

Use a numerical method to find approximate zeros:

#x_1 ~~ -0.908845148702144#

#x_2 ~~ 0.639396468497863#

#x_(3,4) ~~ -0.53194232656453+-1.41826916487156i#

#### Explanation:

By the rational root theorem, any rational zeros of

That means that the only possible rational zeros are:

#+-1/3# ,#+-2/3# ,#+-1# ,#+-4/3# ,#+-2# ,#+-4#

Evaluating

So

The coefficients of

Since this has one change of sign,

The coefficients of

Since this has three changes of sign and we already know about one positive zero, that means that

We could use an algebraic method to find the zeros, but the method is lengthy and the resulting expressions in terms of square and cube roots are horribly messy.

So in this case I would recommend a numeric method.

We could use Durand-Kerner, which gives all four zeros in one go, but I will use Newton's method since we know we have at least

Newton's method starts by choosing an approximation

#a_(i+1) = a_i - (f(a_i))/(f'(a_i)) = a_i - (3x^4+4x^3+6x^2-4)/(12x^3+12x^2+6x)#

For Real zeros we can express this simply in a spreadsheet and find zeros:

#x_1 ~~ -0.908845148702144#

#x_2 ~~ 0.639396468497863#

Then:

#(x-x_1)(x-x_2) ~~ x^2+0.26944868020428 x-0.58111237849157#

Long dividing

#x^2+1.0638846531291x+2.29445006281636#

Then we can use the quadratic formula to find zeros:

#x ~~ -0.53194232656453+-1.41826916487156i#

If you would rather not do the polynomial long division, then a good alternative is to make a more sophisticated spreadsheet with separate columns for Real and imaginary parts expressing the iteration formula above then try some Complex initial approximations.