# How do you find all the zeros of #3x^4+7x^3+x^2-7x-4#?

##### 2 Answers

#### Explanation:

Let f(x) denote the given biquadratic.

Therefore,

Sum of the roots

= -(coefficient of

= #-7/3.

So, the 4th root is

Thus, the zeros are

Zeros:

#### Explanation:

#f(x) = 3x^4+7x^3+x^2-7x-4#

First note that the sum of the coefficients is

That is:

#3+7+1-7-4 = 0#

So

#3x^4+7x^3+x^2-7x-4 = (x-1)(3x^3+10x^2+11x+4)#

Looking at the remaining cubic factor, note that if you reverse the signs of the coefficients of the terms of odd degree, then the sum is

That is:

#-3+10-11+4 = 0#

Hence

#3x^3+10x^2+11x+4 = (x+1)(3x^2+7x+4)#

Looking at the remaining quadratic, reversing the sign of the coefficient of the term of odd degree, the sum of the coefficients is

That is:

#3-7+4 = 0#

Hence

#3x^2+7x+4 = (x+1)(3x+4)#

Finally, from the remaining linear factor, we find the last zero

So the zeros of our quartic are:

#color(blue)(-1)# with multiplicity#2# ,#color(blue)(1)# and#color(blue)(-4/3)# .