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

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Answer 1 · 2016-06-29T05:32:32

#-1, -1, 1 and -4/3#

Let f(x) denote the given biquadratic.

#f(+-1)=0#. So,# +-1# are two of the roots.

#f'=12x^3+21x^2+2x-7 and f(-1)=0#

Therefore, #-1 # is a double root.

Sum of the roots

= -(coefficient of #x^3#)/(coefficient of #x^4#)

= #-7/3.

So, the 4th root is #-7/3-(-1-1+1)=-4/3#

Thus, the zeros are #-1, -1. 1 and -4/3#.

Answer 2 · 2016-06-29T06:15:44

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

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

First note that the sum of the coefficients is #0#.

That is:

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

So #f(1) = 0#, #color(blue)(x=1)# is a zero and #(x-1)# a factor:

#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 #0#.

That is:

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

Hence #color(blue)(x=-1)# is a zero and #(x+1)# a factor:

#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 #0#.

That is:

#3-7+4 = 0#

Hence #color(blue)(x=-1)# is a zero again and #(x+1)# a factor:

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

Finally, from the remaining linear factor, we find the last zero #color(blue)(x=-4/3)#.

So the zeros of our quartic are:

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