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

2 Answers
Jun 29, 2016

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

Explanation:

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.

Jun 29, 2016

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

Explanation:

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).