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 multiplicity2 ,color(blue)(1) andcolor(blue)(-4/3) .