How do you find all the zeros of f(x)=3x4−17x3+33x2−17x−10?

1 Answer
George C.
Apr 16, 2016

Find zeros: −13, 2 and 2±i

Explanation:

By the rational root theorem, any rational zeros of f(x) must be expressible in the form pq for integers p,q with p a divisor of the constant term −10 and q a divisor of the coefficient 3 of the leading term.

That means that the only possible rational zeros are:

±13,±23,±1,±53,±2,±103,±5,±10

Trying each in turn, we find:

f(−13)=381+1727+339+173−10

=1+17+99+153−27027=0

f(2)=48−136+132−34+10=0

So x=−13 and x=2 are zeros and (3x+1) and (x−2) are factors of f(x):

3x4−17x3+33x2−17x−10

=(3x+1)(x3−6x2+13x−10)

=(3x+1)(x−2)(x2−4x+5)

The remaining quadratic factor has negative discriminant, but we can factor it using Complex numbers, completeing the square and the difference of squares identity:

a2−b2=(a−b)(a+b)

as follows:

x2−4x+5=(x−2)2+1=(x−2)2−i2=((x−2)−i)((x−2)+i)=(x−2−i)(x−2+i)

So the remaining zeros are x=2±i

Related questions
See all questions in Zeros
Impact of this question
2345 views around the world
You can reuse this answer
Creative Commons License