# Given f(x)= -4x^3 + 6x^2 -15x + 40, what are the zeros of this function using Rational Zeros Theorem to determine the potential rational zeros. If x=p/q a irreducible rational root then p is a divisor of 40 and q is a divisor of 4

Therefore p is either 1,2,4,5,8,10,20,40, q is either 1,2,4

and GCD p and q is 1

The possibilities are

q=1: +-1/1,+-2/1,+-4/1,+-5/1,+-8/1,+-10/1,+-20/1,+-40/1

q=2: +-1/2,+-5/2

q=4: +-1/4, +-5/4

Remark: 8/4 is not...

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If x=p/q a irreducible rational root then p is a divisor of 40 and q is a divisor of 4

Therefore p is either 1,2,4,5,8,10,20,40, q is either 1,2,4

and GCD p and q is 1

The possibilities are

q=1: +-1/1,+-2/1,+-4/1,+-5/1,+-8/1,+-10/1,+-20/1,+-40/1

q=2: +-1/2,+-5/2

q=4: +-1/4, +-5/4

Remark: 8/4 is not a possibility because it is not an irreducible fraction.

That'ss a lot of possibilities. Notice that for x<0, -4x^3>0, 6x^2>0, -15x>0 therefore f(x)>40. There won't be any negative solution.

Lets evaluate f(x) for the possible positive roots:

f(1)=27 ne 0

f(2)=2

f(4)=-180

f(5)=-385

f(8)=-1744

f(10)=-3510

f(20)=-29860

f(40)=-246960

f(1/2)=33.5

f(5/2)=-60

f(1/4)=36.5625

f(5/4)=22.8125

It appears that this function does'nt have any rational solutions.

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