Question

Yuri thinks that 3/4 is a root of the following function. q(x) = 6x3 + 19x2 – 15x – 28 Explain to Yuri why 3/4 cannot be a root.

Answer 1

According to the rational root theorem, potential rational roots must be in p/q form where p is a factor of the constant term and q is a factor of the leading coefficient.

3 is not a factor of 28, and 4 is not a factor of 6. So, 3/4 not satisfy the rational root theorem.

Substituting 3/4 in for x does not result in 0. The remainder is not 0 when dividing by

x –3/4

Explanation:

Answer 2

If
`(3)/(4)` is a root of the equation, q(x) = 6x3 + 19x2 – 15x – 28 then by the factor formula q(
`(3)/(4)`) = 0; if that is not the case then
`(3)/(4)` cannot be a factor of the function.

Now, q(
`(3)/(4)`) = 6(
`(3)/(4)`)³ + 19(
`(3)/(4)`)² – 15(
`(3)/(4)`) – 28
=
`- (833)/(32)`

Therefore Yuri, since q(
`(3)/(4)`) ≠ 0 then it implies that
`(3)/(4)` is not a root of q(x).

Additionally, a root should be expressed in terms of x, thus the possible root ought to be written as x =
`(3)/(4)`, instead of just
`(3)/(4)`