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

Did you include the following in your response?

According to the rational root theorem, potential rational roots must be in theform p/q 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 does 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:

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Answer 2

Yuri is not correct.

Explanation:

Given expression is q(x) = 6x³ + 19x² - 15x - 28

If 'a' is a root of the given function, then by substituting x = a in the expression, q(a) = 0

Similarly, for x =
`(3)/(4)`,

`q((3)/(4))=6((3)/(4))^3+19((3)/(4))^2-15((3)/(4))-28`

=
`6((27)/(64))+19((9)/(16))-15((3)/(4))-28`

=
`((162)/(64))+((171)/(16))-((45)/(4))-28`

=
`((162)/(64))+((684)/(64))-((720)/(64))-(1792)/(64)`

=
`-(1666)/(64)`

=
`-(833)/(32)` ≠ 0

Therefore, Yuri is not correct. x =
`(3)/(4)` can not be a root of the given expression.