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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.

Respuesta :

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

Now, q([tex] \frac{3}{4} [/tex]) = 6([tex] \frac{3}{4} [/tex])³ + 19([tex] \frac{3}{4} [/tex])² – 15([tex] \frac{3}{4} [/tex]) – 28
                                               = [tex]- \frac{833}{32} [/tex]

Therefore Yuri, since q([tex] \frac{3}{4} [/tex]) ≠ 0 then it implies that [tex] \frac{3}{4} [/tex] 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 = [tex] \frac{3}{4} [/tex], instead of just [tex] \frac{3}{4} [/tex]

Answer:

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

Step-by-step explanation:

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