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

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

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

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

Answer 2

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: