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

Answer:

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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

Step-by-step explanation:

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Answer 2

Answer:

Yuri is not correct.

Step-by-step 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 = $\frac{3}{4}$,

$q(\frac{3}{4})=6(\frac{3}{4})^3+19(\frac{3}{4})^2-15(\frac{3}{4})-28$

       = $6(\frac{27}{64})+19(\frac{9}{16})-15(\frac{3}{4})-28$

       = $(\frac{162}{64})+(\frac{171}{16})-(\frac{45}{4})-28$

       = $(\frac{162}{64})+(\frac{684}{64})-(\frac{720}{64})-\frac{1792}{64}$

       = $-\frac{1666}{64}$

       = $-\frac{833}{32}$ ≠ 0

Therefore, Yuri is not correct. x = $\frac{3}{4}$ can not be a root of the given expression.