Respuesta :
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 = [tex]\frac{3}{4}[/tex],
[tex]q(\frac{3}{4})=6(\frac{3}{4})^3+19(\frac{3}{4})^2-15(\frac{3}{4})-28[/tex]
= [tex]6(\frac{27}{64})+19(\frac{9}{16})-15(\frac{3}{4})-28[/tex]
= [tex](\frac{162}{64})+(\frac{171}{16})-(\frac{45}{4})-28[/tex]
= [tex](\frac{162}{64})+(\frac{684}{64})-(\frac{720}{64})-\frac{1792}{64}[/tex]
= [tex]-\frac{1666}{64}[/tex]
= [tex]-\frac{833}{32}[/tex] ≠ 0
Therefore, Yuri is not correct. x = [tex]\frac{3}{4}[/tex] can not be a root of the given expression.
Answer:
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
Step-by-step explanation:
sample response edgen 2020
