yuri thinks that $\frac{3}{4}$ is a root of the following function. $q(x)=6x^{3}+19x^{2}-15x - 28$. explain…

yuri thinks that $\frac{3}{4}$ is a root of the following function. $q(x)=6x^{3}+19x^{2}-15x - 28$. explain to yuri why $\frac{3}{4}$ cannot be a root.

yuri thinks that $\frac{3}{4}$ is a root of the following function. $q(x)=6x^{3}+19x^{2}-15x - 28$. explain to yuri why $\frac{3}{4}$ cannot be a root.

Answer

Explanation:

Step1: Substitute x = 3/4 into q(x)

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

Step2: Calculate each term

$6(\frac{3}{4})^{3}=6\times\frac{27}{64}=\frac{162}{64}$, $19(\frac{3}{4})^{2}=19\times\frac{9}{16}=\frac{171}{16}=\frac{684}{64}$, $- 15(\frac{3}{4})=-\frac{45}{4}=-\frac{720}{64}$, and $-28=-\frac{1792}{64}$

Step3: Combine the terms

$q(\frac{3}{4})=\frac{162 + 684-720 - 1792}{64}=\frac{846-720 - 1792}{64}=\frac{126-1792}{64}=-\frac{1666}{64}\neq0$

Answer:

Since substituting $x = \frac{3}{4}$ into the function $q(x)$ gives a non - zero result ($q(\frac{3}{4})\neq0$), $\frac{3}{4}$ cannot be a root of the function $q(x)$.