Question

Yuri thinks that 3/4 is a root of the following function.

Answer 1

Given:

The polynomial function is

$q(x)=6x^3+19x^2-15x-28$

Yuri thinks that $\dfrac{3}{4}$ is a root of the given function.

To find:

Why $\dfrac{3}{4}$ cannot be a root?

Solution:

We have,

$q(x)=6x^3+19x^2-15x-28$

If $\dfrac{3}{4}$ is a root, then the value of the function at $\dfrac{3}{4}$ is 0.

Putting $x=\dfrac{3}{4}$ in the given function, we get

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

$q(\dfrac{3}{4})=6(\dfrac{27}{64})+19(\dfrac{9}{16})-\dfrac{45}{4}-28$

$q(\dfrac{3}{4})=3(\dfrac{27}{32})+\dfrac{171}{16}-\dfrac{45}{4}-28$

$q(\dfrac{3}{4})=\dfrac{81}{32}+\dfrac{171}{16}-\dfrac{45}{4}-28$

Taking LCM, we get

$q(\dfrac{3}{4})=\dfrac{81+342-360-896}{32}$

$q(\dfrac{3}{4})=\dfrac{-833}{32}\neq 0$

Since the value of the function at $\dfrac{3}{4}$ is not equal to 0, therefore, $\dfrac{3}{4}$ is not a root of the given function.