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Sagot :
Given:
The polynomial function is
[tex]q(x)=6x^3+19x^2-15x-28[/tex]
Yuri thinks that [tex]\dfrac{3}{4}[/tex] is a root of the given function.
To find:
Why [tex]\dfrac{3}{4}[/tex] cannot be a root?
Solution:
We have,
[tex]q(x)=6x^3+19x^2-15x-28[/tex]
If [tex]\dfrac{3}{4}[/tex] is a root, then the value of the function at [tex]\dfrac{3}{4}[/tex] is 0.
Putting [tex]x=\dfrac{3}{4}[/tex] in the given function, we get
[tex]q(\dfrac{3}{4})=6(\dfrac{3}{4})^3+19(\dfrac{3}{4})^2-15(\dfrac{3}{4})-28[/tex]
[tex]q(\dfrac{3}{4})=6(\dfrac{27}{64})+19(\dfrac{9}{16})-\dfrac{45}{4}-28[/tex]
[tex]q(\dfrac{3}{4})=3(\dfrac{27}{32})+\dfrac{171}{16}-\dfrac{45}{4}-28[/tex]
[tex]q(\dfrac{3}{4})=\dfrac{81}{32}+\dfrac{171}{16}-\dfrac{45}{4}-28[/tex]
Taking LCM, we get
[tex]q(\dfrac{3}{4})=\dfrac{81+342-360-896}{32}[/tex]
[tex]q(\dfrac{3}{4})=\dfrac{-833}{32}\neq 0[/tex]
Since the value of the function at [tex]\dfrac{3}{4}[/tex] is not equal to 0, therefore, [tex]\dfrac{3}{4}[/tex] is not a root of the given function.
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