Respuesta :

kamilmatematik100504

Answer:

[tex]\large \boldsymbol {} 6x^2 -5x-14 =0 \\\\ D=25+14\cdot 4\cdot 6 =361=19^2 \\\\ x_1= \dfrac{5+19}{12} =2 \\\\\\ x_2 =\dfrac{5-19}{12} =-\dfrac{7}{6} \\\\ and \ \ we \ \ \ know \ \ that \\\\\ 6x^2 -5x-14 =6 (x-x_1) ( x- x_ 2) = 6 ( x-(-\frac{7}{6} ))\boldsymbol { ( x-2 ) } \\\\ as \ we \ can \ see \ , \ the \ \ expansion \ is \\\\\ x-2 ; therefore \ \ , \ \ the \ \ answer \ \ is \ \ yes[/tex]

caylus

Answer: Yes: divisible

Step-by-step explanation:

[tex]P(x)=6x^2-5x-14\\\\P(2)=6*2^2-5*2-14=24-10-14=0\\\\P(x)\ is\ exactly\ divisible\ by\ x-2\\[/tex]

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