Find the integral roots of the polynomial f(x) = x3 + 6x2 + 11x + 6.

1.	Find the integral roots of the polynomial f(x) = x3 + 6x2 + 11x + 6.
Answer» We have, f(x) = x³ + 6x² + 11x + 6 Clearly, f (x) is a polynomial with integer coefficient and the coefficient of the highest degree term i.e., the leading coefficient is 1. Therefore, integer root of f (x) are limited to the integer factors of 6, which are: +1,+2,+3,+6 We observe that' f (-1) = (-1)³ + 6 (-1)² + 11 (-1) + 6 = -1 + 6 -11 + 6 = 0 f (-2) = (-2)³ + 6 (-2)² + 11 (-2) + 6 = -8 + 24 – 22 + 6 = 0 f (-3) = (-3)³ + 6 (-3)²+ 11 (-3) + 6 = -27 + 54 – 33 + 6 = 0 Therefore, integral roots of f (x) are -1, -2, -3.

Answer»

We have,

f(x) = x³ + 6x² + 11x + 6

Clearly, f (x) is a polynomial with integer coefficient and the coefficient of the highest degree term i.e., the leading coefficient is 1.

Therefore, integer root of f (x) are limited to the integer factors of 6, which are:

+1,+2,+3,+6

We observe that'

f (-1) = (-1)³ + 6 (-1)² + 11 (-1) + 6

= -1 + 6 -11 + 6

= 0 f (-2) = (-2)³ + 6 (-2)² + 11 (-2) + 6

= -8 + 24 – 22 + 6

= 0

f (-3) = (-3)³ + 6 (-3)²+ 11 (-3) + 6

= -27 + 54 – 33 + 6

= 0

Therefore,

integral roots of f (x) are -1, -2, -3.

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