Find the integral roots of the polynomial f(x) = x³+6x²+11x+6.
Answers
The integral roots of the polynomial f(x) = x³+6x²+11x+6 are (x+1) (x+2) (x+3)
Given,
f(x) = x³+6x²+11x+6
let us take x=-1 to find whether this satisfies f(x).
f(-1) = (-1)^3+6(-1)^2+11(-1)+6
= -1 + 6 - 11 + 6
= -12 + 12
= 0
Hence (x+1) is one of the factor of f(x) = x³+6x²+11x+6
In order to find the remaining factors, we need to divide f(x) = x³+6x²+11x+6 by (x+1).
x+1 )x^3+6x^2+11x+6( x^2+5x+6
x^3+x^2
----------
5x^2+11x
5x^2+5x
----------
6x+6
6x+6
----------
0
we have obtained x^2+5x+6 as the factor
x^2+5x+6 = x^2+2x+3x+6 = x(x+2)+2(x+3) = (x+2)(x+3)
Hence the remaining factors are (x+2)(x+3)
Given : f(x) = x³ + 6x² + 11x + 6.
We see that f(x) is a polynomial with integer coefficients and the coefficient of the highest degree term .
Therefore, integer root of f(x) are limited to the integer factors of 6, which are : ±1 , ±2 , ±3, ±6
Now, on putting x = -1 in f(x) :
f (-1) = (-1)³ + 6 (-1)² + 11 × -1 + 6
= -1 + 6 -11 + 6
= 5 - 5
= 0
On putting x = - 2 in f(x) :
f (-2) = (-2)³ + 6 (- 2)² + 11 × - 2 + 6
= (-2)³ + 6 × 4 + - 22 + 6
= - 8 + 24 - 22 + 6
= 16 - 16
= 0
On putting x = - 3 in f(x) :
f (-3) = (- 3)³ + 6 (- 3)² + 11 (- 3) + 6
= (- 3)³ + 6 × 9 + 11 (- 3) + 6
= -27 + 54 - 33 + 6
= 27 - 27
= 0
Hence, integral roots of f (x) are -1, - 2, - 3.
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