Polynomial Roots - #1
Factorize x 3 − 4 x 2 − 1 1 x + 3 0 .
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3 solutions
that was easy
f ( x ) = x 3 − 4 x 2 − 1 1 x + 3 0 = x 3 − 2 x 2 − 2 x 2 + 4 x − 1 5 x + 3 0 = x 2 ( x − 2 ) − 2 x ( x − 2 ) − 1 5 ( x − 2 ) = ( x − 2 ) ( x 2 − 2 x − 1 5 ) = ( x − 2 ) ( x 2 − 5 x + 3 x − 1 5 ) = ( x − 2 ) { x ( x − 5 ) + 3 ( x − 5 ) } = ( x − 2 ) ( x − 5 ) ( x + 3 ) .
Note: ( x − 2 ) is a factor of f ( x ) because if x = 2 , then f ( x ) = 0 .
Relevant wiki: Rational Root Theorem - Basic
Using rational root theorem x 3 − 4 x 2 − 1 1 x + 3 0 = ( x − 2 ) ( x 2 − 2 x − 1 5 ) = ( x − 2 ) ( x + 3 ) ( x − 5 ) .
**Plugging in values for x gives us 2 has root because 2 3 − 4 ( 2 2 ) − 1 1 ( 2 ) + 3 0 = 0 . We divide x 3 − 4 x 2 − 1 1 x + 3 0 by x − 2 to get x 2 − 2 x − 1 5 . Factoring that we have ( x − 3 ) ( x − 5 ) . So x 3 − 4 x 2 − 1 1 x + 3 0 = ( x − 3 ) ( x − 5 ) ( x − 2 ) . - Source (Aops)