Division of polynomial by linear binomial
Linear binomial. Bezout's theorem.
Linear binomial is a polynomial of the first degree: ax+ b. If to divide a polynomial, containing a
letter x, by a linear binomial x – b, where b is a number ( positive or negative ), then a remainder will be
a polynomial only of zero degree, i.e. some number N , which can be found without finding a quotient. Exactly, this number is equal to the value of the
polynomial, received at x = b. This property is proved by Bezout’s theorem:
a polynomial a_{0 }x^{m} + a_{1 }x^{m}^{}^{1}
+ a_{2} x^{m}^{}^{2} + …+ a_{m} is divided by x – b with a remainder N = a_{0 } b^{m} + a_{1 } b^{m}^{}^{1} + a_{2 } b^{m}^{}^{2} + …+ a_{m }.
The p r o o f . According to the definition of division (see above) we have:
a_{0 }x^{m} + a_{1 }x^{m}^{}^{1}
+ a_{2} x^{m}^{}^{2} + …+ a_{m} = ( x – b )
Q + N ,
where Q is some polynomial, N is some number. Substitute here x = b , then ( x – b ) Q will be missing and we receive:
a_{0 } b^{m}
+ a_{1 } b^{m}^{}^{1} + a_{2 } b^{m}^{}^{2} + …+ a_{m } = N .
The r e m a r k . It is possible, that N = 0 . Then b is a root of the equation:
a_{0 }x^{m} + a_{1 }x^{m}^{}^{1}
+ a_{2} x^{m}^{}^{2} + …+ a_{m} = 0 .
The theorem has been proved.
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