factor_polys

# factor_polys - Factoring polynomials The following two...

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Factoring polynomials The following two theorems are of use in factoring polynomials. The Remainder Theorem When a linear expression = ( ) d x - x a is divided into a polynomial ( ) P x to give a quotient ( ) Q x and remainder R , so that = ( ) P x + ( ) - x a ( ) Q x R ------- (i), then = R ( ) P a , that is, the remainder when a polynomial ( ) P x is divided by = ( ) d x - x a , is the value of ( ) P x at = x a , namely ( ) P a . The Factor Theorem A linear expression of the form = ( ) d x - x a is a factor of a polynomial ( ) P x exactly when = ( ) P a 0. Example 1 : Factor the polynomial = ( ) P x + - - x 3 5 x 2 2 x 24, and find the zeros of ( ) P x . Solution : Suppose that = ( ) P x ( ) + x p ( ) + x q ( ) + x r , where p , q and r are integers. Then = p q r - 24 and p , q and r are integer factors of the constant term - 24 in ( ) P x . The associated zeros - p , - q and - r are also integer factors of - 24. If ( ) P x has only one linear factor = ( ) A x + x p , with p an integer and a companion quadratic factor = ( ) B x + + x 2 m x n , with no integer zeros, such that = ( ) P x ( ) A x ( ) B x = ( ) + x p ( ) + + x 2 m x n , we must have = p n - 24, so that p , or equivalently - p , is still an integer factor of - 24. It is therefore reasonable to look for a zero of ( ) P x which is an integer factor of the constant term - 24. The possiblities to consider for such an integer zero of ( ) P x are: + 1, + 2, + 3, + 4, + 6, + 8, + 12, + 24. It is probably easier to check whether + 1 and + 2 are zeros by direct substitution rather than by using synthetic division. = ( ) P 1 + - - 1 5 2 24 = - 20 0, = ( ) P - 1 - + + - 1 5 2 24 = - 18 0 , = ( ) P 2 + - - 8 20 4 24 = 0. Since = x 2 is a zero of ( ) P x , it follows by the Factor Theorem that - x 2 is a factor of ( ) P x . We can check this by synthetic division, and also find the quotient ( ) Q x when ( ) P x is divided by - x 2. 2 | 1 5 - 2 - 24 2 14 24 ________________________ 1 7 12 | 0 = + - - x 3 5 x 2 2 x 24 ( ) - x 2 ( ) + + x 2 7 x 12 . Hence, by factoring the residual quadratic + + x 2 7 x 12, we see that = ( ) P x ( ) - x 2 ( ) + x 3 ( ) + x 4 . ____________ The corresponding zeros of ( ) P x are 2, - 3 and - 4. Example 2 : Factor the polynomial = ( ) P x + - + x 3 2 x 2 13 x 10, and find the zeros of ( ) P x . Solution : The integer factors of 10 provide candidates for zeros of ( ) P x , that is, + 1, + 2, + 5, + 10. Straight away = ( ) P 1 + - + 1 2 13 10 = 0, so it follows by the Factor Theorem that - x 1 is a factor of ( ) P x . We can find the quotient when ( ) P x is divided by - x 1 by using synthetic division. 1 | 1 2 - 13 10 1 3 - 10 ________________________ 1 3 - 10 | 0 = + - + x 3 2 x 2 13 x 10 ( ) - x 1 ( ) + - x 2 3 x 10 = ( ) - x 1 ( ) - x 2 ( ) + x 5 __________ The zeros of ( ) P x are 1, 2 and - 5.

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Example 3 : Factor the polynomial = ( ) P x + + + + x 4 11 x 3 41 x 2 61 x 30, and find the zeros of ( ) P x . Solution
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factor_polys - Factoring polynomials The following two...

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