factor_poly

# factor_poly - Factoring polynomials Common factors Consider...

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Factoring polynomials Common factors . Consider the trinomial + - 3 x 2 y 3 15 x y 4 21 x 3 y 2 . All three terms have 3 x y 2 as a factor. 3 x y 2 is a common factor of the three terms. In fact = + - 3 x 2 y 3 15 x y 4 21 x 3 y 2 3 x y 2 ( ) + - x y 5 y 2 7 x 2 . Always look for a possible common factor before attempting any other methods of factorisation. Example 1 : Find a factorisation of the following polynomials by obtaining a common factor that leaves the remainding terms with no common factor. (i) - + 10 a 3 b c 3 15 a 2 b 2 c 3 20 a 2 b c 4 (ii) + - 5 p 4 q 3 r 10 p 2 q 5 r 5 p 2 q 3 r 3 (iii) + + a b 2 2 a b c a c 2 Solution : (i) - + 10 a 3 b c 3 15 a 2 b 2 c 3 20 a 2 b c 4 = 5 a 2 b c 3 ( ) - + 2 a 3 b 4 c (ii) = + - 5 p 4 q 3 r 10 p 2 q 5 r 5 p 2 q 3 r 3 5 p 2 q 3 r ( ) + - p 2 2 q 2 r 2 (iii) = a ( ) + + b 2 2 b c c 2 a ( ) + b c 2 Example 2 : Factor the expression + + + x 4 y x 3 y 2 x 2 y 3 x y 4 completely. Solution : The four terms of this polynomial in the variables x and y have x y as a common factor. Hence = + + + x 4 y x 3 y 2 x 2 y 3 x y 4 x y ( ) + + + x 3 x 2 y x y 2 y 3 . Note that the first two terms of + + + x 3 x 2 y x y 2 y 3 have x 2 as a common factor and the last two terms have y 2 as a common factor. Hence = + + + x 3 x 2 y x y 2 y 3 + x 2 ( ) + x y y 2 ( ) + x y . The two terms + x 2 ( ) + x y y 2 ( ) + x y have + x y as a common factor hence = + x 2 ( ) + x y y 2 ( ) + x y ( ) + x 2 y 2 ( ) + x y . Hence = + + + x 4 y x 3 y 2 x 2 y 3 x y 4 x y ( ) + + + x 3 x 2 y x y 2 y 3 . = x y ( ) + x 2 ( ) + x y y 2 ( ) + x y = x y ( ) + x y ( ) + x 2 y 2 . ________ Factoring trinomials of the form + + x 2 b x c . We start by considering trinomials in a single variable x of the form + + x 2 b x c , where b and c are integers and look for a factorisation of the form = + + x 2 b x c ( ) + x p ( ) + x q ------- (i). where p and q are integers.

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Now = ( ) + x p ( ) + x q + + + x 2 x q p x p q = + + x 2 ( ) + p q x p q . Hence a factorisation of the form (i) will be possible if we can find integers p and q such that { = + p q b = p q c ------- (ii). For example, we can factor + + x 2 5 x 6, by observing that { = + 2 3 5 = 2 . 3 6 . It follows that + + x 2 5 x 6 = ( ) + x 2 ( ) + x 3 . Example 3 : Factor each of the following trinomials: (a) - - x 2 x 6 (b) - + x 2 9 x 20 (c) - + x 2 37 x 322 (d) + - 3 x 2 6 x y 105 y 2 Solution : (a) Since { = + 3 ( ) - 2 1 = 3 . ( ) - 2 - 6 , it follows that = - - x 2 x 6 ( ) + x 3 ( ) - x 2 . (b) Since { = - + 4 ( ) - 5 - 9 = ( ) - 4 . ( ) - 5 20 , it follows that = - + x 2 9 x 20 ( ) - x 4 ( ) - x 5 . (c) We need to find two integers p and q such that { = + p q - 37 = p q 322 . Because the product p q is positive while the sum + p q i s negative we see that p and q must both be negative integers . The corresponding positive integers
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factor_poly - Factoring polynomials Common factors Consider...

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