(2x^2-5x-12 )/ (x-4)=
(10x^2+17x+6) / (2x+1)=
ALGUIEN ME PUEDE AYUDAR
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May this help.......
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First of all I am not getting what you want, so secondly I am writing down all the possible answers one can get...
So the first thing we need to do is factorise both the equations...
(a).(2x²-5x-12)=(2x²-8x+3x-12)={2x(x-4)+3(x-4)}=(2x+3)(x-4)
(b).(10x²+17x+6)=(10x²+5x+12x+6)={5x(2x+1)+6(2x+1)}=(5x+1)(2x+1)
So now because the factorisation is done we proceed to the question...
1. Is the question asks for quotients, then the answer is as follows...
(a).{(2x+3)(x-4)}/(x-4)=(2x+3)
(b).{(5x+1)(2x+1)}/(2x+1)=(5x+1)
2. If the question asks for the remainders, then as you see, the dividends are factors of the equations then the remainders would be 0...
(a). The remainder of 0
(b). The remainder too is 0
So the first thing we need to do is factorise both the equations...
(a).(2x²-5x-12)=(2x²-8x+3x-12)={2x(x-4)+3(x-4)}=(2x+3)(x-4)
(b).(10x²+17x+6)=(10x²+5x+12x+6)={5x(2x+1)+6(2x+1)}=(5x+1)(2x+1)
So now because the factorisation is done we proceed to the question...
1. Is the question asks for quotients, then the answer is as follows...
(a).{(2x+3)(x-4)}/(x-4)=(2x+3)
(b).{(5x+1)(2x+1)}/(2x+1)=(5x+1)
2. If the question asks for the remainders, then as you see, the dividends are factors of the equations then the remainders would be 0...
(a). The remainder of 0
(b). The remainder too is 0
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