DEAR EXPERTS ,
I SOLVED A FACTORISATION QUESTION ON 19-7-15 . THE QUESTION IS :
FACTORISE : 2y- 4y​2 -2y +4 
I SOLVED THE COMPLTE ANSWER ON MY OWN . DURING DIVISION I 
DIVIDED THE POLYNOMIAL BY y+1 
AND GOT THE SAME ANSWER (ANSWER : (y+1) (2y-2) (y-2) ) . WHEN I CHECKED THE SOLUTION , THE POLYNOMIAL WAS DIVIDED BY y-1 .  BUT THE ANSWER WAS SAME AS MY ANSWER >
EXPERTS !!! PLS TELL ME DIVISION WTH Y+1 IS CORRECT OR Y-1 ?
IN EXAM IF THAT QUESTION COMES, I SHOULD DIVIDE WITH Y+1 OR Y-1 ??? WHICH IS APPROPRIATE ?? LOOKING FOR ANSWER SOON......

FIRST METHOD :Let py = 2y3 - 4y2 - 2y + 4Now, p1 = 213 - 412 - 21 + 4 = 2 - 4 - 2 + 4 = 0y-1 is a factor of py.Now, we will divide py by y-1 to obtain the remaining factors.y - 1)  2y3 - 4y2 - 2y + 4(2y2-2y-4             2y3 - 2y2                    - 2y2 - 2y                    - 2y2 + 2y                                - 4y + 4                                - 4y + 4                                         0py = y-12y2-2y-4=y-1. 2y+1y-2=2y-2y+1y-2SECOND METHOD :Let py = 2y3 - 4y2 - 2y + 4Now, p-1 = 2-13 - 4-12 - 2-1 + 4  = 0y+1 is a factor of py.Now, we will divide py by y+1 to obtain the remaining factors.On division, we get 2y2-6y+4 as quotient and 0 as remainder.So, py = y+12y2-6y+4=y+1. 2y-1y-2=2y-2y+1y-2Hence, both the methods are correct.

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2y3 - 4y-2y + 4 = 2y(y-2)-2(y-2)
                             =(2y- 2) (y-2)
                             =2(y2 - 1)(y-2)
                              =2{(y)2-(1)2}(y-2)
            [Using the identity:a2-b2=(a+b)(a-b)]  
                               Here, a=y and b=1
=2(y+1)(y-1)(y-2)
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that will depend what you are using for P(y)
if you are putting y = 1 to show that P(1) = 0, then divide by y - 1
and if you are putting y = -1, then divide by y + 1

both will give same answer:

also, the answer that you have got (y + 1) (2y - 2) (y - 2)
write it as 2(y+1)(y-1)(y-2)  //by taking 2 common in the second term 2y-2
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