DEAR EXPERTS , I SOLVED A FACTORISATION QUESTION ON 19-7-15 . THE QUESTION IS : FACTORISE : 2y3 - 4y2 -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...... Share with your friends Share 0 Manbar Singh answered this FIRST METHOD :Let py = 2y3 - 4y2 - 2y + 4Now, p1 = 213 - 412 - 21 + 4 = 2 - 4 - 2 + 4 = 0⇒y-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 = 0⇒y+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. 0 View Full Answer Avinash answered this 2y3 - 4y2 -2y + 4 = 2y2 (y-2)-2(y-2) =(2y2 - 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) 1 Swagat Swargari answered this 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 0