Answer :

Let p(x) = 2x4 – 7x3 13x2 + 63x – 48 be the dividend. Arranging p(x) according to the according descending powers of x.


p(x) = 2x4 – 7x3 13x2 + 63x – 48


Divisor, q(x) = 2x – 1


To find out Zero of the divisor –


q(x) = 0


2x – 1 = 0


x =


zero of divisor is .


And, p(x) = 2x4 – 7x3 13x2 + 63x – 48


Put zero for the first entry in the 2nd row.



Quotient = 2x3 – 6x2 – 16x + 55


p(x) = (Quotient)×q(x) + remainder.


So, 2x4 – 7x3 13x2 + 63x – 48


= (x – )( 8x3 – 2x2 x + ) + ()


= (2x – 1)(2x3 – 6x2 – 16x + 55)


Thus, the Quotient = (2x3 – 6x2 – 16x + 55) = (x3 – 3x2 – 8x + ) and remainder is .


Hence, when p(x) is divided by (2x – 1) the quotient is (x3 – 3x2 – 8x + ) and remainder is .


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