Find the quotient and remainder using synthetic division.

(2x⁴ – 7x³ – 13x² + 63x – 48) ÷ (2x – 1)

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 2^nd 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 .