Question

if 4x^4-3x^3-3x^2+x-7 is divided by 1-2x. then remainder will bee

Answer 1

Answer: 1-2x=0

1 = 2x

1/2 = x

P(x) = 4x^4-3x^3-3x^2+x-7

= 4(1/2)^4-3(1/2)^3-3(1/2)^2+(1/2)-7

= 4(1/16)-3(1/8)-3(1/4)+(1/2)-7

= (1/4)-(3/8)-(3/4)+(1/2)-7

= 2-3-6+4-56

8

= -59 will be the remainder.

8

Answer 2

Step-by-step explanation:

1st step :

Find the zero of 1-2x

=》 1 - 2x = 0

=》 2x = 1

=》 x = 1/2

2nd step :

Replace the value of x by 1/2

$\mathsf{p(x) = {4x}^{4} - {3x}^{3} + x - 7}$

$\mathsf{p( \frac{1}{2} ) = 4( \frac{1}{2} {)}^{4} - 3( \frac{1}{2} ) {}^{3} - 3( \frac{1}{2} ) {}^{2} + ( \frac{1}{2}) - 7 }$

p(x) = 4(1/16) - 3(1/8) -3(1/4) + (1/2) -7

p(x) = 4/16 - 3/8 - 3/4 + 1/2 -7/1

LCM of 16 , 8 , 4 , 2 = 16

p(x) = 4-6-12+8-112/16

p(x) = -118/16

p(x) = -59/8