Question

by actual division, find the remainder.if

${x}^{4} - 2 {x}^{2} + 6x + 3$
Divide by actual division method by
$x - 2$

Answer 1

AHOY USER !!

Given ,

p(x) = $x ^4 - 2x ^2 + 6x + 3$

g(x) = $x - 2$

Now , find the zero of g(x)..

$x - 2 = 0$

$x = 0+2$

$x = 2$

Now,,

$p(2) = 2 ^4 - 2*2 ^2 + 6*2 + 3$

$= 16 - 2*4 + 12 + 3$

$= 16 - 8 + 15$

$= 31 - 8$

$= 28...$

So, when $x ^4 - 2x ^2 + 6x + 3$ is divided by $x - 2$ then the remainder will be 28....

☺✌

Answer 2

Hey...!!

p(x) = x ^4 - 2x ^2 + 6x + 3
g(x) = x - 2
..
x - 2 = 0
x = 2

p(2) = 2 ^4 - 2*2 ^2 + 6*2 + 3
= 16 - 2*4 + 12 + 3
= 16 - 8 + 15
= 31 - 8
= 28

Hope it helps u.....