Question

using
remainder​ theorem x square - 3 x + 4
divided by x minus 3
​

Answer 1

hope this helps u

given p(X)x^2-3x+4

g(x) x-3

X=3

sub X= 3 in p(X)

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

=9-9+4

=0+4

p(3)= 4

remainder = 4

Answer 2

Your Question is:

${x}^{2} - 3x + 4 \div x - 3$

$x - 3){x}^{2} - 3x + 4($

$x - 3) {x}^{2} - 3x + 3(x \\ \: \: {x}^{2} - 3 + 4 \\ \: \: \: \: \: 4$

Therefore the remainder is 4.