When P(x) and Q(x) are divided by x^(2)-2x+3 the remainders are(x-2) and (x+3) respectively?

P(x)/(x^(2)-2x+3)=f(x)+(x-2)
Q(x)/(x^(2)-2x+3)=g(x)+(x+3)
Solve x^(2)-2x+3=0
x1=1+sqrt(2)i
X2=1-sqrt (2)i
P(1+sqrt (2)i)=x-2 =-1+sqrt(2)i
P(1-sqrt (2)i)=x-2 =-1-sqrt(2)i
Q(1+sqrt(2)i)=x+3=4+sqrt(2)i
Q(1-sqrt(2)i)=x+3=4-sqrt(2)i
Remainder of p(x)*Q(x) is a linear equation because the divisor is a quadratic equation
P(1+sqrt(2)i)*Q(1+sqrt(2)i=ax+b
P (-1-sqrt(2)i)*Q(-1-sqrt(2)i=ax+b
Then from this system of equation you can find answer
a=3
b=-9
3x-9 is the Remainder
Good luck????

Let f(x) and g(x) be the quotients when P(x) and Q(x) are divided by x² - 2x + 3.

When P(x) is divided by (x² - 2x + 3), the quotient is f(x) and the remainder is (x - 2).
P(x) = (x² - 2x + 3)*f(x) + (x - 2)

When Q(x) is divided by (x² - 2x + 3), the quotient is g(x) and the remainder is (x + 3).
Q(x) = (x² - 2x + 3)*g(x) + (x + 3)

Then, P(x)*Q(x)
= [(x² - 2x + 3)*f(x) + (x - 2)] [(x² - 2x + 3)*g(x) + (x + 3)]
= (x² - 2x + 3)²*f(x)*g(x) + (x² - 2x + 3)*f(x)*(x+3) + (x² - 2x + 3)*g(x)*(x+2) + (x - 2)(x + 3)
= (x² - 2x + 3)²*f(x)*g(x) + (x² - 2x + 3)*f(x)*(x+3) + (x² - 2x + 3)*g(x)*(x + 2) + x² + x - 6
= (x² - 2x + 3)²*f(x)*g(x) + (x² - 2x + 3)*f(x)*(x+3) + (x² - 2x + 3)*g(x)*(x + 2) + (x² - 2x + 3) + (3x - 9)
= (x² - 2x + 3)*[(x² - 2x + 3)*f(x)*g(x) + f(x)*(x+3) + g(x)*(x + 2) + 1] + (3x - 9)

Hence, when P(x)*Q(x) is divided by (x² - 2x + 3), the remainder = 3x - 9

Answer