Question

p(x)=x^5+2(x)^3-(x)^²+4x+³ g(x)=x+3 by remainder theorem and long division​

Answer 1

Given:-

→p(x)=x⁵+2x³-x²+4x+3

→g(x)=x+3

$\rule{200}{1}$

To Find:-

→The remainder-[r(x)]?

$\rule{200}{1}$

AnsWer:-

★Using Euclid's Formula★

→p(x)=q(x)×g(x)+r(x)

→$\frac{p(x)}{g(x)}$=q(x)+r(x)

♦Putting the Values♦

→$\frac{x^{5}+2x^{3}-x^{2}+4x+3}{x+3}$=q(x)+r(x)

→$\frac{\cancel{x^{5}+2x^{3}-x^{2}+4x+3}}{\cancel{x+3}}$=q(x)+r(x)

→r(x)=-315

→q(x)=x⁴-3x³+11x²-34x+106

$\rule{200}{2}$

Answer 2

$\huge\mathbb\green{Answer }$