Question

P(x)=2x3-3x2+6x+7divided by g(x)=x2-4x+8

Answer 1

Step-by-step explanation:

by doing long division method I got it

but

it is not making the remainder as 0

for Better understanding see the figure

plz mark as brainliest

Answer 2

Given:-

→p(x)=2x³-3x²+6x+7

→g(x)=x²-4x+8

$\rule{200}{1}$

To Find:-

→The Remainder–[r(x)] and The Quotient–[q(x)]?

$\rule{200}{1}$

AnsWer:-

•Using Formula•

→p(x)=q(x)×g(x)+r(x)

→$\frac{p(x)}{g(x)}$=q(x)+r(x)

★Putting the Values★

→$\frac{2x^{3}-3x^{2}+6x+7}{x^{2}-4x+8}$=q(x)+r(x)

→$\frac{\cancel{2x^{3}-3x^{2}+6x+7}}{\cancel{x^{2}-4x+8}}$=q(x)+r(x)

→q(x)=2x+5

→r(x)=10x-33

$\rule{200}{2}$