Question

It's maths plz help !¡!!!​

Answer 1

Answer:

Quotient is $x^2 -x+3$ and remainder is 3

Step-by-step explanation:

Answer 2

Answer: $=\frac{2x^3-9x^2+12x-7}{\left(x-2\right)^2}$

Step-by-step explanation:

$\frac{d}{dx}\left(\frac{x^3-3x^2+5x-3}{x-2}\right)$

$\mathrm{Apply\:the\:Quotient\:Rule}:\quad \left(\frac{f}{g}\right)^'=\frac{f\:'\cdot g-g'\cdot f}{g^2}$

$=\frac{\frac{d}{dx}\left(x^3-3x^2+5x-3\right)\left(x-2\right)-\frac{d}{dx}\left(x-2\right)\left(x^3-3x^2+5x-3\right)}{\left(x-2\right)^2}$

$\frac{d}{dx}\left(x^3-3x^2+5x-3\right)$

$\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'$

$=\frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(3x^2\right)+\frac{d}{dx}\left(5x\right)-\frac{d}{dx}\left(3\right)$

$=3x^2-6x+5-0$

$\frac{d}{dx}\left(x-2\right)$

$(\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g' )$

1-0 =1

$\frac{\left(3x^2-6x+5\right)\left(x-2\right)-1\cdot \left(x^3-3x^2+5x-3\right)}{\left(x-2\right)^2}$

$=\frac{2x^3-9x^2+12x-7}{\left(x-2\right)^2}$