Question

PLEASE ANSWER ASAP! SHOW ALL WORK! WILL GIVE BRAINLIEST!

Answer 1

1) $x^2=x\cdot x$, and $x(x+3)=x^2+3x$. Subtracting this from the numerator gives a remainder of

$(x^2-13x-48)-(x^2+3x)=-16x-48$

$-16x=-16\cdot x$, and $-16(x+3)=-16x-48$. Subtracting this from the previous remainder gives a new remainder of

$(-16x-48)-(-16x-48)=0$

This means that

$\dfrac{x^2-13x-48}{x+3}=x-16$

2) $3x^3=3x^2\cdot x$, and $3x^2(x+2)=3x^3+6x^2$. Subtracting this from the numerator gives a remainder of

$(3x^3-x^2-7x+6)-(3x^3+6x^2)=-7x^2-7x+6$

$-7x^2=-7x\cdot x$, and $-7x(x+2)=-7x^2-14x$. Subtracting this from the previous remainder gives a new remainder of

$(-7x^2-7x+6)-(-7x^2-14x)=7x+6$

$7x=7\cdot x$, and $7(x+2)=7x+14$. Subtracting this from the previous remainder gives a new remainder of

$(7x+6)-(7x+14)=-8$

This means that

$\dfrac{3x^3-x^2-7x+6}{x+2}=3x^2-7x+7-\dfrac8{x+2}$

3) $x+2$ will be a factor of $x^3+3x^2-10x-24$ if dividing the latter by $x+2$ leaves a remainder of 0.

$x^3=x^2\cdot x$, and $x^2(x+2)=x^3+2x^2$. Subtracting this from the numerator gives a remainder of

$(x^3+3x^2-10x-24)-(x^3+2x^2)=x^2-10x-24$

$x^2=x\cdot x$, and $x(x+2)=x^2+2x$. Subtracting this from the previous remainder gives a new remainder of

$(x^2-10x-24)-(x^2+2x)=-12x-24$

$-12x=-12\cdot x$, and $-12(x+2)=-12x-24$. Subtracting this from the previous remainder gives a new remainder of

$(-12x-24)-(-12x-24)=0$

and since the remainder is 0, $x+2$ is indeed a factor of $x^3+3x^2-10x-24$.