Question

Consider the following division of polynomials.

Answer 1

Answer:

Step-by-step explanation:

A) Long division :

x²+2x+8) x⁴+ x³+ 7x²-6x+8 ( x²-x+1

               x⁴+2x³+8x²

                        -x³ -x² -6x

                        -x³ -x² -8x

                              x² +2x + 8

                              x² +2x + 8

                                   NIL      

B)  By multiplying (x² + 2x + 8) and quotient (x²-x+1) We should get the expression (x⁴+x³+7x²-6x+8) as of numerator.

(x²+2x+8) × (x²-x+1)

= x²(x²-x+1) + 2x(x²-x+1)+8(x²-x+1)

= x⁴-x³+x²+2x³-2x²+2x+8x²-8x+8

= (x⁴+x³+7x²-6x+8)

Hence proved.

Answer 2

$x^4=x^2\cdot x^2$. Multiplying the denominator by $x^2$ gives

$x^2(x^2+2x+8)=x^4+2x^3+8x^2$

Subtracting this from the numerator gives a remainder of

$(x^4+x^3+7x^2-6x+8)-(x^4+2x^3+8x^2)=-x^3-x^2-6x+8$

$-x^3=-x\cdot x^2$. Multiplying the denominator by $-x$ gives

$-x(x^2+2x+8)=-x^3-2x^2-8x$

and subtracting this from the previous remainder gives a new remainder of

$(-x^3-x^2-6x+8)-(-x^3-2x^2-8x)=x^2+2x+8$

This last remainder is exactly the same as the denominator, so $x^2+2x+8$ divides through it exactly and leaves us with 1.

What we showed here is that

$\dfrac{x^4+x^3+7x^2-6x+8}{x^2+2x+8}=x^2-\dfrac{x^3+x^2+6x-8}{x^2+2x+8}$

$=x^2-x+\dfrac{x^2+2x+8}{x^2+2x+8}$

$=x^2-x+1$

and this last expression is the quotient.

To verify this solution, we can simply multiply this by the original denominator:

$(x^2+2x+8)(x^2-x+1)=x^2(x^2-x+1)+2x(x^2-x+1)+8(x^2-x+1)$

$=(x^4-x^3+x^2)+(2x^3-2x^2+2x)+(8x^2-8x+8)$

$=x^4+x^3+7x^2-6x+8$

which matches the original numerator.