Arithmetic with Polynomials and Rational Expressions: Rewriting Rational Expressions (CCSS.A-APR.6)

Divide $2x^3 + 3x^2 - x + 5$ by $x^2 - 1$. First term: $(2x^3)/(x^2)=2x$. Subtract $(2x^3-2x)$ to get $3x^2 + x + 5$. Next term: $(3x^2)/(x^2)=3$. Subtract $(3x^2-3)$ to get remainder $x+8$. Thus $$\frac{2x^3 + 3x^2 - x + 5}{x^2 - 1}=2x+3+\frac{x+8}{x^2-1},$$ and $\deg(x+8)=1<2$. The other choices have a remainder with too high degree, a sign error, or an arithmetic slip.

Use synthetic division with $2$: coefficients $1,-4,5,-7$ give quotient $x^2-2x+1$ and remainder $-5$. So $$\frac{x^3 - 4x^2 + 5x - 7}{x - 2}=x^2 - 2x + 1 - \frac{5}{x-2}.$$ The remainder has degree \$0<1$. Other choices show a sign error on the remainder, a remainder with degree too high, or an arithmetic slip in the quotient.

Long divide by $x^2+x$: $(3x^4)/(x^2)=3x^2$; subtract $(3x^4+3x^3)$ to get $-4x^3+2x^2-5x+4$. Next $(-4x^3)/(x^2)=-4x$; subtract $(-4x^3-4x^2)$ to get $6x^2-5x+4$. Next $(6x^2)/(x^2)=6$; subtract $(6x^2+6x)$ to get remainder $-11x+4$. Thus $$\frac{3x^4 - x^3 + 2x^2 - 5x + 4}{x^2 + x}=3x^2 - 4x + 6 + \frac{-11x + 4}{x^2 + x}.$$ The remainder's degree \$1<2$. Other options have a sign error, a quotient slip, or a remainder with too high degree.

Divide: $(4x^2)/(2x)=2x$; subtract $(4x^2+6x)$ to get $x-1$. Then $(x)/(2x)=\tfrac{1}{2}$; subtract $(x+\tfrac{3}{2})$ to get remainder $-\tfrac{5}{2}$. Thus $$\frac{4x^2 + 7x - 1}{2x + 3}=2x + \frac{1}{2} + \frac{-5/2}{2x+3},$$ with $\deg r=0<1$. Other choices have a sign error on the remainder, an incorrect quotient constant, or the wrong remainder magnitude.

Divide: $(x^4)/(x^2)=x^2$; subtract $(x^4+2x^2)$ to get $-2x^2-1$. Next $(-2x^2)/(x^2)=-2$; subtract $(-2x^2-4)$ to get remainder $3$. So $$\frac{x^4 - 1}{x^2 + 2}=x^2 - 2 + \frac{3}{x^2 + 2},$$ and $\deg(3)=0<2$. Distractors include a remainder with degree too high, a sign error, or a quotient error.

Use synthetic division with 2: coefficients $2, -3, 5, -7$. Bring down 2; $2\cdot 2=4$ add to get 1; $1\cdot 2=2$ add to get 7; $7\cdot 2=14$ add to get 7. Quotient $2x^2 + x + 7$, remainder $7$. So $\dfrac{2x^3 - 3x^2 + 5x - 7}{x - 2} = 2x^2 + x + 7 + \dfrac{7}{x-2}$. The remainder is constant, so $\deg r = 0 < \deg b = 1$.

Synthetic division with $-1$: coefficients $3, 1, -4, 6$. Bring down 3; $3(-1)=-3$ add to get $-2$; $-2(-1)=2$ add to get $-2$; $-2(-1)=2$ add to get $8$. Quotient $3x^2 - 2x - 2$, remainder $8$. Thus $\dfrac{3x^3 + x^2 - 4x + 6}{x + 1} = 3x^2 - 2x - 2 + \dfrac{8}{x+1}$ with $\deg r = 0 < \deg b = 1$.

Long division (or factoring) gives $(x^4 - 1) = (x^2 - 1)(x^2 + 1) + 0$. So $\dfrac{x^4 - 1}{x^2 - 1} = x^2 + 1$ with remainder 0, which satisfies $\deg r < \deg b$.

Divide: $4x^2/(2x)=2x$, multiply to get $4x^2+2x$, subtract to get $6x-5$. Next $6x/(2x)=3$, multiply to get $6x+3$, subtract to get $-8$. So $\dfrac{4x^2 + 8x - 5}{2x + 1} = 2x + 3 + \dfrac{-8}{2x+1} = 2x + 3 - \dfrac{8}{2x+1}$. Here $\deg r = 0 < \deg b = 1$.

Divide: $(6x^3)/(3x^2)=2x$, multiply to get $6x^3+4x$, subtract from $6x^3- x + 10$ to get $-5x + 10$. Since $\deg(-5x + 10)=1 < \deg(3x^2 + 2)=2$, we stop. So $\dfrac{6x^3 - x + 10}{3x^2 + 2} = 2x + \dfrac{-5x + 10}{3x^2 + 2}$.