Question

find the quotient for each equation. remember to rearrange the terms in descending order. write any remainders as fractions.

1. $\frac{15a^{2}-11a-14}{3a+2} = $
2. $\frac{8x^{3}-27}{2x-3} = $
3. $\frac{-a^{3}-4a+4}{a+2} = $
4. $\frac{-4a+1+8a^{3}}{2a-1} = $
5. $\frac{-2a^{3}+2a+3}{1-a} = $
6. $\frac{1-81x^{4}}{1+3x} = $
7. $\frac{x^{4}-4x^{3}+2x^{2}+4x+1}{x^{2}-2x-1} = $
8. $\frac{a^{3}-7a+6}{a+3} = $

Explanation:

1. Problem 1:

Step1: Divide leading terms

$\frac{15a^2}{3a} = 5a$

Step2: Multiply divisor by result

$5a(3a+2) = 15a^2 + 10a$

Step3: Subtract from dividend

$(15a^2-11a-14)-(15a^2+10a) = -21a-14$

Step4: Divide new leading terms

$\frac{-21a}{3a} = -7$

Step5: Multiply divisor by result

$-7(3a+2) = -21a-14$

Step6: Subtract to find remainder

$(-21a-14)-(-21a-14) = 0$

2. Problem 2:

Step1: Recognize difference of cubes

$8x^3-27=(2x)^3-3^3$

Step2: Apply difference of cubes formula

$(2x-3)(4x^2+6x+9)$

Step3: Cancel common factor

$\frac{(2x-3)(4x^2+6x+9)}{2x-3}=4x^2+6x+9$

3. Problem 3:

Step1: Rearrange dividend terms

$-a^3 + 0a^2 -4a +4$

Step2: Divide leading terms

$\frac{-a^3}{a} = -a^2$

Step3: Multiply divisor by result

$-a^2(a+2) = -a^3-2a^2$

Step4: Subtract from dividend

$(-a^3+0a^2-4a+4)-(-a^3-2a^2)=2a^2-4a+4$

Step5: Divide new leading terms

$\frac{2a^2}{a}=2a$

Step6: Multiply divisor by result

$2a(a+2)=2a^2+4a$

Step7: Subtract from current polynomial

$(2a^2-4a+4)-(2a^2+4a)=-8a+4$

Step8: Divide new leading terms

$\frac{-8a}{a}=-8$

Step9: Multiply divisor by result

$-8(a+2)=-8a-16$

Step10: Subtract to find remainder

$(-8a+4)-(-8a-16)=20$

4. Problem 4:

Step1: Rearrange dividend terms

$8a^3 + 0a^2 -4a +1$

Step2: Divide leading terms

$\frac{8a^3}{2a}=4a^2$

Step3: Multiply divisor by result

$4a^2(2a-1)=8a^3-4a^2$

Step4: Subtract from dividend

$(8a^3+0a^2-4a+1)-(8a^3-4a^2)=4a^2-4a+1$

Step5: Divide new leading terms

$\frac{4a^2}{2a}=2a$

Step6: Multiply divisor by result

$2a(2a-1)=4a^2-2a$

Step7: Subtract from current polynomial

$(4a^2-4a+1)-(4a^2-2a)=-2a+1$

Step8: Divide new leading terms

$\frac{-2a}{2a}=-1$

Step9: Multiply divisor by result

$-1(2a-1)=-2a+1$

Step10: Subtract to find remainder

$(-2a+1)-(-2a+1)=0$

5. Problem 5:

Step1: Rearrange dividend terms

$-2a^3 + 0a^2 +2a +3$

Step2: Rewrite divisor

$1-a=-(a-1)$

Step3: Divide leading terms

$\frac{-2a^3}{-a}=2a^2$

Step4: Multiply divisor by result

$2a^2(1-a)=2a^2-2a^3$

Step5: Subtract from dividend

$(-2a^3+0a^2+2a+3)-(2a^2-2a^3)=-2a^2+2a+3$

Step6: Divide new leading terms

$\frac{-2a^2}{-a}=2a$

Step7: Multiply divisor by result

$2a(1-a)=2a-2a^2$

Step8: Subtract from current polynomial

$(-2a^2+2a+3)-(2a-2a^2)=3$

6. Problem 6:

Step1: Recognize difference of squares

$1-81x^4=1^2-(9x^2)^2$

Step2: Factor first difference of squares

$(1-9x^2)(1+9x^2)$

Step3: Factor second difference of squares

$(1-3x)(1+3x)(1+9x^2)$

Step4: Cancel common factor

$\frac{(1-3x)(1+3x)(1+9x^2)}{1+3x}=(1-3x)(1+9x^2)$

Step5: Expand the product

$1+9x^2-3x-27x^3=-27x^3+9x^2-3x+1$

7. Problem 7:

Step1: Divide leading terms

$\frac{x^4}{x^2}=x^2$

Step2: Multiply divisor by result

$x^2(x^2-2x-1)=x^4-2x^3-x^2$

Step3: Subtract from dividend

$(x^4-4x^3+2x^2+4x+1)-(x^4-2x^3-x^2)=-2x^3+3x^2+4x+1$

Step4: Divide new leading terms

$\frac{-2x^3}{x^2}=-2x$

Step5: Multiply divisor by result

$-2x(x^2-2x-1)=-2x^3+4x^2+2x$

Step6: Subtract from current polynomial

$(-2x^3+3x^2+4x+1)-(-2x^3+4x^2+2x)=-x^2+2x+1$

Step7: Divide new leading terms

$\frac{-x^2}{x^2}=-1$

Step8: Multiply divisor by result

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

Step9: Subtract to find remainder

$(-x^2+2x+1)-(-x^2+2x+1)=0$

8. Problem 8:

Step1: Factor the dividend

$a^3-7a+6=(a-1)(a-2)(a+3)$

Step2: Cancel common factor

$\frac{(a-1)(a-2)(a+3)}{a+3}=(a-1)(a-2)$

Step3: Expand the product

$a^2-3a+2$

Answer:

1. $5a - 7$
2. $4x^2 + 6x + 9$
3. $-a^2 + 2a - 8 + \frac{20}{a+2}$
4. $4a^2 + 2a - 1$
5. $2a^2 + 2a + \frac{3}{1-a}$
6. $-27x^3 + 9x^2 - 3x + 1$
7. $x^2 - 2x - 1$
8. $a^2 - 3a + 2$