Question

names: ______
3.5.2 practice problems
period: ____
find each quotient using long division.

1. $(k^3 - 10k^2 + k + 1) \div (k - 1)$
2. $(x^2 - 16x + 63) \div (x - 9)$
3. $(y^4 - y^2 + 8y + 5) \div (y + 1)$
4. $(3a^4 + 4a^3 + 15a + 10) \div (3a + 4)$

Explanation:

Problem 1: $(k^3 -10k^2 +k +1)\div(k-1)$

Step1: Divide leading terms

$\frac{k^3}{k}=k^2$

Step2: Multiply divisor by $k^2$

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

Step3: Subtract from dividend

$(k^3 -10k^2 +k +1)-(k^3 -k^2)=-9k^2 +k +1$

Step4: Divide new leading term

$\frac{-9k^2}{k}=-9k$

Step5: Multiply divisor by $-9k$

$-9k(k-1)=-9k^2 +9k$

Step6: Subtract the result

$(-9k^2 +k +1)-(-9k^2 +9k)=-8k +1$

Step7: Divide new leading term

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

Step8: Multiply divisor by $-8$

$-8(k-1)=-8k +8$

Step9: Subtract to get remainder

$(-8k +1)-(-8k +8)=-7$

Problem 2: $(x^2 -16x +63)\div(x-9)$

Step1: Divide leading terms

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

Step2: Multiply divisor by $x$

$x(x-9)=x^2 -9x$

Step3: Subtract from dividend

$(x^2 -16x +63)-(x^2 -9x)=-7x +63$

Step4: Divide new leading term

$\frac{-7x}{x}=-7$

Step5: Multiply divisor by $-7$

$-7(x-9)=-7x +63$

Step6: Subtract to get remainder

$(-7x +63)-(-7x +63)=0$

Problem 3: $(y^4 -y^2 +8y +5)\div(y+1)$

Step1: Divide leading terms

$\frac{y^4}{y}=y^3$

Step2: Multiply divisor by $y^3$

$y^3(y+1)=y^4 +y^3$

Step3: Subtract from dividend

$(y^4 -y^2 +8y +5)-(y^4 +y^3)=-y^3 -y^2 +8y +5$

Step4: Divide new leading term

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

Step5: Multiply divisor by $-y^2$

$-y^2(y+1)=-y^3 -y^2$

Step6: Subtract the result

$(-y^3 -y^2 +8y +5)-(-y^3 -y^2)=8y +5$

Step7: Divide new leading term

$\frac{8y}{y}=8$

Step8: Multiply divisor by $8$

$8(y+1)=8y +8$

Step9: Subtract to get remainder

$(8y +5)-(8y +8)=-3$

Problem 4: $(3a^4 +4a^3 +15a +10)\div(3a+4)$

Step1: Divide leading terms

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

Step2: Multiply divisor by $a^3$

$a^3(3a+4)=3a^4 +4a^3$

Step3: Subtract from dividend

$(3a^4 +4a^3 +15a +10)-(3a^4 +4a^3)=15a +10$

Step4: Divide new leading term

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

Step5: Multiply divisor by $5$

$5(3a+4)=15a +20$

Step6: Subtract to get remainder

$(15a +10)-(15a +20)=-10$

Answer:

1. $k^2 -9k -8 -\frac{7}{k-1}$
2. $x -7$
3. $y^3 -y^2 +8 -\frac{3}{y+1}$
4. $a^3 +5 -\frac{10}{3a+4}$