Solved on Feb 24, 2024

Divide polynomials using long division: a. (k3k2k2)÷(k2)\left(k^{3}-k^{2}-k-2\right) \div(k-2) b. (6a3+20a215a+9)÷(a+4)\left(6 a^{3}+20 a^{2}-15 a+9\right) \div(a+4) c. (n4+10n3+21n2+6n8)÷(n+2)\left(n^{4}+10 n^{3}+21 n^{2}+6 n-8\right) \div(n+2) d. (8v5+32v4+5v+20)÷(v+4)\left(8 v^{5}+32 v^{4}+5 v+20\right) \div(v+4)

STEP 1

Assumptions
1. We are performing polynomial long division.
2. There is no remainder unless explicitly found during the division process.
3. The divisor is a binomial in each case.
4. The dividend is a polynomial of higher degree than the divisor in each case.

STEP 2

Start with the first polynomial division (k3k2k2)÷(k2)\left(k^{3}-k^{2}-k-2\right) \div(k-2).

STEP 3

Divide the first term of the dividend k3k^3 by the first term of the divisor kk to get the first term of the quotient.
k3k=k2\frac{k^3}{k} = k^2

STEP 4

Multiply the divisor (k2)(k-2) by the first term of the quotient k2k^2 and subtract it from the dividend.
k3k2(k2)=k3k3+2k2=2k2k^3 - k^2(k-2) = k^3 - k^3 + 2k^2 = 2k^2

STEP 5

Bring down the next term of the dividend to get a new sub-dividend.
2k2k2k^2 - k

STEP 6

Divide the first term of the new sub-dividend 2k22k^2 by the first term of the divisor kk to get the next term of the quotient.
2k2k=2k\frac{2k^2}{k} = 2k

STEP 7

Multiply the divisor (k2)(k-2) by the new term of the quotient 2k2k and subtract it from the new sub-dividend.
2k2k2k(k2)=2k22k2+4kk=3k2k^2 - k - 2k(k-2) = 2k^2 - 2k^2 + 4k - k = 3k

STEP 8

Bring down the next term of the dividend to get the next new sub-dividend.
3k23k - 2

STEP 9

Divide the first term of the next new sub-dividend 3k3k by the first term of the divisor kk to get the next term of the quotient.
3kk=3\frac{3k}{k} = 3

STEP 10

Multiply the divisor (k2)(k-2) by the new term of the quotient 33 and subtract it from the next new sub-dividend.
3k23(k2)=3k3k+62=43k - 2 - 3(k-2) = 3k - 3k + 6 - 2 = 4

STEP 11

Since the degree of the remainder 44 is less than the degree of the divisor (k2)(k-2), we stop the division process. The remainder is 44.

STEP 12

Write the complete quotient for the first division with the remainder.
k3k2k2k2=k2+2k+3+4k2\frac{k^{3}-k^{2}-k-2}{k-2} = k^2 + 2k + 3 + \frac{4}{k-2}

STEP 13

Move on to the second polynomial division (6a3+20a215a+9)÷(a+4)\left(6 a^{3}+20 a^{2}-15 a+9\right) \div(a+4).

STEP 14

Divide the first term of the dividend 6a36a^3 by the first term of the divisor aa to get the first term of the quotient.
6a3a=6a2\frac{6a^3}{a} = 6a^2

STEP 15

Multiply the divisor (a+4)(a+4) by the first term of the quotient 6a26a^2 and subtract it from the dividend.
6a3+20a26a2(a+4)=6a36a3+24a220a2=4a26a^3 + 20a^2 - 6a^2(a+4) = 6a^3 - 6a^3 + 24a^2 - 20a^2 = 4a^2

STEP 16

Bring down the next term of the dividend to get a new sub-dividend.
4a215a4a^2 - 15a

STEP 17

Divide the first term of the new sub-dividend 4a24a^2 by the first term of the divisor aa to get the next term of the quotient.
4a2a=4a\frac{4a^2}{a} = 4a

STEP 18

Multiply the divisor (a+4)(a+4) by the new term of the quotient 4a4a and subtract it from the new sub-dividend.
4a215a4a(a+4)=4a24a216a+15a=a4a^2 - 15a - 4a(a+4) = 4a^2 - 4a^2 - 16a + 15a = -a

STEP 19

Bring down the next term of the dividend to get the next new sub-dividend.
a+9-a + 9

STEP 20

Divide the first term of the next new sub-dividend a-a by the first term of the divisor aa to get the next term of the quotient.
aa=1\frac{-a}{a} = -1

STEP 21

Multiply the divisor (a+4)(a+4) by the new term of the quotient 1-1 and subtract it from the next new sub-dividend.
a+9(1)(a+4)=a+a4+9=5-a + 9 - (-1)(a+4) = -a + a - 4 + 9 = 5

STEP 22

Since the degree of the remainder 55 is less than the degree of the divisor (a+4)(a+4), we stop the division process. The remainder is 55.

STEP 23

Write the complete quotient for the second division with the remainder.
6a3+20a215a+9a+4=6a2+4a1+5a+4\frac{6 a^{3}+20 a^{2}-15 a+9}{a+4} = 6a^2 + 4a - 1 + \frac{5}{a+4}

STEP 24

Move on to the third polynomial division (n4+10n3+21n2+6n8)÷(n+2)\left(n^{4}+10 n^{3}+21 n^{2}+6 n-8\right) \div(n+2).

STEP 25

Repeat the long division process for the third polynomial as done in previous steps, dividing the terms one by one and subtracting the products from the sub-dividends until the degree of the remainder is less than the degree of the divisor.

STEP 26

Write the complete quotient for the third division.
n4+10n3+21n2+6n8n+2=n3+8n2+5n4\frac{n^{4}+10 n^{3}+21 n^{2}+6 n-8}{n+2} = n^3 + 8n^2 + 5n - 4

STEP 27

Move on to the fourth polynomial division (8v5+32v4+5v+20)÷(v+4)\left(8 v^{5}+32 v^{4}+5 v+20\right) \div(v+4).

STEP 28

Repeat the long division process for the fourth polynomial as done in previous steps, dividing the terms one by one and subtracting the products from the sub-dividends until the degree of the remainder is less than the degree of the divisor.

STEP 29

Write the complete quotient for the fourth division.
8v5+32v4+5v+20v+4=8v42v+1+4v+4\frac{8 v^{5}+32 v^{4}+5 v+20}{v+4} = 8v^4 - 2v + 1 + \frac{4}{v+4}
The solutions to the polynomial divisions are:
a. k3k2k2k2=k2+2k+3+4k2\frac{k^{3}-k^{2}-k-2}{k-2} = k^2 + 2k + 3 + \frac{4}{k-2}
b. 6a3+20a215a+9a+4=6a2+4a1+5a+4\frac{6 a^{3}+20 a^{2}-15 a+9}{a+4} = 6a^2 + 4a - 1 + \frac{5}{a+4}
c. n4+10n3+21n2+6n8n+2=n3+8n2+5n4\frac{n^{4}+10 n^{3}+21 n^{2}+6 n-8}{n+2} = n^3 + 8n^2 + 5n - 4
d. 8v5+32v4+5v+20v+4=8v42v+1+4v+4\frac{8 v^{5}+32 v^{4}+5 v+20}{v+4} = 8v^4 - 2v + 1 + \frac{4}{v+4}

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