Solved on Jan 16, 2024

Use long division to show $a-5$ is a factor of $a^3 - 5a^2 - 6a + 30$ .

STEP 1

Assumptions
1. We are given a polynomial $a^{3}-5 a^{2}-6 a+30$ .
2. We need to use long division to show that $a-5$ is a factor of the given polynomial.

STEP 2

To show that $a-5$ is a factor of the polynomial $a^{3}-5 a^{2}-6 a+30$ , we will perform polynomial long division, dividing the given polynomial by $a-5$ .

STEP 3

Set up the long division by writing $a-5$ as the divisor and $a^{3}-5 a^{2}-6 a+30$ as the dividend.

STEP 4

Determine how many times $a-5$ goes into the first term of the dividend, $a^{3}$ . This is done by dividing the leading term of the dividend, $a^{3}$ , by the leading term of the divisor, $a$ . The result is $a^{2}$ .

STEP 5

Multiply the entire divisor $a-5$ by the result from the previous step, $a^{2}$ , and write this product underneath the first two terms of the dividend.
$a^{2}(a-5) = a^{3} - 5a^{2}$

STEP 6

Subtract the product from the first two terms of the dividend. To do this, change the signs of the product and add it to the dividend.
$\begin{array}{r} a^{3} - 5a^{2} \\ -(a^{3} - 5a^{2}) \\ \hline 0a^{2} \end{array}$

STEP 7

Bring down the next term of the dividend, which is $-6a$ , to the remainder from the previous step.
$0a^{2} - 6a$

STEP 8

Repeat the process: Determine how many times $a-5$ goes into the new term $-6a$ . Divide the leading term of the new dividend, $-6a$ , by the leading term of the divisor, $a$ . The result is $-6$ .

STEP 9

Multiply the entire divisor $a-5$ by $-6$ and write this product underneath the corresponding terms of the dividend.
$-6(a-5) = -6a + 30$

STEP 10

Subtract this product from the corresponding terms of the dividend. Change the signs of the product and add it to the dividend.
$\begin{array}{r} -6a \\ -(-6a + 30) \\ \hline -30 \end{array}$

STEP 11

Bring down the last term of the dividend, which is $+30$ , to the remainder from the previous step.
$-30 + 30$

STEP 12

Add the terms to find the new remainder.
$-30 + 30 = 0$

STEP 13

Since the remainder is $0$ , this confirms that $a-5$ is a factor of the polynomial $a^{3}-5 a^{2}-6 a+30$ .
The long division process shows that the polynomial can be divided evenly by $a-5$ , with no remainder, thus verifying that $a-5$ is indeed a factor of the polynomial.

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Use long division to show $a-5$ is a factor of $a^3 - 5a^2 - 6a + 30$ .

STEP 1

Assumptions
1. We are given a polynomial $a^{3}-5 a^{2}-6 a+30$ .
2. We need to use long division to show that $a-5$ is a factor of the given polynomial.

STEP 2

To show that $a-5$ is a factor of the polynomial $a^{3}-5 a^{2}-6 a+30$ , we will perform polynomial long division, dividing the given polynomial by $a-5$ .

STEP 3

Set up the long division by writing $a-5$ as the divisor and $a^{3}-5 a^{2}-6 a+30$ as the dividend.

STEP 4

Determine how many times $a-5$ goes into the first term of the dividend, $a^{3}$ . This is done by dividing the leading term of the dividend, $a^{3}$ , by the leading term of the divisor, $a$ . The result is $a^{2}$ .

STEP 5

Multiply the entire divisor $a-5$ by the result from the previous step, $a^{2}$ , and write this product underneath the first two terms of the dividend.
$a^{2}(a-5) = a^{3} - 5a^{2}$

STEP 6

Subtract the product from the first two terms of the dividend. To do this, change the signs of the product and add it to the dividend.
$\begin{array}{r} a^{3} - 5a^{2} \\ -(a^{3} - 5a^{2}) \\ \hline 0a^{2} \end{array}$

STEP 7

Bring down the next term of the dividend, which is $-6a$ , to the remainder from the previous step.
$0a^{2} - 6a$

STEP 8

Repeat the process: Determine how many times $a-5$ goes into the new term $-6a$ . Divide the leading term of the new dividend, $-6a$ , by the leading term of the divisor, $a$ . The result is $-6$ .

STEP 9

Multiply the entire divisor $a-5$ by $-6$ and write this product underneath the corresponding terms of the dividend.
$-6(a-5) = -6a + 30$

STEP 10

Subtract this product from the corresponding terms of the dividend. Change the signs of the product and add it to the dividend.
$\begin{array}{r} -6a \\ -(-6a + 30) \\ \hline -30 \end{array}$

STEP 11

Bring down the last term of the dividend, which is $+30$ , to the remainder from the previous step.
$-30 + 30$

STEP 12

Add the terms to find the new remainder.
$-30 + 30 = 0$

STEP 13

Since the remainder is $0$ , this confirms that $a-5$ is a factor of the polynomial $a^{3}-5 a^{2}-6 a+30$ .
The long division process shows that the polynomial can be divided evenly by $a-5$ , with no remainder, thus verifying that $a-5$ is indeed a factor of the polynomial.

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Use long division to show a−5a-5a−5 is a factor of a3−5a2−6a+30a^3 - 5a^2 - 6a + 30a3−5a2−6a+30.

STEP 1

Assumptions1. We are given a polynomial a3−5a2−6a+30a^{3}-5 a^{2}-6 a+30a3−5a2−6a+30.2. We need to use long division to show that a−5a-5a−5 is a factor of the given polynomial.

STEP 2

To show that a−5a-5a−5 is a factor of the polynomial a3−5a2−6a+30a^{3}-5 a^{2}-6 a+30a3−5a2−6a+30, we will perform polynomial long division, dividing the given polynomial by a−5a-5a−5.

STEP 3

Set up the long division by writing a−5a-5a−5 as the divisor and a3−5a2−6a+30a^{3}-5 a^{2}-6 a+30a3−5a2−6a+30 as the dividend.

STEP 4

Determine how many times a−5a-5a−5 goes into the first term of the dividend, a3a^{3}a3. This is done by dividing the leading term of the dividend, a3a^{3}a3, by the leading term of the divisor, aaa. The result is a2a^{2}a2.

STEP 5

Multiply the entire divisor a−5a-5a−5 by the result from the previous step, a2a^{2}a2, and write this product underneath the first two terms of the dividend.a2(a−5)=a3−5a2a^{2}(a-5) = a^{3} - 5a^{2}a2(a−5)=a3−5a2

STEP 6

Subtract the product from the first two terms of the dividend. To do this, change the signs of the product and add it to the dividend.a3−5a2−(a3−5a2)0a2 \begin{array}{r} a^{3} - 5a^{2} \\ -(a^{3} - 5a^{2}) \\ \hline 0a^{2} \end{array} a3−5a2−(a3−5a2)0a2​​

STEP 7

Bring down the next term of the dividend, which is −6a-6a−6a, to the remainder from the previous step.0a2−6a0a^{2} - 6a0a2−6a

STEP 8

Repeat the process: Determine how many times a−5a-5a−5 goes into the new term −6a-6a−6a. Divide the leading term of the new dividend, −6a-6a−6a, by the leading term of the divisor, aaa. The result is −6-6−6.

STEP 9

Multiply the entire divisor a−5a-5a−5 by −6-6−6 and write this product underneath the corresponding terms of the dividend.−6(a−5)=−6a+30-6(a-5) = -6a + 30−6(a−5)=−6a+30

STEP 10

Subtract this product from the corresponding terms of the dividend. Change the signs of the product and add it to the dividend.−6a−(−6a+30)−30 \begin{array}{r} -6a \\ -(-6a + 30) \\ \hline -30 \end{array} −6a−(−6a+30)−30​​

STEP 11

Bring down the last term of the dividend, which is +30+30+30, to the remainder from the previous step.−30+30-30 + 30−30+30

STEP 12

Add the terms to find the new remainder.−30+30=0-30 + 30 = 0−30+30=0

STEP 13

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Use long division to show $a-5$ is a factor of $a^3 - 5a^2 - 6a + 30$ .

Assumptions
1. We are given a polynomial $a^{3}-5 a^{2}-6 a+30$ .
2. We need to use long division to show that $a-5$ is a factor of the given polynomial.

To show that $a-5$ is a factor of the polynomial $a^{3}-5 a^{2}-6 a+30$ , we will perform polynomial long division, dividing the given polynomial by $a-5$ .

Set up the long division by writing $a-5$ as the divisor and $a^{3}-5 a^{2}-6 a+30$ as the dividend.

Determine how many times $a-5$ goes into the first term of the dividend, $a^{3}$ . This is done by dividing the leading term of the dividend, $a^{3}$ , by the leading term of the divisor, $a$ . The result is $a^{2}$ .

Multiply the entire divisor $a-5$ by the result from the previous step, $a^{2}$ , and write this product underneath the first two terms of the dividend.
$a^{2}(a-5) = a^{3} - 5a^{2}$

Subtract the product from the first two terms of the dividend. To do this, change the signs of the product and add it to the dividend.
$\begin{array}{r} a^{3} - 5a^{2} \\ -(a^{3} - 5a^{2}) \\ \hline 0a^{2} \end{array}$

Bring down the next term of the dividend, which is $-6a$ , to the remainder from the previous step.
$0a^{2} - 6a$

Repeat the process: Determine how many times $a-5$ goes into the new term $-6a$ . Divide the leading term of the new dividend, $-6a$ , by the leading term of the divisor, $a$ . The result is $-6$ .

Multiply the entire divisor $a-5$ by $-6$ and write this product underneath the corresponding terms of the dividend.
$-6(a-5) = -6a + 30$

Subtract this product from the corresponding terms of the dividend. Change the signs of the product and add it to the dividend.
$\begin{array}{r} -6a \\ -(-6a + 30) \\ \hline -30 \end{array}$

Bring down the last term of the dividend, which is $+30$ , to the remainder from the previous step.
$-30 + 30$

Add the terms to find the new remainder.
$-30 + 30 = 0$