Solved on Mar 01, 2024

Use remainder theorem to determine if $x=2$ is a zero of $p(x)=x^{3}+21 x^{2}+74 x-240$ . Find quotient and remainder.

STEP 1

Assumptions
1. The polynomial is given by $p(x) = x^{3} + 21x^{2} + 74x - 240$ .
2. We need to check if $x = 2$ is a zero of the polynomial.
3. To use the remainder theorem, we will evaluate $p(2)$ .
4. If $p(2) = 0$ , then $x = 2$ is a zero of the polynomial.
5. If $x = 2$ is a zero, then $(x - 2)$ is a factor of $p(x)$ , and the quotient will be a polynomial of degree 2.

STEP 2

Use the remainder theorem to evaluate $p(2)$ .
$p(2) = 2^{3} + 21 \cdot 2^{2} + 74 \cdot 2 - 240$

STEP 3

Calculate the value of $p(2)$ .
$p(2) = 8 + 21 \cdot 4 + 74 \cdot 2 - 240$

STEP 4

Continue the calculation by performing the multiplications.
$p(2) = 8 + 84 + 148 - 240$

STEP 5

Add and subtract the numbers to find the value of $p(2)$ .
$p(2) = 8 + 84 + 148 - 240 = 240 - 240 = 0$

STEP 6

Since $p(2) = 0$ , $x = 2$ is indeed a zero of the polynomial. Now we need to find the quotient by performing polynomial division of $p(x)$ by $(x - 2)$ .

STEP 7

Set up the polynomial division.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \end{array}$

STEP 8

Divide the first term of the polynomial $x^{3}$ by the first term of the divisor $x$ to find the first term of the quotient.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$

STEP 9

Multiply the divisor $x - 2$ by the first term of the quotient $x^{2}$ and subtract from the polynomial.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$
$x^{3} - 2x^{2}$

STEP 10

Perform the subtraction to find the new polynomial to divide.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$
$x^{3} + 21x^{2} - (x^{3} - 2x^{2}) = 23x^{2}$

STEP 11

Bring down the next term of the polynomial to continue the division.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$
$23x^{2} + 74x$

STEP 12

Divide the new term $23x^{2}$ by $x$ to find the next term of the quotient.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array}$

STEP 13

Multiply the divisor $x - 2$ by the new term of the quotient $23x$ and subtract from the new polynomial.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array}$
$23x^{2} + 74x - (23x^{2} - 46x) = 120x$

STEP 14

Bring down the last term of the polynomial to continue the division.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array}$
$120x - 240$

STEP 15

Divide the term $120x$ by $x$ to find the final term of the quotient.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x + 120 & \\ \end{array}$

STEP 16

Multiply the divisor $x - 2$ by the final term of the quotient $120$ and subtract from the last polynomial.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x + 120 & \\ \end{array}$
$120x - 240 - (120x - 240) = 0$

STEP 17

Since the remainder is $0$ , the quotient is $x^{2} + 23x + 120$ and the remainder is $0$ .
The correct answer is A. Yes, $x=2$ is a zero of the polynomial. The quotient is $x^{2} + 23x + 120$ , and the remainder is $0$ .

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Use remainder theorem to determine if $x=2$ is a zero of $p(x)=x^{3}+21 x^{2}+74 x-240$ . Find quotient and remainder.

STEP 1

STEP 2

Use the remainder theorem to evaluate $p(2)$ .
$p(2) = 2^{3} + 21 \cdot 2^{2} + 74 \cdot 2 - 240$

STEP 3

Calculate the value of $p(2)$ .
$p(2) = 8 + 21 \cdot 4 + 74 \cdot 2 - 240$

STEP 4

Continue the calculation by performing the multiplications.
$p(2) = 8 + 84 + 148 - 240$

STEP 5

Add and subtract the numbers to find the value of $p(2)$ .
$p(2) = 8 + 84 + 148 - 240 = 240 - 240 = 0$

STEP 6

Since $p(2) = 0$ , $x = 2$ is indeed a zero of the polynomial. Now we need to find the quotient by performing polynomial division of $p(x)$ by $(x - 2)$ .

STEP 7

Set up the polynomial division.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \end{array}$

STEP 8

Divide the first term of the polynomial $x^{3}$ by the first term of the divisor $x$ to find the first term of the quotient.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$

STEP 9

Multiply the divisor $x - 2$ by the first term of the quotient $x^{2}$ and subtract from the polynomial.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$
$x^{3} - 2x^{2}$

STEP 10

Perform the subtraction to find the new polynomial to divide.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$
$x^{3} + 21x^{2} - (x^{3} - 2x^{2}) = 23x^{2}$

STEP 11

Bring down the next term of the polynomial to continue the division.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$
$23x^{2} + 74x$

STEP 12

Divide the new term $23x^{2}$ by $x$ to find the next term of the quotient.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array}$

STEP 13

Multiply the divisor $x - 2$ by the new term of the quotient $23x$ and subtract from the new polynomial.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array}$
$23x^{2} + 74x - (23x^{2} - 46x) = 120x$

STEP 14

Bring down the last term of the polynomial to continue the division.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array}$
$120x - 240$

STEP 15

Divide the term $120x$ by $x$ to find the final term of the quotient.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x + 120 & \\ \end{array}$

STEP 16

Multiply the divisor $x - 2$ by the final term of the quotient $120$ and subtract from the last polynomial.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x + 120 & \\ \end{array}$
$120x - 240 - (120x - 240) = 0$

STEP 17

Since the remainder is $0$ , the quotient is $x^{2} + 23x + 120$ and the remainder is $0$ .
The correct answer is A. Yes, $x=2$ is a zero of the polynomial. The quotient is $x^{2} + 23x + 120$ , and the remainder is $0$ .

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Use remainder theorem to determine if x=2x=2x=2 is a zero of p(x)=x3+21x2+74x−240p(x)=x^{3}+21 x^{2}+74 x-240p(x)=x3+21x2+74x−240. Find quotient and remainder.

STEP 1

STEP 2

Use the remainder theorem to evaluate p(2) p(2) p(2).p(2)=23+21⋅22+74⋅2−240 p(2) = 2^{3} + 21 \cdot 2^{2} + 74 \cdot 2 - 240 p(2)=23+21⋅22+74⋅2−240

STEP 3

Calculate the value of p(2) p(2) p(2).p(2)=8+21⋅4+74⋅2−240 p(2) = 8 + 21 \cdot 4 + 74 \cdot 2 - 240 p(2)=8+21⋅4+74⋅2−240

STEP 4

Continue the calculation by performing the multiplications.p(2)=8+84+148−240 p(2) = 8 + 84 + 148 - 240 p(2)=8+84+148−240

STEP 5

Add and subtract the numbers to find the value of p(2) p(2) p(2).p(2)=8+84+148−240=240−240=0 p(2) = 8 + 84 + 148 - 240 = 240 - 240 = 0 p(2)=8+84+148−240=240−240=0

STEP 6

Since p(2)=0 p(2) = 0 p(2)=0, x=2 x = 2 x=2 is indeed a zero of the polynomial. Now we need to find the quotient by performing polynomial division of p(x) p(x) p(x) by (x−2) (x - 2) (x−2).

STEP 7

Set up the polynomial division.x−2x3+21x2+74x−240 \begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \end{array} x−2​x3+21x2+74x−240​

STEP 8

Divide the first term of the polynomial x3 x^{3} x3 by the first term of the divisor x x x to find the first term of the quotient.x−2x3+21x2+74x−240x2 \begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array} x−2x2​x3+21x2+74x−240​​

STEP 9

Multiply the divisor x−2 x - 2 x−2 by the first term of the quotient x2 x^{2} x2 and subtract from the polynomial.x−2x3+21x2+74x−240x2 \begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array} x−2x2​x3+21x2+74x−240​​x3−2x2 x^{3} - 2x^{2} x3−2x2

STEP 10

STEP 11

Bring down the next term of the polynomial to continue the division.x−2x3+21x2+74x−240x2 \begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array} x−2x2​x3+21x2+74x−240​​23x2+74x 23x^{2} + 74x 23x2+74x

STEP 12

Divide the new term 23x2 23x^{2} 23x2 by x x x to find the next term of the quotient.x−2x3+21x2+74x−240x2+23x \begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array} x−2x2+23x​x3+21x2+74x−240​​

STEP 13

STEP 14

Bring down the last term of the polynomial to continue the division.x−2x3+21x2+74x−240x2+23x \begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array} x−2x2+23x​x3+21x2+74x−240​​120x−240 120x - 240 120x−240

STEP 15

Divide the term 120x 120x 120x by x x x to find the final term of the quotient.x−2x3+21x2+74x−240x2+23x+120 \begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x + 120 & \\ \end{array} x−2x2+23x+120​x3+21x2+74x−240​​

STEP 16

STEP 17

Since the remainder is 0 0 0, the quotient is x2+23x+120 x^{2} + 23x + 120 x2+23x+120 and the remainder is 0 0 0.The correct answer is A. Yes, x=2 x=2 x=2 is a zero of the polynomial. The quotient is x2+23x+120 x^{2} + 23x + 120 x2+23x+120, and the remainder is 0 0 0.

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Use remainder theorem to determine if $x=2$ is a zero of $p(x)=x^{3}+21 x^{2}+74 x-240$ . Find quotient and remainder.

Use the remainder theorem to evaluate $p(2)$ .
$p(2) = 2^{3} + 21 \cdot 2^{2} + 74 \cdot 2 - 240$

Calculate the value of $p(2)$ .
$p(2) = 8 + 21 \cdot 4 + 74 \cdot 2 - 240$

Continue the calculation by performing the multiplications.
$p(2) = 8 + 84 + 148 - 240$

Add and subtract the numbers to find the value of $p(2)$ .
$p(2) = 8 + 84 + 148 - 240 = 240 - 240 = 0$

Since $p(2) = 0$ , $x = 2$ is indeed a zero of the polynomial. Now we need to find the quotient by performing polynomial division of $p(x)$ by $(x - 2)$ .

Set up the polynomial division.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \end{array}$

Divide the first term of the polynomial $x^{3}$ by the first term of the divisor $x$ to find the first term of the quotient.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$

Multiply the divisor $x - 2$ by the first term of the quotient $x^{2}$ and subtract from the polynomial.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$
$x^{3} - 2x^{2}$

Bring down the next term of the polynomial to continue the division.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} & \\ \end{array}$
$23x^{2} + 74x$

Divide the new term $23x^{2}$ by $x$ to find the next term of the quotient.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array}$

Bring down the last term of the polynomial to continue the division.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x & \\ \end{array}$
$120x - 240$

Divide the term $120x$ by $x$ to find the final term of the quotient.
$\begin{array}{r|l} x - 2 & x^{3} + 21x^{2} + 74x - 240 \\ \hline x^{2} + 23x + 120 & \\ \end{array}$

Since the remainder is $0$ , the quotient is $x^{2} + 23x + 120$ and the remainder is $0$ .
The correct answer is A. Yes, $x=2$ is a zero of the polynomial. The quotient is $x^{2} + 23x + 120$ , and the remainder is $0$ .