Answer :
Sure! Let's solve these questions step-by-step using synthetic division to determine if the given value [tex]\( k \)[/tex] is a zero of the polynomial, or if not, find the value of the polynomial at [tex]\( k \)[/tex].
### Question 11:
Polynomial: [tex]\( p(x) = 6x^5 - 23x^4 - 95x^3 + 70x^2 + 204x - 72 \)[/tex]
Given [tex]\( k \)[/tex]: [tex]\( k = 1 \)[/tex]
Synthetic Division Steps:
1. Write down the coefficients: [tex]\( [6, -23, -95, 70, 204, -72] \)[/tex].
2. Start synthetic division with [tex]\( k = 1 \)[/tex]:
- Bring down the leading coefficient: [tex]\( 6 \)[/tex].
- Multiply [tex]\( 6 \)[/tex] by [tex]\( 1 \)[/tex] and add to the next coefficient [tex]\(-23\)[/tex]: [tex]\( 6 \times 1 = 6 \)[/tex], [tex]\(-23 + 6 = -17\)[/tex].
- Multiply [tex]\(-17\)[/tex] by [tex]\( 1 \)[/tex] and add to [tex]\(-95\)[/tex]: [tex]\(-17 \times 1 = -17\)[/tex], [tex]\(-95 + (-17) = -112\)[/tex].
- Multiply [tex]\(-112\)[/tex] by [tex]\( 1 \)[/tex] and add to [tex]\(70\)[/tex]: [tex]\(-112 \times 1 = -112\)[/tex], [tex]\(70 + (-112) = -42\)[/tex].
- Multiply [tex]\(-42\)[/tex] by [tex]\( 1 \)[/tex] and add to [tex]\(204\)[/tex]: [tex]\(-42 \times 1 = -42\)[/tex], [tex]\(204 + (-42) = 162\)[/tex].
- Multiply [tex]\(162\)[/tex] by [tex]\( 1 \)[/tex] and add to [tex]\(-72\)[/tex]: [tex]\(162 \times 1 = 162\)[/tex], [tex]\(-72 + 162 = 90\)[/tex].
3. The remainder, [tex]\( p(1) \)[/tex], is [tex]\( 90 \)[/tex].
Since the remainder is not zero, [tex]\( k = 1 \)[/tex] is not a zero of the polynomial. [tex]\( p(1) = 90 \)[/tex].
### Question 12:
Polynomial: [tex]\( p(x) = 48x^4 + 10x^3 - 51x^2 - 10x + 3 \)[/tex]
Given [tex]\( k \)[/tex]: [tex]\( k = \frac{1}{6} \)[/tex]
Synthetic Division Steps:
Using the process similar to question 11, proceed with:
- Write the coefficients: [tex]\( [48, 10, -51, -10, 3] \)[/tex].
Continue with synthetic division using each step:
1. Calculate the intermediate results as per the division process.
Finally:
- Since the remainder is [tex]\( 0 \)[/tex], [tex]\( k = \frac{1}{6} \)[/tex] is indeed a zero of the polynomial.
### Question 13:
Polynomial: [tex]\( p(x) = 18x^5 - 87x^4 + 110x^3 - 28x^2 - 16x + 3 \)[/tex]
Given [tex]\( k \)[/tex]: [tex]\( k = \frac{2}{3} \)[/tex]
Synthetic Division Steps:
1. Write the coefficients: [tex]\( [18, -87, 110, -28, -16, 3] \)[/tex].
Carry out the synthetic division process:
1. Perform each step using the value [tex]\( \frac{2}{3} \)[/tex] to multiply and add.
After completing these steps:
- The remainder, [tex]\( p\left(\frac{2}{3}\right) \)[/tex], is approximately [tex]\(-2.33\)[/tex].
Since the remainder is not zero, [tex]\( k = \frac{2}{3} \)[/tex] is not a zero of the polynomial. [tex]\( p\left(\frac{2}{3}\right) \approx -2.33 \)[/tex].
These step-by-step solutions help determine whether the given values for [tex]\( k \)[/tex] are zeros and the respective evaluations of polynomials at those values.
### Question 11:
Polynomial: [tex]\( p(x) = 6x^5 - 23x^4 - 95x^3 + 70x^2 + 204x - 72 \)[/tex]
Given [tex]\( k \)[/tex]: [tex]\( k = 1 \)[/tex]
Synthetic Division Steps:
1. Write down the coefficients: [tex]\( [6, -23, -95, 70, 204, -72] \)[/tex].
2. Start synthetic division with [tex]\( k = 1 \)[/tex]:
- Bring down the leading coefficient: [tex]\( 6 \)[/tex].
- Multiply [tex]\( 6 \)[/tex] by [tex]\( 1 \)[/tex] and add to the next coefficient [tex]\(-23\)[/tex]: [tex]\( 6 \times 1 = 6 \)[/tex], [tex]\(-23 + 6 = -17\)[/tex].
- Multiply [tex]\(-17\)[/tex] by [tex]\( 1 \)[/tex] and add to [tex]\(-95\)[/tex]: [tex]\(-17 \times 1 = -17\)[/tex], [tex]\(-95 + (-17) = -112\)[/tex].
- Multiply [tex]\(-112\)[/tex] by [tex]\( 1 \)[/tex] and add to [tex]\(70\)[/tex]: [tex]\(-112 \times 1 = -112\)[/tex], [tex]\(70 + (-112) = -42\)[/tex].
- Multiply [tex]\(-42\)[/tex] by [tex]\( 1 \)[/tex] and add to [tex]\(204\)[/tex]: [tex]\(-42 \times 1 = -42\)[/tex], [tex]\(204 + (-42) = 162\)[/tex].
- Multiply [tex]\(162\)[/tex] by [tex]\( 1 \)[/tex] and add to [tex]\(-72\)[/tex]: [tex]\(162 \times 1 = 162\)[/tex], [tex]\(-72 + 162 = 90\)[/tex].
3. The remainder, [tex]\( p(1) \)[/tex], is [tex]\( 90 \)[/tex].
Since the remainder is not zero, [tex]\( k = 1 \)[/tex] is not a zero of the polynomial. [tex]\( p(1) = 90 \)[/tex].
### Question 12:
Polynomial: [tex]\( p(x) = 48x^4 + 10x^3 - 51x^2 - 10x + 3 \)[/tex]
Given [tex]\( k \)[/tex]: [tex]\( k = \frac{1}{6} \)[/tex]
Synthetic Division Steps:
Using the process similar to question 11, proceed with:
- Write the coefficients: [tex]\( [48, 10, -51, -10, 3] \)[/tex].
Continue with synthetic division using each step:
1. Calculate the intermediate results as per the division process.
Finally:
- Since the remainder is [tex]\( 0 \)[/tex], [tex]\( k = \frac{1}{6} \)[/tex] is indeed a zero of the polynomial.
### Question 13:
Polynomial: [tex]\( p(x) = 18x^5 - 87x^4 + 110x^3 - 28x^2 - 16x + 3 \)[/tex]
Given [tex]\( k \)[/tex]: [tex]\( k = \frac{2}{3} \)[/tex]
Synthetic Division Steps:
1. Write the coefficients: [tex]\( [18, -87, 110, -28, -16, 3] \)[/tex].
Carry out the synthetic division process:
1. Perform each step using the value [tex]\( \frac{2}{3} \)[/tex] to multiply and add.
After completing these steps:
- The remainder, [tex]\( p\left(\frac{2}{3}\right) \)[/tex], is approximately [tex]\(-2.33\)[/tex].
Since the remainder is not zero, [tex]\( k = \frac{2}{3} \)[/tex] is not a zero of the polynomial. [tex]\( p\left(\frac{2}{3}\right) \approx -2.33 \)[/tex].
These step-by-step solutions help determine whether the given values for [tex]\( k \)[/tex] are zeros and the respective evaluations of polynomials at those values.
