To solve the problem, we need to determine the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] in the polynomial [tex]\(p(x) = ax^2 + bx + c\)[/tex] such that the remainders of the polynomial divided by certain linear factors match the given conditions.
Given:
- [tex]\(p(0) = 7\)[/tex] (remainder from division by [tex]\(x\)[/tex])
- [tex]\(p(1) = 9\)[/tex] (remainder from division by [tex]\(x-1\)[/tex])
- [tex]\(p(2) = 49\)[/tex] (remainder from division by [tex]\(x-2\)[/tex])
Let’s write these conditions as equations:
1. [tex]\(p(0) = a(0)^2 + b(0) + c = c = 7\)[/tex]
2. [tex]\(p(1) = a(1)^2 + b(1) + c = a + b + c = 9\)[/tex]
3. [tex]\(p(2) = a(2)^2 + b(2) + c = 4a + 2b + c = 49\)[/tex]
We now have the following system of linear equations:
1. [tex]\(c = 7\)[/tex]
2. [tex]\(a + b + 7 = 9\)[/tex]
3. [tex]\(4a + 2b + 7 = 49\)[/tex]
First, solve the second equation for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[a + b + 7 = 9 \implies a + b = 2 \tag{1}\][/tex]
Next, solve the third equation for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[4a + 2b + 7 = 49 \implies 4a + 2b = 42 \implies 2a + b = 21 \tag{2}\][/tex]
Now we have a system of two equations:
1. [tex]\(a + b = 2 \tag{1}\)[/tex]
2. [tex]\(2a + b = 21 \tag{2}\)[/tex]
Subtract equation (1) from equation (2):
[tex]\[
(2a + b) - (a + b) = 21 - 2 \implies a = 19
\][/tex]
Using the value of [tex]\(a\)[/tex] in equation (1):
[tex]\[
19 + b = 2 \implies b = 2 - 19 \implies b = -17
\][/tex]
We already know from the first set of equations:
[tex]\[c = 7\][/tex]
Now, we need to find the value of [tex]\(3a + 5b + 2c\)[/tex]:
[tex]\[
3a + 5b + 2c = 3(19) + 5(-17) + 2(7)
\][/tex]
[tex]\[
3 \times 19 = 57
\][/tex]
[tex]\[
5 \times (-17) = -85
\][/tex]
[tex]\[
2 \times 7 = 14
\][/tex]
[tex]\[
57 - 85 + 14 = -14
\][/tex]
So, the value of [tex]\(3a + 5b + 2c\)[/tex] is [tex]\(-14\)[/tex].
Given the options:
(a) -5
(b) 5
(c) 2
None of the provided options is correct based on the calculation. However, based on our correct calculated result, the answer would be [tex]\(-14\)[/tex].
