Certainly! Let's solve the problem step by step:
We are given two polynomials:
[tex]\[ P_1(x) = 2x^3 + ax^2 + 3x - 5 \][/tex]
[tex]\[ P_2(x) = x^3 + x^2 - 2x + a \][/tex]
We need to find the value of [tex]\(a\)[/tex] such that both polynomials leave the same remainder when divided by [tex]\((x-2)\)[/tex].
1. Evaluate the remainders when [tex]\(x = 2\)[/tex]:
We use the fact that the remainder when a polynomial [tex]\(P(x)\)[/tex] is divided by [tex]\(x - c\)[/tex] is [tex]\(P(c)\)[/tex]. So, we need to find the remainders of [tex]\(P_1(x)\)[/tex] and [tex]\(P_2(x)\)[/tex] at [tex]\(x = 2\)[/tex].
- For [tex]\(P_1(x)\)[/tex]:
[tex]\[ P_1(2) = 2(2)^3 + a(2)^2 + 3(2) - 5 \][/tex]
[tex]\[ P_1(2) = 2 \cdot 8 + a \cdot 4 + 6 - 5 \][/tex]
[tex]\[ P_1(2) = 16 + 4a + 6 - 5 \][/tex]
[tex]\[ P_1(2) = 4a + 17 \][/tex]
- For [tex]\(P_2(x)\)[/tex]:
[tex]\[ P_2(2) = (2)^3 + (2)^2 - 2(2) + a \][/tex]
[tex]\[ P_2(2) = 8 + 4 - 4 + a \][/tex]
[tex]\[ P_2(2) = 8 + a \][/tex]
2. Set the remainders equal to each other:
Since both polynomials leave the same remainder when divided by [tex]\((x-2)\)[/tex], we equate the remainders:
[tex]\[ 4a + 17 = 8 + a \][/tex]
3. Solve for [tex]\(a\)[/tex]:
[tex]\[ 4a + 17 = 8 + a \][/tex]
[tex]\[ 4a - a = 8 - 17 \][/tex]
[tex]\[ 3a = -9 \][/tex]
[tex]\[ a = -3 \][/tex]
4. Find the remainders using the value of [tex]\(a\)[/tex]:
Substitute [tex]\(a = -3\)[/tex] back into the expressions we found for the remainders:
- For [tex]\(P_1(x)\)[/tex]:
[tex]\[ P_1(2) = 4a + 17 \][/tex]
[tex]\[ P_1(2) = 4(-3) + 17 \][/tex]
[tex]\[ P_1(2) = -12 + 17 \][/tex]
[tex]\[ P_1(2) = 5 \][/tex]
- For [tex]\(P_2(x)\)[/tex]:
[tex]\[ P_2(2) = 8 + a \][/tex]
[tex]\[ P_2(2) = 8 - 3 \][/tex]
[tex]\[ P_2(2) = 5 \][/tex]
So, the value of [tex]\(a\)[/tex] is [tex]\(-3\)[/tex], and the remainder in each case is 5.
