Polynomial remainder theorem?
- Thread starter momoko
- Start date
You will have two equations with two unknowns to solve for a and b.
\(\displaystyle \frac{x^{3}+ax^{2}+bx-3}{x-2}=\frac{\overbrace{4a+2b+5}^{\text{remainder}}}{x-2}+x^{2}+(2+a)x+2a+b+4\)
\(\displaystyle \frac{x^{3}+ax^{2}+bx-3}{x+1}=\frac{\overbrace{a-b-4}^{\text{remainder}}}{x+1}+x^{2}+(a-1)x-a+b+1\)
\(\displaystyle 4a+2b+5=27\)
\(\displaystyle a-b-4=3\)
Solve this for a and b.
Then, using those values, sub them into the remainder here:
\(\displaystyle \frac{x^{3}+ax^{2}+bx-3}{x-1}=\frac{\overbrace{a+b-2}^{\text{remainder}}}{x-1}+x^{2}+x(a+1)+a+b+1\)
The terms to the right aren't really needed. We are looking at the remainder.
\(\displaystyle \frac{x^{3}+ax^{2}+bx-3}{x-2}=\frac{\overbrace{4a+2b+5}^{\text{remainder}}}{x-2}+x^{2}+(2+a)x+2a+b+4\)
\(\displaystyle \frac{x^{3}+ax^{2}+bx-3}{x+1}=\frac{\overbrace{a-b-4}^{\text{remainder}}}{x+1}+x^{2}+(a-1)x-a+b+1\)
\(\displaystyle 4a+2b+5=27\)
\(\displaystyle a-b-4=3\)
Solve this for a and b.
Then, using those values, sub them into the remainder here:
\(\displaystyle \frac{x^{3}+ax^{2}+bx-3}{x-1}=\frac{\overbrace{a+b-2}^{\text{remainder}}}{x-1}+x^{2}+x(a+1)+a+b+1\)
The terms to the right aren't really needed. We are looking at the remainder.
Hello, momoko!
Sounds like you never heard of the Remainder Theorem . . .
From the statements about division and remainders, we have:
. . \(\displaystyle \begin{array}{ccccccccc}f(2) \:=\:27 & 2^3 + a(2^2) + b(2) - 3 &=& 27 & \Rightarrow & 4a + 2b &=& 22 & [1] \\ \\[-3mm] f(\text{-}1) \:=\:3 & (\text{-}1)^3 + a(\text{-}1)^2 + b(\text{-}1) - 3 &=& 3 & \Rightarrow & a - b &=& 7 & [2] \end{array}\)
. . \(\displaystyle \begin{array}{cccccccc}\text{Divide [1] by 2:} & 2a + b &=& 11 \\ \text{Add [2]:} & a - b &=& 7 \end{array}\)
We have: .\(\displaystyle 3a \:=\:18 \quad\Rightarrow\quad a \:=\:6\)
Substitute into [2]: .\(\displaystyle 6 - b \:=\:7 \quad\Rightarrow\quad b \:=\:\text{-}1\)
Hence:. . \(\displaystyle f(x) \:=\:x^3 + 6x^2 - x - 3\)
Therefore: .\(\displaystyle f(1) \:=\:1^2 + 6(1^2) - 1 - 3 \;=\;\boxed{3}\)
Sounds like you never heard of the Remainder Theorem . . .
From the statements about division and remainders, we have:
. . \(\displaystyle \begin{array}{ccccccccc}f(2) \:=\:27 & 2^3 + a(2^2) + b(2) - 3 &=& 27 & \Rightarrow & 4a + 2b &=& 22 & [1] \\ \\[-3mm] f(\text{-}1) \:=\:3 & (\text{-}1)^3 + a(\text{-}1)^2 + b(\text{-}1) - 3 &=& 3 & \Rightarrow & a - b &=& 7 & [2] \end{array}\)
. . \(\displaystyle \begin{array}{cccccccc}\text{Divide [1] by 2:} & 2a + b &=& 11 \\ \text{Add [2]:} & a - b &=& 7 \end{array}\)
We have: .\(\displaystyle 3a \:=\:18 \quad\Rightarrow\quad a \:=\:6\)
Substitute into [2]: .\(\displaystyle 6 - b \:=\:7 \quad\Rightarrow\quad b \:=\:\text{-}1\)
Hence:. . \(\displaystyle f(x) \:=\:x^3 + 6x^2 - x - 3\)
Therefore: .\(\displaystyle f(1) \:=\:1^2 + 6(1^2) - 1 - 3 \;=\;\boxed{3}\)
