Answer:
The interpolating polynomial is [tex]p(x) = 1-4x+3x^2[/tex].
Step-by-step explanation:
We want to find a quadratic polynomial [tex]p(x)[/tex] such that [tex]p(-1)=8[/tex], [tex]p(1)=0[/tex] and [tex]p(2)=5[/tex]. In order to do this let us write [tex]p(x) = a_0+a_1x+a_3x^2[/tex].
Now, evaluating the polynomial in the points -1, 1 and 2 we get
[tex]\begin{cases} 8 = p(-1) &= a_0-a_1+a_2\\ 0 = p(1) &= a_0+a_1+a_2\\ 5 = p(2) &= a_0+2a_1+4a_2\end{cases}[/tex]
This relations give us a linear system of equations:
[tex]\begin{cases} 8 &= a_0-a_1+a_2\\ 0 &= a_0+a_1+a_2\\ 5&= a_0+2a_1+4a_2\end{cases}[/tex]
where the [tex]a_0[/tex], [tex]a_1[/tex] and [tex]a_2[/tex] are the unknowns.
The augmented matrix of the system is
[tex]\begin{pmatrix}1 & -1 & 1 & 8\\ 1 & 1 & 1 & 0\\ 1 & 2 & 4 & 5\end{pmatrix}[/tex]
In this matrix it is easy to eliminate the 1's of the first column and get
[tex]\begin{pmatrix} 1 & -1 & 1 & 8\\ 0 & 2 & 0 & -8\\ 0 & 3 & 3 & -3\end{pmatrix}[/tex]
From this matrix we can find the values of each unknown. Notice that the second row gives us [tex]2a_2=-8[/tex] that yields [tex]a_1=-4[/tex].
Then, the third row means [tex]3a_1+3a_2=-3[/tex] that gives [tex]-12+3a_2=-3[/tex]. So, [tex]a_2=3[/tex].
Finally, the first row is [tex]a_0-a_1+a_2=8[/tex] and substituting is [tex]a_0+7=8[/tex] that yields [tex]a_0=1[/tex].
Therefore, the interpolating polynomial is
[tex]p(x) = 1-4x+3x^2[/tex].
