Polynomials through points
Task number: 2637
Determine all polynomials of degree four whose graph passes through given points:
Variant 1
\([-1{,}3]\), \([0,-3]\), \([1{,}3]\) and \([2{,}15]\).
Hint
Denote \(p(x) = ax^4+bx^3+cx^2+dx+e\). The given points will yield a system of linear equations.
Resolution
The system is \[\begin{array}{rrrrrrr} a & -b & +c & -d & +e &=& 3 \\ & & & & e &=& -3 \\ a & +b & +c & +d & +e &=& 3 \\ 16a & +8b & +4c & +2d & +e &=& 15 \\ \end{array}\]
The second equation yields \(e=-3\), we substitute it to the remaining ones. We get the matrix: \[ \left( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 6 \\ 1 & -1 & 1 & -1 & 6 \\ 16 & 8 & 4 & 2 & 18 \\ \end{array} \right) \sim \left( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 6 \\ 2 & 0 & 2 & 0 & 12 \\ 12 & 8 & 0 & 2 & -6 \\ \end{array} \right) \sim \left( \begin{array}{cccc|c} 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 6 \\ 12 & 6 & 0 & 0 & -6 \\ \end{array} \right) \]
Now we se that \(b=-d\), \(c=-a+6\), \(b=-2a-1\).
Result
The desired polynomials are \(p(x) = a x^4 - (1+2a) x^3 + (6-a) x^2 + (1+2a) x -3\), where \(a \in \mathbb R \setminus \{0\}\).
Variant 2
\([-1,-3]\), \([0,-8]\), \([1{,}5]\) and \([2{,}6]\).
Resolution
We get a system \[\begin{array}{rrrrrrr} a & -b & +c & -d & +e &=& -3 \\ & & & & e &=& -8 \\ a & +b & +c & +d & +e &=& 5 \\ 16a & +8b & +4c & +2d & +e &=& 6 \\ \end{array}\] that after elimination of \(e=-8\) leads to a matrix: \[ \left( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 13 \\ 1 & -1 & 1 & -1 & 5 \\ 16 & 8 & 4 & 2 & 14 \\ \end{array} \right) \sim \left( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 13 \\ 2 & 0 & 2 & 0 & 18 \\ 12 & 8 & 0 & 2 & -22 \\ \end{array} \right) \sim \left( \begin{array}{cccc|c} 0 & 1 & 0 & 1 & 4 \\ 1 & 0 & 1 & 0 & 9 \\ 12 & 6 & 0 & 0 & -30 \\ \end{array} \right) \]
Now \(b=4-d\), \(c=-a+9\), \(b=-2a-5\).
Result
The desired polynomils are \(p(x) = a x^4 - (5+2a) x^3 + (9-a) x^2 + (9+2a) x -8\), kde \(a \in \mathbb R \setminus \{0\}\).
