The polynomial f(x) has degree 3. If f(-1)=15, f(0)=0, f(1)=5 and f(2)=12, then what are the x-intercepts of the graph of f?
+6248 \(\text{here may be some clever way to attack this but if so I'm not aware of it}\\ p(x) = a x^3 + b x^2 + c x + d\\ \text{using the function values give we can form the equations}\\ -a+b-c+d=15\\ d=0\\ a+b+c+d=5\\ 8a+4b+2c+d=12\\ \text{I would dump these into a matrix and solve it like that}\\ \begin{pmatrix}-1 &1 &-1 &1\\0 &0 &0 &1 \\1 &1 &1 &1 \\8 &4 &2 &1\end{pmatrix} \begin{pmatrix}a \\ b\\ c \\ d\end{pmatrix} = \begin{pmatrix}15 \\0 \\5 \\12 \end{pmatrix}\)
Solving this using the method of you choice we find
\(a=-3,~b=10,~c=-2,~d=0 \\ p(x) = -3x^3 + 10x^2 -2x\)
The x-intercepts occur where p(x)=0, factoring we get
\(p(x)=-3x^3+10x^2-2x = \\ -x(3x^2 - 10x+2)\\ \text{and this has zeros at } x=0 \text{ and}\\ x = \dfrac{10\pm \sqrt{100-24}}{6} = \dfrac 1 3\left(5 \pm \sqrt{19}\right)\)
+6248 \(\text{here may be some clever way to attack this but if so I'm not aware of it}\\ p(x) = a x^3 + b x^2 + c x + d\\ \text{using the function values give we can form the equations}\\ -a+b-c+d=15\\ d=0\\ a+b+c+d=5\\ 8a+4b+2c+d=12\\ \text{I would dump these into a matrix and solve it like that}\\ \begin{pmatrix}-1 &1 &-1 &1\\0 &0 &0 &1 \\1 &1 &1 &1 \\8 &4 &2 &1\end{pmatrix} \begin{pmatrix}a \\ b\\ c \\ d\end{pmatrix} = \begin{pmatrix}15 \\0 \\5 \\12 \end{pmatrix}\)
Solving this using the method of you choice we find
\(a=-3,~b=10,~c=-2,~d=0 \\ p(x) = -3x^3 + 10x^2 -2x\)
The x-intercepts occur where p(x)=0, factoring we get
\(p(x)=-3x^3+10x^2-2x = \\ -x(3x^2 - 10x+2)\\ \text{and this has zeros at } x=0 \text{ and}\\ x = \dfrac{10\pm \sqrt{100-24}}{6} = \dfrac 1 3\left(5 \pm \sqrt{19}\right)\)
