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Question Number 202795 by mathlove last updated on 03/Jan/24

$$iff(-1)=f\left(0\right)=f\left(2\right)=0andf\left(1\right)=6$$$$thenfindf\left(x\right)=?$$

Answered by mr W last updated on 03/Jan/24

$$f(-1)=f\left(0\right)=f\left(2\right)=0:$$$$\Rightarrow f\left(x\right)=x(x+1)(x-2)p\left(x\right)$$$$f\left(1\right)=6:$$$$\Rightarrow f\left(1\right)=1\times (1+1)(1-2)p\left(1\right)=6$$$$\Rightarrow p\left(1\right)=-3$$$$\Rightarrow p\left(x\right)=(x-4)q\left(x\right)withq\left(1\right)=1$$$$solution:$$$$f\left(x\right)=x(x+1)(x-2)(x-4)q\left(x\right)$$$$withq\left(x\right)=anyfunctionandq\left(1\right)=1.$$

Commented by mathlove last updated on 03/Jan/24

$$thanksmrW$$

Commented by Rasheed.Sindhi last updated on 03/Jan/24

$$f\left(x\right)=x(x+1)(x-2)\stackrel{?}{(x-4)}q\left(x\right)$$$$\mathcal{T}hanxinadvancesir!$$

Commented by mr W last updated on 03/Jan/24

$$suchthatf\left(1\right)=6$$

Commented by Rasheed.Sindhi last updated on 03/Jan/24

$$\mathcal{T}han\mathcal{X}again\mathit{s}\mathit{i}\mathit{r}!$$

Answered by a.lgnaoui last updated on 03/Jan/24

$$-1;0\mathrm{et}2\mathrm{racines}\mathrm{de}\mathrm{f}\left(\mathrm{x}\right)$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}(\mathrm{x}+1)(\mathrm{x}-2)$$$$\mathrm{f}\left(1\right)=\mathrm{a}\times 2\times (-1)=-2\mathrm{a}=6$$$$\Rightarrow \mathrm{a}=-\frac{6}{2}=-3$$$$\mathrm{alors}\mathbf{f}\left(\mathbf{x}\right)=-3\mathbf{x}(\mathbf{x}+1)(\mathbf{x}-2)$$$$$$$$\mathbf{f}\left(\mathbf{x}\right)=-3\mathbf{x}({\mathbf{x}}^2-\mathbf{x}-2)$$

Answered by Frix last updated on 04/Jan/24

$$\mathrm{Usually}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{looking}\mathrm{for}\mathrm{the}\mathrm{polynome}$$$$\mathrm{of}\mathrm{minimal}\mathrm{degree}.\mathrm{With}n\mathrm{equations}\mathrm{we}$$$$\mathrm{get}\mathrm{a}\mathrm{polynome}\mathrm{of}\mathrm{degree}n-1\Rightarrow \mathrm{with}4$$$$\mathrm{equations}\mathrm{we}\mathrm{get}\mathrm{a}\mathrm{polynome}\mathrm{of}\mathrm{degree}3.$$$$\mathrm{The}\mathrm{usual}\mathrm{method}\mathrm{is}\mathrm{this}:$$$$$$$${ax}^3+{bx}^2+cx+d=f\left(x\right)$$$$f\left(0\right)=0\Rightarrow d=0\left(1\right)$$$$f(-1)=0$$$$-a+b-c=0\left(2\right)$$$$f\left(2\right)=0$$$$8a+4b+2c=0\left(3\right)$$$$f\left(1\right)=6$$$$a+b+c=6\left(4\right)$$$$\left(2\right)\Rightarrow b=a+c$$$$\left(3\right)12a+6c=0\Rightarrow c=-2a\Rightarrow b=-a$$$$\left(4\right)-2a=6\Rightarrow a=-3\Rightarrow b=3\wedge c=6$$$$\Rightarrow$$$$f\left(x\right)=-3x^3+3x^2+6x=-3x(x-2)(x+1)$$$$$$$$\mathrm{But}\mathrm{with}f(-1)=f\left(0\right)=f\left(2\right)=0\mathrm{we}\mathrm{directly}$$$$\mathrm{get}$$$$f\left(x\right)=a(x+1)x(x-2)$$$$f\left(1\right)=6\Rightarrow a\times 2\times 1\times (-1)=6\Rightarrow a=-3$$$$\Rightarrow$$$$f\left(x\right)=-3x(x-2)(x+1)$$

Commented by mathlove last updated on 04/Jan/24

$$thanks$$

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