Mathematics, 24.11.2019 02:31 JeonTaekook
Find a polynomial of degree 3 with real cofficients and zeros of -3,-1,4 for which f(-2)=24
Answers: 1
![\bf \textit{zeros at } \begin{cases} x = -3\implies &x+3=0\\ x = -1\implies &x+1=0\\ x = 4\implies &x-4=0 \end{cases}\qquad \implies (x+3)(x+1)(x-4)=\stackrel{y}{0} \\\\\\ (x^2+4x+3)(x-4)=0\implies x^3~~\begin{matrix}+ 4x^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+3x~~\begin{matrix} -4x^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-16x-12=0 \\\\\\ x^3-13x-12=0](/tpl/images/0388/4608/09859.png)
![\bf 4(x^3-13x-12)=y\implies \stackrel{x = -2}{4[~~(-2)^3-13(-2)-12~~]}=y \\\\\\ 4[~~-8+26-22~~]=y\implies 4[6]=y\implies 24=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill 4x^3-52x-48=y~\hfill](/tpl/images/0388/4608/ce89e.png)
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Find a polynomial of degree 3 with real cofficients and zeros of -3,-1,4 for which f(-2)=24...
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