MATH SOLVE

4 months ago

Q:
# find a polynomial of degree 3 with real cofficients and zeros of -3,-1,4 for which f(-2)=24β

Accepted Solution

A:

[tex]\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[/tex]we know that f(-2) = 24, namely when x = -2, y = 24, let's see if that's true[tex]\bf x^3-13x-12=y\implies \stackrel{x = -2}{(-2)^3-13(-2)-12}=y \\\\\\ -8+26-22=y \implies 6=y[/tex]darn!! no dice.... hmmmm wait a second.... 4 * 6 = 24, if we could just use a common factor of 4 on the function, that common factor times 6 will give us 24, let's check.[tex]\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[/tex]