Question

How do I obtain the real number solutions to the polynomial function f(x)=x^3 - 4x^2 + x +6?

Answer 1

$\bf f(x)=x^3-4x^2+x+6\\\\ -----------------------------\\\\ \begin{array}{l|lrrrrllllll} 3&&1&-4&1&6\\ &&&3&-3&-6\\ --&&--&--&--&--\\ &&1&-1&-2&0 \end{array}\impliedby x=3\implies (x-3)=0 \\\\\\ \textit{that give us the factors of } \\\\\\ f(x)=(x-3)(x^2-x-2)\impliedby \textit{factoring the quadratic, we get} \\\\\\ f(x)=(x-3)(x-2)(x+1)\impliedby \textit{setting f(x) to 0} \\\\\\ 0=(x-3)(x-2)(x+1)\implies \begin{cases} x-3=0\to &x=3\\ x-2=0\to &x=2\\ x+1=0\to &x=-1 \end{cases}$

so.... notice, using the p/q factors, we ended up using 3/1 or just 3, so the root of x = 3, which gives us the factor of x-3 then

then you factor the quadratic, then set f(x) to 0, and solve each for "x"