Question

Use synthetic division to find all the real zeros of the polynomial.
f(x) = x³ + x² - 8x - 6

Answer 1

so hmmm we look at hmm x³ + x² - 8x - 6 so how do squeak out a factor out of it, well, using the rational root test and using our p/q checking for factors, we find that likely factors can be ± 6/1 and ±(3, 2, 1) / 1, anyhow so we check some of those and hopefully without boring you to death we find that -3 works, we can always use the remainder theorem to check if that's true, by simply plugin in f(-3) and if it's indeed a factor, that'd give us 0, well, f(-3) = -27 + 9 +24 +6 = 0, holly smokes!! is a factor.

well, all that jazz means that -3 is a factor, namely the factor is really x = -3 or x + 3 = 0 so the factor is (x+3) and for our synthetic division we'll use the -3 version, Check the picture below.

so we end up with the factors of (x+3)(x²-2x-2), now for the 2nd factor what we can do to break it up is use the quadratic formula

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{-2}x\stackrel{\stackrel{c}{\downarrow }}{-2} \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}`

`x= \cfrac{ - (-2) \pm \sqrt { (-2)^2 -4(1)(-2)}}{2(1)} \implies x = \cfrac{ 2 \pm \sqrt { 4 +8}}{ 2 } \\\\\\ x= \cfrac{ 2 \pm \sqrt { 12 }}{ 2 }\implies x=\cfrac{2\pm√(2^2\cdot 3)}{2}\implies x=\cfrac{2\pm 2√(3)}{2}\implies x=1\pm√(3) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\Large \begin{array}{llll} (x+3)(x-1+√(3))(x-1-√(3)) \end{array}}~\hfill`

now, just to clarify, all those are real roots, and the last two come from simply setting x = 1 ± √3 to 0 to get the factors.