Algebra Know-It-ALL

688 Worked-Out Solutions to Exercises: Chapters 21 to 29

We can multiply through by −1 to get

105 x^3 + 211 x^2 + 112 x+ 12 = 0

Either of the last two equations is a legitimate expression of the polynomial standard form

for the cubic. The second one more “sign-friendly.”

4. Here are the real roots we found in the chapter text for the equation stated back at the
   beginning of Prob. 3:

x=−2/3 or x=−6/5 or x=−1/7

When we plug x=−2/3 into the final polynomial equation we found in solution to Prob. 3

and then grind through the arithmetic, the process goes like this, step-by-step:

105 x^3 + 211 x^2 + 112 x+ 12 = 0

105 × (−2/3)^3 + 211 × (−2/3)^2 + 112 × (−2/3)+ 12 = 0

105 × (−8/27)+ 211 × (4/9) + 112 × (−2/3)+ 12 = 0

−840/27+ 844/9 − 224/3 + 12 = 0

−840/27+ 2,532/27 − 2,016/27 + 324/27 = 0

(− 840 + 2,532 − 2,016 + 324) / 27 = 0

− 840 + 2,532 − 2,016 + 324 = 0

0 = 0

To check the root x=−6/5, we go through this sequence of calculations:

105 x^3 + 211 x^2 + 112 x+ 12 = 0

105 × (−6/5)^3 + 211 × (−6/5)^2 + 112 × (−6/5)+ 12 = 0

105 × (−216/125)+ 211 × (36/25) + 112 × (−6/5)+ 12 = 0

−22,680/125+ 7,596/25 − 672/5 + 12 = 0

−22,680/125+ 37,980/125 − 16,800/125 + 1,500/125 = 0

(−22,680+ 37,980 − 16,800 + 1,500) / 125 = 0

−22,680+ 37,980 − 16,800 + 1,500 = 0

0 = 0

To check the root x=−1/7, we go through this arithmetic, step-by-step:

105 x^3 + 211 x^2 + 112 x+ 12 = 0

105 × (−1/7)^3 + 211 × (−1/7)^2 + 112 × (−1/7)+ 12 = 0

105 × (−1/343)+ 211 × (1/49) + 112 × (−1/7)+ 12 = 0

−105/343+ 211/49 − 112/7 + 12 = 0

−105/343+ 1,477/343 − 5,488/343 + 4,116/343 = 0

(− 105 + 1,477 − 5,488 + 4,116) / 343 = 0

− 105 + 1,477 − 5,488 + 4,116 = 0

0 = 0