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How do I solve this with Wolfram Alpha?
I'm not looking for the answer, but rather how I can get the answer with Wolfram Alpha...
The following function is given.
f(x)=x^3 - 5x^2 - 4x + 20
A. List all rational zeros that are possible according to the Rational Zero Theorem.
B. Use synthethic division to test several possible rational zeros in order to identify one actual zero.
It's an online class, so good thing I don't have to actually show work. I figured it out. The possible rational zeroes is all the factors of the constant and the leading coefficient. And the actual rational zeros Wolfram Alpha can give me.
3 Answers
- Randy PLv 77 years ago
I don't think that's going to work.
A. The possible roots are +-p/q where p is a factor of 20 and q is a factor of 1. So it's just +-1, +-2, ..., etc, all the factors of 20 with a +- in front of them. Just write down the list, should take 5 seconds.
B. This is asking you not only to divide f(x) by (x + 1), (x- 1), (x + 2), (x - 2), etc. but to SHOW THE PROCESS OF SYNTHETIC DIVISION. Wolfram Alpha can do the division, but it's not going to show the process to you. Afraid you'll have to put some effort into your own homework.
- plays_poorly...Lv 77 years ago
WolframAlpha comes up with the same three roots, whether you copy it out of the Y-Answers page and paste it in directly, or copy everything except the f(x) part and put in an "=0" on the right side.
You may wish to put WolframAlpha in your bookmarks, as I have, since I expect you will use it often. Good luck!
x = -2, 2, and 5.
- la consoleLv 77 years ago
By factorization
x³ - 5x² - 4x + 20 = 0
(x³ - 5x²) - (4x - 20) = 0
x²(x - 5) - 4(x - 5) = 0
(x - 5)(x² - 4) = 0
(x - 5)(x² - 2²) = 0
(x - 5)(x + 2)(x - 2) = 0 → the roots are: - 2 ; 2 ; 5
Directly with Wolfram Alpha
http://www.wolframalpha.com/input/?i=x%C2%B3+-+5x%...
→ the roots are: - 2 ; 2 ; 5
You have to find an obvious root between an easy value between (- 2 ; - 1 ; 1 ; 2)
Let's try with x = 1
= (1)³ - 5(1)² - 4(1) + 20
= 1 - 5 - 4 + 20
= 12 ≠ 0 → so 1 is not a root
Let's try with x = - 1
= (- 1)³ - 5(- 1)² - 4(- 1) + 20
= - 1 - 5 + 4 + 20
= 18 ≠ 0 → so 1 is not a root
Let's try with x = 2
= (2)³ - 5(2)² - 4(2) + 20
= 8 - 20 - 8 + 20
= 0 → so you can factorize (x - 2)
= (x - 2)(x² + ax + b) → you expand
= x³ + ax² + bx - 2x² - 2ax - 2b → you group
= x³ + x²(a - 2) + x(b - 2a) - 2b → you compare to the initial (x³ - 5x² - 4x + 20)
- 2b = 20 → b = - 10
(b - 2a) = - 4 → 2a = b + 4 → 2a = - 6 → a = - 3
(a - 2) = - 5 → a = 2 - 5 → a = - 3 of course
= (x - 2)(x² + ax + b) → you substitute a and b by their value
= (x - 2)(x² - 3x - 10)
= (x - 2)[x² - (5x - 2x) - 10]
= (x - 2)[x² - 5x + 2x - 10]
= (x - 2)[(x² - 5x) + (2x - 10)]
= (x - 2)[x(x - 5) + 2(x - 5)]
= (x - 2)[(x - 5)(x + 2)]
= (x - 2)(x - 5)(x + 2) → the roots are: - 2 ; 2 ; 5
