Pre-Cal Find all real zeros and prove it?

Question

I need help with a pre-cal question
i've looked for help but i can solve it

p(x) = x^5 - 4x^4 - x^3 + 10x^2 + 2x - 4

Find all real zeros of the polynomial and prove each zero by syntheric division, factoring or the quadratic formula.

Answer 1

By the Rational Root Theorem, the only possible rational roots of p(x) = 0 are factors of the constant term -4 (since the leading coefficient is 1).



By trial and error, we find that x = -1 is a zero of p(x).

So, x - (-1) = x + 1 is a factor by the Factor Theorem.



Using synthetic division,

-1|1...-4...-1...10...2...-4

..........-1...5....-4...-6....0

....----------------------------

.....1..-5....4.....6...-4....0; the last entry is the remainder.



So, p(x) = (x + 1)(x^4 - 5x^3 + 4x^2 + 6x - 4).

------

Again by trial and error, we find that x = -1 is a zero of x^4 - 5x^3 + 4x^2 + 6x - 4.



Using synthetic division,

-1|1...-5...4....6...-4

..........-1...6..-10...4

....----------------------------

.....1..-6...10..-4...0



So, p(x) = (x + 1)^2 (x^3 - 6x^2 + 10x - 4).

----------

Next, we find that x = 2 is a zero.



2|1....-6...10...-4

..........2....-8...4

...-------------------

...1....-4.....2...0.



So, p(x) = (x + 1)^2 (x - 2) (x^2 - 4x + 2).

---------

By the quadratic formula, the final factor has zeros x = 2 ± √2.



Hence, p(x) has five real zeros (counting multiplicity):

x = -1, -1, 2, 2 ± √2.



I hope this helps!

Answer 2

The Rational Root Theorem offers the following candidates: -4, -2, -1, 1, 2, 4



If it's allowed, I would recommend plotting p(x) for promising candidates; otherwise, you will have to test all six candidates (which means computing p for each candidate and noting any candidates that give p = 0). Here is the plot: http://www.wolframalpha.com/input/?i=plot+x%5E5+-+4x%5E4+-+x%5E3+%2B+10x%5E2+%2B+2x+-+4



An additional aid given by the plot is that it suggests that x = -1 is not merely a root, but a repeated root (since the graph of p appears to be tangent to the x-axis). Those non-integer crossings of the x-axis by p are zeros, but they are irrational since the Rational Root Theorem did not offer any non-integer candidates. One way or another, you will find that -1 is a double root and 2 is a single root. Therefore, p(x) = (x + 1)²(x - 2)(x² - 4x + 2). You can use the quadratic formula to find the two irrational roots.



Synthetic division:





.....1...-4...-1...10....2...-4

2.... ....2...-4..-10....0....4

------------------------------------

.....1...-2...-5.... 0....2...0

-1...... -1....3.....2...-2

-----------------------------------

.....1...-3...-2.....2....0

-1.......-1....4... -2

----------------------------------

.....1...-4....2.... 0

Answer 3

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