# Determining the zeros of x^4 -(x^3) -(31x^x) + x 30. EASY PRECAL PROBLEM

• Sep 15th 2012, 10:18 AM
hunnyelle
Determining the zeros of x^4 -(x^3) -(31x^x) + x 30. EASY PRECAL PROBLEM
• Correct answer is 1, -1, -5, and 6. SOLVE IT THE SAME WAY AS A PRECAL WOULD -.i.e. using descartes rule of signs, synthetric division,upper bound and lower bound test, rational zero theorem, narrowing down choices using a ti-84 or ti-89 calcutior, or a combination of these methods. IMPORTANT NOTE: Our teacher we can't exclusively use the graphing calculator to determine the zeros of this function. These are notes my teachers used to teach us this type of problem: http://images.pcmac.org/SiSFiles/Sch...0(day%201).pdf

My work so far:
-plugged in 1 using synthetic devision, determined 1 was a zero. Then got decompressed function on bottom of synthetic division, x^3 -31x + 30.

• Sep 15th 2012, 11:17 AM
skeeter
Re: Determining the zeros of x^4 -(x^3) -(31x^x) + x 30. EASY PRECAL PROBLEM
Quote:

Determining the zeros of x^4 -(x^3) -(31x^x) + x 30
fyi, this is not a polynomial ... fix it.
• Sep 15th 2012, 11:31 AM
HallsofIvy
Re: Determining the zeros of x^4 -(x^3) -(31x^x) + x 30. EASY PRECAL PROBLEM
I presume you mean $\displaystyle x^4- x^3- 31x^2+ x+ 30$.

Okay, "Descarte's rule of signs" tells you that this polynomial has either 0 or 2 positive zeros and the "rational zero theorem" tells you that any rational zero must be an integer divisor of 30. Certainly, x= 1 is a possibility and testing it by directly substituting or by "synthetic division" you find it is a root. And then, perhaps by using synthetic division as you say, we know that $\displaystyle x^4- x^3- 31x^2+ x+ 30= (x- 1)(x^3- 31x- 30)$ (you have a sign wrong). Now "Descarte's rule of signs" tells us this has one real root (of course- having found one positive root, we know the original polynomial had 2 positive zeros which leaves one for this. Again the "rational zero theorem" tells us any rational root must be an integer divisor of 30. Checking some of those, we eventually find that 6 is also a root. Do another "synthetic division" to see that and that $\displaystyle x^4- x^3- 31x^2+ x+ 30= (x- 1)(x^3- 31x- 30)= (x- 1)(x- 6)(x^2+ 6x+ 5)$.

And, now, it should be easy to factor $\displaystyle x^2+ 6x+ 5$ (or note that Descarte's rule of signs says there are NO positive roots so either 0 or 2 negative roots and any rational root must be an integer divisor of 5- i.e. -1 or -5).
• Sep 15th 2012, 11:57 AM
hunnyelle
Re: Determining the zeros of x^4 -(x^3) -(31x^x) + x 30. EASY PRECAL PROBLEM
You're right, I incorrectly wrote down the function. @HallsofIvy: Thank you sooooo much !
• Sep 15th 2012, 12:09 PM
Soroban
Re: Determining the zeros of x^4 -(x^3) -(31x^x) + x 30. EASY PRECAL PROBLEM
Hello, hunnyelle!

Don't put the problem in the title line ... especialy with those typos.

Quote:

$\displaystyle \text{Find the zeros of: }\:f(x) \:=\:x^4-3x^3 - 31x^2 + x + 30$

Your work is correct: $\displaystyle \boxed{x = 1}$ is one of the zeros.

Then: .$\displaystyle (x^4-3x^2-3x^2+x+30) \div (x-1) \:=\:x^3-31x - 30$

We see that $\displaystyle \boxed{x = \text{-}1}$ is another zero.

Then: .$\displaystyle (x^3 - 31x - 30) \div (x+1) \:=\:x^2-x-30$

Factor: .$\displaystyle x^2-x-30 \:=\:(x+5)(x-6)$

And we have two more zeros: .$\displaystyle \boxed{x = \text{-}5},\;\boxed{x=6}$