Rational zero test and finding solutions

**aikenfan** · December 16th 2007, 06:24 PM

I have two problems on my review that I just cannot seem to remember how to work...If anyone could please help, I would really appreciate it!
1. Use the Rational Zero Test to determine all possible rational zeros of the function and then find all the actual zeros of f(x) = x^3 + 2x^2 - 21x + 18
2. Find all the solutions to f(x) = -8x^4 + 128x^2

**Soroban** · December 16th 2007, 06:46 PM

Hello, aikenfan!

1. Use the Rational Zero Test to determine all possible rational zeros
of the function and then find all the actual zeros of: . $f(x) \:= \:x^3 + 2x^2 - 21x + 18$

For this problem, the only rational zeros are factors of 18: . $\pm1,\:\pm2,\;\pm3,\;\pm6,\:\pm18$

We're in luck . . . The first one works!
. . $f(1) \:=\:1^3 + 2\!\cdot\!1^2 - 21\!\cdot\!1 + 18 \:=\:0$
Hence, $(x-1)$ is a factor of $f(x).$

We find that: . $f(x)\;=\;(x-1)(x^2+3x-18) \;=\;(x-1)(x-3)(x+6)$

. . Therefore, the zeroes are: . $\boxed{x \;=\;1,\,3,\,-6}$

2. Find all the solutions to: . $f(x) \:= \:-8x^4 + 128x^2$

Factor: . $-8x^2(x^2-16) \:=\:0\quad\Rightarrow\quad-8x^2(x-4)(x+4) \:=\:0$

Therefore, the solutions are: . $\boxed{x \;=\;0,\,\pm4}$

**aikenfan** · December 16th 2007, 06:58 PM

Thank you for your help! Just one more quick question about the Rational Zero Test...how exactly does it work? Like, if I were to do another similar problem?

**badgerigar** · December 17th 2007, 03:03 AM

if $f(x) = a_nx^n+a_{n-1}x^{n-1}+...+a_0$
and all of the coefficients are integers then all rational roots of f can be written as $\frac{p}{q}$ where $p|a_0$ and $q|a_n$

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