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OpenStudy (darkbluechocobo):

Use Descartes' Rule of Signs to determine the number of negative real zeros of the following function: f(t) = -5t^7 + 8t^2 − 2

OpenStudy (jdoe0001):

http://ededu.net/Main/Math/175precal/175Q2/175Q24A1.gif <-- use descartes rule of signs then any ideas on how many real positive ones?

OpenStudy (darkbluechocobo):

1 ? @jdoe0001

OpenStudy (jdoe0001):

hmmm well.... how did you get 1 though?

OpenStudy (darkbluechocobo):

Ohh wait the -5 well it would be 2 because the change from - to + is 1 then back to - is 2?

OpenStudy (jdoe0001):

one sec

OpenStudy (jdoe0001):

yeap checking f( x ) f(t) = -5t^7 + 8t^2 − 2 ^ ^ ^ yes yes so 2 OR 2-2 = 0 real positive solutions or zeros checking now for f( -x ) f(t) = +5t^7 + 8t^2 − 2 ^ ^ ^ nope yes so 1 real negative zero or solution

OpenStudy (darkbluechocobo):

Alright i despise that one mistake messes it up

OpenStudy (darkbluechocobo):

soh for positive you just look at the signs and for negative you switch them?

OpenStudy (darkbluechocobo):

be right back

OpenStudy (jdoe0001):

well...for negative roots.... what you do is replace the "x" argument with "-x" so say \(\large { f(-t) = -5({\color{brown}{ -t}})^7 + 8({\color{brown}{ -t}})^2 - 2\\ \quad \\ \begin{cases} (-t)^7\to -t\cdot -t\cdot -t\cdot -t\cdot -t\cdot -t\cdot -t\to -t^7\\ (-t)^2\to -t\cdot -t\to +t \end{cases} \\ \quad \\ f(-t) = +5{\color{brown}{ -t}}^7 + 8{\color{brown}{ t}}^2 - 2 }\) thus, usually when the exponent of the variable is EVEN, then the sign remains the same and when it's ODD, the sign changes since - * - * - = - and - * - = +

OpenStudy (darkbluechocobo):

back sorry

OpenStudy (darkbluechocobo):

Hokay i understand can we do a couple more jsut to make sure

OpenStudy (jdoe0001):

sure

OpenStudy (darkbluechocobo):

Use Descartes' Rule of Signs to determine the number of positive real zeros of the following function. f(x) = -3x^4 + 7x^2 − x + 5

OpenStudy (darkbluechocobo):

Well this equals 3

OpenStudy (jdoe0001):

well..you're asked for only the real positive ones so you don't have to check for f( -x ) just for f(x) which is given thus f(x) = -3x^4 + 7x^2 − x + 5 ^ ^ ^ ^ yes yes yes 3 changes....thus real POSITIVE ONES will be 3 OR 3-2 = 1

OpenStudy (jdoe0001):

recall that you'd usually subtract any values greater than 1 by 2 so say if we have had say 6 changes then it'd be 6 or 6 -2, or 6-2-2, or 6-2-2-2 so 6 or 4 or 2 or 0 if it were 5 then 5 or 5-2, or 5-2-2 or 5 or 3 or 1

OpenStudy (darkbluechocobo):

soh for this one it would be 1 or 3?

OpenStudy (jdoe0001):

yeap

OpenStudy (darkbluechocobo):

hokay. Use Descartes' Rule of Signs to determine the number of negative real zeros of the following function. f(x) = -9x^4 + 2x^3 + x^2 + x − 1 soh this one has 2 changes before using -x

OpenStudy (jdoe0001):

yeap but if you use f( -x ) recall... if the exponent is EVEN, it remains the same sign if ODD, it changes

OpenStudy (darkbluechocobo):

hokay soh -9x^4 -2^3 +x^2 - x -1?

OpenStudy (darkbluechocobo):

2 changes

OpenStudy (jdoe0001):

yeap so 2 OR 2-2 = 0 real negative roots

OpenStudy (darkbluechocobo):

Use Descartes' Rule of Signs to determine the number of negative real zeros of the following function. f(x) = 2x^4 − 5x^3 + 8x^2 − 3x + 1 Soh the changes before -x is 3?

OpenStudy (darkbluechocobo):

I lie its 4

OpenStudy (jdoe0001):

so the real positive ones are 4 or 4-2 or 4-2-2

OpenStudy (darkbluechocobo):

ok soh 2x^4 +5x^3 +8x^2 + 3x + 1 There is no change for negative

OpenStudy (jdoe0001):

yeap so 0 negative ones... it doesn't have any real negative solutions or roots

OpenStudy (darkbluechocobo):

Thank you for your help soooo much

OpenStudy (jdoe0001):

yw

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