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\[f(x) = 3x^3-10x^2-81x+28 \] this?
if we need to find 0's then we need the rational root test.
The Rational Roots Test (also known as Rational Zeros Theorem) allows us to find all possible rational roots of a polynomial. Suppose "a" is root of the polynomial P(x) that means P(a) = 0. In other words, if we substitute "a" into the polynomial P(x) and get zero it means that the input value is a root. so we take the last term, 28, and find all factors of 28 which is 1,2,4,7,14,28. We also have to take the negative versions as well. So, 1,-1,2,-2,4,-4,7,-7,14,-14,28,-28 just plug these numbers into the equation and stop once you have a 0 there could be more than one root, just one, or none. If we don't have any roots, the test fails.
is the answer 7, -4, 1/3
but the last root lies somewhere between 0 and 1 and I think 1/3 is pretty close
answer choices 7,-4,1/3 7,-4,-1/3 7,4,1/3 7,4,-1/3
hmmm.... let x = 7,4, or -4 and plug it into the equation ... one of them has to produce a zero .we may have to go further like long division to grab the remaining roots
im pretty sure it is the last choice
did you test it out?
because there's a negative root.. and that number is not -1/3
so that knocks out two choices.
that means it is the third choice because after you test it out it gives you the third choice
negative root is x=-4
oops i think i know the answer for this question can you help me with another
yeah sure. you actually had the right answer 10 minutes ago XD
Which of the following is a polynomial with roots 5, 4i, and −4i
f(x) = x^3-5x^2+20x-16 f(x) = x^3-5x^2+16x-80 f(x)=x^3-20x^2+5x-16 f(x) = x^3-16x^2+80x-5
wow... it's like we have to use the roots given and plugging in it into the function having x = 5 will make it easier though.
so plug 5 for each one
try let x = 5 for the second function
yeah for each one, but I think I've weeded it out x=5 for the second function
i think i can take out the 2nd one and also the the first one
first one is nasty... x.x
so that is definitely eliminated
oh yeah... not to mention that the real root is x=1... not what we need.
the answer is the third one
umm are you sure?
im pretty sure
the real root I'm getting for that third choice is really bad ... same with the imaginary... the real root is approx.. 20.. we need x = 5,4i,-4i cross that out
im feeling that my work shows that it is the answer
the last one
try plugging x = 5 for the second function you have and tell me if your result is 0 f(x) = x^3-5x^2+16x-80
then that's the function. since the result became 0... for x = 5... then the roots x =4i, -4i work as well.
you're welcome :)