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Mathematics 14 Online
OpenStudy (anonymous):

PLEASE HELP WILL GIVE MEDAL!! Find all the zeros of each equation. picture down bellow! ( i only need help on question #3)

OpenStudy (anonymous):

A and D are out, because if \(i\) is a zero, then so is \(-i\)

OpenStudy (anonymous):

now you can check if \(3\) is a zero or not, by substitution

OpenStudy (anonymous):

actually none of those answers are right, i suspect you made a typo

OpenStudy (anonymous):

i mean for the first problem you wrote none of those answers are correct

OpenStudy (anonymous):

\[x^5 – 3x^4 – 15x^3 – 45x^2 – 16x + 48 = 0\] right?

OpenStudy (anonymous):

oh i see, it was a typo it should have been \(+45x^2\) not \(-45x^2\)

OpenStudy (anonymous):

ohh i thought u meant in the answer choices

OpenStudy (anonymous):

in that case the zeros are \(3,-4,4,i,-i\)

OpenStudy (anonymous):

thank u so much! do u know the other ones?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[x^3-x^2+5x-5=0\] first thing to check is that \(1\) is a solution, because it is, and it is easy to check \[1-1+5-5=0\]

OpenStudy (anonymous):

then you know you can factor it as \[(x-1)(something)\] and you can find the something by division (synthetic division is easiest)

OpenStudy (anonymous):

when you factor it, you should get \[(x-1)(x^2+5)\] and so if \(x^2+5=0\) then \(x=\pm\sqrt{5}i\)

OpenStudy (anonymous):

okk..then what do i do?

OpenStudy (anonymous):

then you are done with that oneo

OpenStudy (anonymous):

solutions are \(1,\sqrt5 i, -\sqrt5 i\)

OpenStudy (anonymous):

is that one of the options or did u put it in a different order? haha im trying to figure it out lol

OpenStudy (anonymous):

i don't know i didn't look at the options

OpenStudy (anonymous):

oh i see, it is option A no one writes \(i\sqrt5\) though, looks weird

OpenStudy (anonymous):

ohhh ok yeah haha! can u help me with the last one?

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