Here's the question you clicked on:
evy15
Find all zeros 5x^5-20x^4-40x^3-16x^2-45x+180=0
What do you mean by "Find all zeros" ?
Use rational root theorem to find first zero. Then use synthetic division to factor it out.
hi @evy15 are you in numerical analysis? the root are -2.07316, 1.2355, 5.5573 and two complex roots check it for yourself if they are true
@evy though the equation is 5x^5-20x^4-40x^3-16x^2-45x+180=0 correct bu t may be i feel u have typed in -16X^2 instead of 160x^2
can someone please help me and explain to me how to begin this problem
Have you done numerical methods in your course? This is not a problem that can be solved by factorization, or rearrangements. I suggest you triple check the question.
its correct can you tell me how to begin
It can be solved numerically, or graphically. Are these expected methods to use in your course?
yes the thing is i missed a day in class and now i dont know how to work this out
Are you working on factorization, Descartes rule of signs, etc.? I suggest you check your course outline to see what you're expected to know, or check with a friend to see what has been done that day. The prof would also be pleased to tell you what has been covered that day. This problem requires numerical methods which are techniques completely different from your previous problems.
can u help me and explain it to me
i have tried to use the book but i keep getting it wrong
What is the name of your math course?
Then there is probably another typo in the question.
there isn't, i already checked
I believe you, but your prof made a typo! Check in the index of your textbook to see if Newton's method is in there. What is the title of your textbook?
Holt McDougal Larson Algebra 2
Can you find "Newton's method" in the index?
On which chapter are you working?
Is it paper home-work or online?
its a take home test and i couldnt find it
As matricked pointed out, IF -16x^2 had been +160x^2, then there are rational roots obtainable by grouping. 5x^5-20x^4-40x^3+160x^2-45x+180 =5(x^5-4x^4-8x^3+32x^2-9x+36) \(= 5( x^4(x-4) -8x^2(x-4) -9(x-4)) \) \(= 5(x-4)(x^4-8x^2-9) \) \(= 5(x-4)(x^2-9)(x^2+1) \) \(= 5(x-4)(x+3)(x-3)(x^+1) \)
*last factor is \( (x^2+1) \)
ok do you solve for them now
hold on a second, im sorry