Question

find all the zeros of the equations. please show work
x^3-5x^2+6x=0
x^4-8x^2-9=0
x^3-125=0

Answer 1

x^3-5x^2+6x=0

x(x^2 - 5x + 6) = 0

x(x - 3)(x - 2) = 0

I'll let you finish up

Answer 2

x^4-8x^2-9=0

z^2-8z-9=0 ... let z = x^2, so z^2 = x^4

(z - 9)(z - 1) = 0

(x^2 - 9)(x^2 - 1) = 0

(x-3)(x+3)(x-1)(x+1) = 0

again I'll stop here to let you finish

Answer 3

x^3-125=0

x^3-5^3=0

(x-5)(x^2 + 5x + 25) = 0 ... use the difference of cubes factoring rule

stopping here

Answer 4

what would i have to do to finsh it up?

Answer 5

you would use the zero product property to go from

x(x - 3)(x - 2) = 0

to

x = 0, x-3=0 or x-2=0

then you would solve each equation for x

Answer 6

im not sure with the zero product proterty

Answer 7

the zero product property is the idea that if A*B = 0 then either A = 0 or B = 0 (or both could be zero)

Answer 8

what would i need to do to solve for x?

Answer 9

x(x - 3)(x - 2) = 0

x = 0, x-3=0 or x-2=0

x=0, x=3, or x = 2

Answer 10

see how I'm getting this?

Answer 11

I really appreciate your help. I am really trying hard to understand, but it is so confusing.

Answer 12

basically if I told you two numbers multiply to 0

then at least one of them must be zero

Answer 13

so that's how something like x times (x-3) = 0 turns into x = 0 or x-3 = 0

Answer 14

ok thank you for your help

Answer 15

np

Answer 16

could you help me with another problem please?

Answer 17

Determine the number and type of complex solutions and possible real solutions for 2x^4+x^2-x+6=0?

Answer 18

have you ever heard of descartes rule of signs?

Answer 19

no i have not, sorry

Answer 20

ok have a look at this page

http://www.purplemath.com/modules/drofsign.htm

and tell me if it helps or not

Answer 21

i read the page and i really didn't understand anything at all

Answer 22

ok I'll try my best to explain the idea

Answer 23

2x^4+x^2-x+6=0

is the same as

+2x^4+1x^2-1x+6=0

notice from the first term +2x^4 to the second term +1x^2, the coefficients are +2 and +1 respectively

Answer 24

thank you so much, im failing my alg class

Answer 25

so there is NO sign change in the coefficients

Answer 26

However...

the jump from 1x^2 to -1x has a sign change happening since you go from a positive coefficient (+1) to a negative coefficient (-1)

Answer 27

see the sign change I'm pointing out?

Answer 28

yeah i see the sign change

Answer 29

ok great

there is another sign change: as you go from -1x to +6, there is a sign change from negative to positive

Answer 30

in total, there are 2 sign changes

Answer 31

which means that there are a maximum of two positive roots

Answer 32

I'm just using the rule described on that page I sent you

Rule: if there are n sign changes, then there are a maximum of n positive roots

Answer 33

with me so far?

Answer 34

yes

Answer 35

ok let f(x) = 2x^4+x^2-x+6

we need to find f(-x) now


f(x) = 2x^4+x^2-x+6

f(-x) = 2(-x)^4+(-x)^2-(-x)+6 ... replace every 'x' with '-x'

f(-x) = 2x^4+x^2+x+6 ... simplify


Notice how there are NO sign changes in f(-x) = 2x^4+x^2+x+6 since everything is positive and there are no negative terms

Answer 36

so this means that there are NO negative real roots

Answer 37

to sum up so far

f(x) = 2x^4+x^2-x+6 has at max two real positive roots

f(x) = 2x^4+x^2-x+6 has NO real negative roots

Answer 38

Now if f(x) = 2x^4+x^2-x+6 had 2 real positive roots, then you would need 2 more complex roots (because you need 4 roots total)


if f(x) = 2x^4+x^2-x+6 had 1 real positive root, then you would need 3 more complex roots (because you need 4 roots total)...but complex roots ALWAYS come in pairs...so this scenario isn't possible


if f(x) = 2x^4+x^2-x+6 had 0 real positive roots, then you would need 4 more complex roots (because you need 4 roots total)

Answer 39

it's definitely a lot to take in, so read it over and ask about anything that stumps you

Answer 40

I guess you could sum it up like this

f(x) = 2x^4+x^2-x+6 has at max two real positive roots
f(x) = 2x^4+x^2-x+6 has NO real negative roots
f(x) = 2x^4+x^2-x+6 has at most 4 complex roots


The number of real and complex roots must add to 4

Answer 41

thank you so much

Answer 42

sure thing