Question

please hellp i'm begging
Find all of the zeros of the equations

4) x3 – 5x2 + 6x = 0
5) x4 – 8x2 – 9 = 0
6) 5x3 – 5x = 0

Answer 1

(4) has x in each term of the left side, so factor out x. What's left? How would you go about factoring x^2 - 5x + 6?

Answer 2

what is your question exactly

Answer 3

I'm asking you to factor x out of x^ – 5x^2 + 6x = 0. You'll be left with a factor that's a quadratic expression. How would you factor that expression?

Answer 4

um i don't know :'( i take cyber school and math is the hardest thing ever online

Answer 5

\[x^3-5x^2+6x = 0\]All of the terms have \(x\) in them, so you can factor out \(x\). What do you get when you do that?

Answer 6

(x-3)(x-2)? i have no idea

Answer 7

What is \(x^3/x\)? What is \(-5x^2/x\)? What is \(6x/x\)?

Answer 8

whpalmer will help you pkay

Answer 9

wait how do i figure that out?

Answer 10

ask him.. i have no idea etheir

Answer 11

\[\frac{ x^3−5x^2+6x=0 }{ x }\] may be a little easier to understand.

Answer 12

If you have \(xa + xb + xc\), that's the same as \(x(a+b+c)\), right?

Answer 13

Here you are asked to divide each term of the equation by x.

\[\frac{ x ^{3} }{ x }=x ^2\]. This is precisely what whpalmer4 is asking you to do.

Answer 14

exponents confuse me

Answer 15

You can write that polynomial as\[x*x^2 - x*5x+x*6\]or\[x(x^2-5x+6)\]

exponents are just a shorthand. \[x^2 = x*x\]\[x^3 = x*x*x\]\[x^1 = x\]The only moderately tricky one is \[x^0 = 1, \text{ as long as }x\ne0\]

Answer 16

So the division of \[\frac{x^3}{x} = x^2\] is the same as \[\frac{x*x*x}{x} = \frac{\cancel{x*}x*x}{\cancel{x}} = x*x = x^2\]

Answer 17

ohh okay

Answer 18

So, we can factor out that \(x\) from the polynomial:
\[x^3-5x^2 + 6x = x(x^2-5x+6)\]

Now we want to factor \(x^2-5x+6\)

We'll have two product terms:
\[(x+a)(x+b) = x*x + b*x + a*x + a*b = x^2 + (a+b)x + a*b\]

Now we need to pick two numbers \(a,b\) to make \[x^2-5x+6\]match with \[x^2 + (a+b)x + a*b\]

For that to be true, we need \(a+b = -5\) and \(a*b = 6\)

Can you think of a pair of numbers that makes that true?

Answer 19

that negative throws me off because it could be a= 1 b=-6 but then when you multiply it wouldn't it make 6 negative?

Answer 20

here are the ways we can get a*b = 6

1*6
2*3
3*2
6*1
-1*-6
-2*-3
-3*-2
-6*-1

do any of those pairs add up to -5?

Answer 21

the -2 and -3 ?

Answer 22

i'm sorry it is midnight here and my brain is not working haha

Answer 23

right, you got it!

\[(x-2)(x-3) = x*x - 2x - 3x -2(-3) = x^2 - 5x + 6\]
So the first equation factors to
\[x(x-2)(x-3) = 0\]
What does that mean? What are the values of \(x\) that can make that true?

Answer 24

hmm

Answer 25

Well, if I tell you that I've multiplied two numbers together, and the result is 0, does that suggest anything about either number?

Answer 26

one of them is 0

Answer 27

Bingo!!!

So, if we have three numbers multiplied together, and the result is 0, that means that any one of the 3 might be 0, right?

Answer 28

WHP: Great work!!

Answer 29

\[x(x-2)(x-3)= 0\]means that either \[x=0\]\[x-2=0\]\[x-3=0\]What are the values of \(x\) (the zeros) which make those equations true? The first one is pretty easy :-)

Answer 30

wait so i'm just plugging in numbers to make them equal zero?

Answer 31

or solving.

\[x = 0\]Well, that's easy to solve :-)
\[x - 2 = 0\]add 2 to both sides\[x-2+2 = 0+2\]\[x = 2\]
\[x-3=0\]add 3 to both sides\[x-3+3 = 0+3\]\[x=3\]

So x = 0, x = 2, x = 3 are all zeros of that function, because they make the value of the function be 0.

For example:
\[x^3-5x^2+6x = 0\]Plug in x = 0:\[(0)^3-5(0)^2+6(0) = 0-0 + 0 = 0\]Plug in x = 2:\[(2)^3-5(2)^2 + 6(2) = 2*2*2-5*2*2+6*2 \]\[= 8 - 20 + 12 = 8+12 - 20 = 20 - 20 = 0\]
Plug in x = 3:\[(3)^3 - 5(3)^2+6(3) = 3*3*3 - 5*3*3 + 6*3 = 27-45+18 = 0\]

So those are the zeros of this function. Make sense?

Answer 32

pretty much yeah

Answer 33

you are like the smartest person ever haha

Answer 34

I'm guessing you might need help with the other two...

\[x^4 – 8x^2 – 9 = 0\]
Here we need two numbers that multiply to -9, and add to -8. What could they be?

Answer 35

I guess you called it a night. Sounds like a good plan.