Question

find the zeros of f(x)=3x^3-x^2-2x ,f(x)=X^3-25x ,f(x)=7(x-3)^2(x+5)^4

Answer 1

1. Factor
2. Set f(x) = 0
3. Solve

0 = x(x^2 - 25)
0 = x(x + 5)(x - 5)

0 = x
0 = x + 5
0 = x - 5

0 = 7(x - 3)^2(x + 5)^4
0 = (x - 3)^2(x + 5)^4

0 = (x - 3)^2
0 = (x + 5)^4

0 = x - 3
0 = x + 5

Answer 2

Basically, I helped you with the last two

Answer 3

By the way, just in case you were wondering
The square root of zero is zero
The quad root of zero is zero

Answer 4

thanks a lot you are really my hero :D

Answer 5

For the first one:

0 = x(3x^2 - x - 2)
0 = x(3x^2 - 3x + 2x - 2)
0 = x(3x(x - 1)+2(x - 1))
0 = x((x - 1)(3x + 2))
0 = x(x - 1)(3x + 2)

Answer 6

Once you have completely factored, you can set each factor equal to zero, then solve for x.

Answer 7

on the second one what happen to the 7?

Answer 8

We have an equation. In an equation if you have something like

0 = 7x

You can divide both sides by 7 to get
x = 0

If you have something like
0 = 7(x + 4)(x - 3)

It is still legal to divide both sides by 7 to get
0 = (x + 4)(x - 3)

Answer 9

Remember, 0/7 = 0

Answer 10

You can divide zero by any number other than zero and you will get zero. What you cannot do is divide by zero.

Answer 11

I hope that doesn't confuse you

Answer 12

ohhh ok thanks and nop im not confused, yet

Answer 13

Okay good.

Answer 14

thanks a lot you are truly a hero