MATHHH help?

Question

DETERNMINE ZEROS?
whattt is this and how do you do these


y=x(power of 5)-9x³






y=(x²+2x-15)(6x²-4x-2)

Answer 1

Zeros means the numbers you put in for x to make y be zero. You find them by factoring.

y = x^5 - 9x^3

= x^3(x^2 - 9)

= x^3 (x+3)(x-3)

so the zeros are 0, -3, and 3

Answer 2

Okay when they ask for the zeros they mean that you take y=0 and solve for x.

Ex. In the first problem you have x^5 - 9x^3 = 0

We can see that if we factor out x^3 from both we get

x^3(x^2 - 9) = 0

thus we have x^3 =0, (x-3)(x+3) = 0 and thus we can see that the zeros are 0, -3, +3

In the next problem you can consider each expression in the two separate parenthesis as almost two different problems.

You first take x^2 +2x -15 which can be factored into

(x+5)(x-3) = 0 so the zeros are x =-5, and x = +3

The other part is 6x^2 -4x -2, this is a bit more tricky but still the same idea... we can factor this into (6x + 2)(x - 1) = 0 and the zeros are 6x = -2 or -1/3 and x =+1

The final answer is thus the combination of both these seperate problems so you the zeros as x = -5, +3, -1/3 , 1

Please check my math in these but I hope you understand...Also I don't know if you understand what I mean when I factor...remember to check your factors you do F O I L!!

Answer 3

"Zeros" are the values of x that will make y = 0.

For the first one y = x^5 - 9x^3
To find the zero, x^5 - 9x^3 = 0

Notice you can factor out x^3 from both terms (it's like distributing backwards).

x^3 (x^2 - 9) = 0
Since (x^2 - 9) is a difference of squares, it equals (x-3)(x+3), so
x^3(x-3)(x+3) = 0

Notice this is read "x^3 times (x-3) times (x+3) equals 0".
If I am multiplying three numbers together and the answer is zero, the only way that is possible is if at least one of the three numbers is equal to zero.

Therefore
If x^3 = 0, then x =0. This is one solution.
If x - 3 = 0, then x = 3. This is a second solution.
If x + 3 = 0, then x = -3. This is a third solution.

It's almost bedtime. I'll let the good people of Y! Answers give you the solution to the second one.

Answer 4

The idea is that the expression on the right side of this equation is the product of several numbers. In this particular case, they want you to find the numbers which when multiplied together, make the equation equal to 0. That's why they are called zeros.

Factor out x³. Then y=x(power of 5)-9x³ = 0 = x³(x² - 9) = x³(x + 3)(x - 3).

Now, set each factor equal to 0, because any one of them could be equal to 0 to make the whole equation equal to 0.

That implies x³ = 0 ---> x = 0, or
(x + 3 ) = 0 ---> x = -3, or
(x - 3) = 0 ---> x = 3.

These are the three zeros of this equation. Any of these values will make the entire equation equal to 0. To verify, plug any one of them in for x and perform the algebra.

Answer 5

To determine the zeros of a function, set y=0 and solve for x.

1. y=x^5-9x^3
x^3(x^2-9)=0
x^3(x+3)(x-3)=0
x=0, -3, 3

Because this is a fifth-power equation, it technically has five zeros. However, 0 is a "triple zero", because x-0 (or simply x) appears as a factor three times.

2. y=(x²+2x-15)(6x²-4x-2)

(x^2+2x-15)(6x^2-4x-2)=0
(x+5)(x-3) * 2(3x^2-2x-1)=0
(x+5) (x-3) (3x+1) (x-1) = 0
x=-5, 3, -1/3, 1

P.S. How do you get the superscript 2 and 3? I know you have to hold down the Alt key and type a four digit number on the number pad, but what are they? Thanks :)

Answer 6

Factor, factor factor

y=x^5-9x^3 = x^3 (x^2-9) = x^3 (x-3)(x+3)

zero is a triple root, and 3 & -3 are also roots.

y=(x^2+2x-15)(6x^2-4x-2)
A little trickier. 15 can be divisible by1,3,5,&15. And factor out a two from the second polynomial. So,

y=2(x+5)(x-3)(3x^2-2x-1)
The second polynomial you need to play around with factors.

y=2(x+5)(x-3)(3x+1)(x+1) roots are -5,3,1/3, and -1

If you ever get into a bind remember the quadratic formula
[-b+/-sqrt(b^2-4ac)]/(2a)

Answer 7

y=x^3(x^2-9)=x^3(x-3)(x+3)

solutions:x= -3,0,3

y=(x+5)(x-3)(3x+1)(2x-2)

solutions: x=-5,-1/3,1,3