find the zeros of this function?

Question

Given the function f defined by f(x)= 2x^3 - 3x^2- 12x +20; Find the zeros of f

I don't know how do do this can someone please help me?

Answer 1

Use the rational root theorem which states that if a polynomial of degree k with the form f(x) = px^k+...+q has any rational roots, then they must of the form a/b where a is a factor of q (the constant) and b is a factor of p (the leading coefficient).

So, all the factors of 20 (the q) (both positive and negative) are
-20, -10, -5, -4, -2, -1, 1, 2, 4, 5, 10, 20
All the factors of 2 (the p) are
-2, -1, 1, 2

Which means that the list of all the possible rational roots includes
-20, -10, -5, -4, -5/2, -2, -1, -1/2, 1/2, 1, 2, 5/2, 4, 5, 10, 20

So you simply plug all of them in one by one and see which one gives you a zero. If all three of the roots are (distinct) rational roots, then you can find them here. Otherwise, what we usually do is find only one root, divide it by the factor after which you will have a quadratic equation, and then you simply use the quadratic formula. So that is the standard method. Reduce the polynomial's degree by one (if it is still more than 2, then you have to do the whole cycle again of finding the rational roots and pluggin them in, you cannot use the same numbers, because the possibilities might change, the roots obviously won't change) and then when you have a quadratic, use the quadratic formula.

So for your polynomial, only two of the possibilities are going to work, x=2 or x=-5/2. Now we know for sure that one of these two roots repeats but we can't tell which one. So we divide (by both of the roots together because we already know two of them)
2x^3-3x^2-12x+20
---------------------- = x-2
(x-2)(2x+5)
which means that the third root is x=2.


P.S.Tom has it wrong underneath.
f(-1)=27
f(1/2)=27/2
They are not the roots.
The roots for f(x) are x=-5/2, 2, 2.
Which means that f(x)=(x-2)^2(2x+5).
If you don't believe me then just expand the factorization and you are going to get your original f(x).

Answer 2

When finding the roots of any polynomial of the form (a_0)*x^n + (a_1)*x^(n-1) + ... + (a_(n-1))*x + (a_n), where (a_i) is some constant in front of some power of x, each root must be a factor of (a_n) divided by a factor of (a_0). In your case, each root must be a a factor of 20 (which factors into 20, 10, 5, 4, 2, 1)divided by a factor of 2 (which factors into 2 and 1). Since you know each polynomial of degree 3 (leading term is x^3) has 3 roots, you can simply plug and chug using +/- of each of the numbers 20, 10, 5, 4, 2, 1, 5/2, 1/2. Hint: once you find one root, you can use synthetic division to divide the polynomial by that root. Then, you will be looking for 2 more factors of a polynomial of degree 2. For instance, first I found that f(x) was divisible by 2. So, I divided f(x) by (x-2) to get 2x^2 + x -10. I then used the quadratic formula to find the last two roots of f(x). Be careful of double roots; in your case, f(x) had a double root of 2. Your final answer is that 2 is a double root and -5/2 is the other root. (You may have to read through what I said a couple times to understand it all; sorry it's so jumbled- I'm in a hurry.)

Answer 3

Zeroes are: 2, -1, 1/2

You calculate the zeroes by finding values of x such that f(x) = 0. You can get clues to these by examing the constant term, in this case 20. Start with positive and negative factors.

2x^3 - 3x^2- 12x +20 = (x-2)(2x^2+x-10)

You can find the roots of (2x^2+x-10) by using the quadratic formula.

Answer 4

Wow, busy work from Alg 2. You know, the funny thing is, any other time you'll just use a calc for this. Obviously, they want to see if you can factor a cubic. There's a formula for factoring a cubic function somewhere in your book. However, beyond alg 2, nobody ever uses it. So it's busy work. Here's the short cut; grab your graphing calculator and look for when the function hits the x-axis.

The x-intercepts ARE the ZEROs. Another name for 'zeros' is 'the solutions'.

Answer 5

The zeroes are where f(x) equals zero, not where x equals zero, so brenmore is off track but Tom is leading you in the right direction.

I factored completely and got (x-2)(2x+5)(x-2) for 2, -5/2, and 2 again

Answer 6

I think this is what you are looking for:

f(x)= 2x^3 - 3x^2- 12x +20
When the function = 0, then:
f(0) = 2(0)^3 - 3(0)²- 12(0) +20
f(0) = 20.

Answer 7

by trial and error f(2) = 0
dividing f(x) by x-2 (use synthetic division)

f(x) = (x-2)(2x^2+x-10)
think of two numbers whose product is {constant*coefficient of x^2}(2*-10) that is (-20) and sum is 1{coefficient of x}
numbers are -4 and 5. use them to split the x term

f(x) = (x-2) [ 2x^2 -4x + 5x -10]

f(x) = (x-2) [2x(x-2)+5(x-2) ]

f(x) = (x-2)(x-2)(2x+5)

zero's are 2, 2, (5/2)

Answer 8

the easiest way to do this is to plug it into a graphing calculator and find the zeros visually.

Answer 9

2,1/2 and -1