# Thread: Real Zeros of a Function

1. ## Real Zeros of a Function

Well, I have searched all over the internet for a simple explanation for how to

"find all real zeros of the function: f(x) = 2x^3 + 4x^2 - 2x - 4"

but to no avail. I knew how to do this at some point, and I don't remember it being that hard, but I think my mind erased it. Help please?

2. $\displaystyle \begin{array}{rcl} 2x^3 + 4x^2 - 2x - 4 & = & 0 \\ 2x^2 \left( {x + 2} \right) - 2\left( {x + 2} \right) & = & 0 \\ 2\left( {x + 2} \right)\left( {x^2 - 1} \right) & = & 0 \\ \end{array}$.

3. Hello, iGuess!

Find all real zeros of the function: .$\displaystyle f(x) \:= \:2x^3 + 4x^2 - 2x - 4$
The procedure is simple: .Solve $\displaystyle f(x) = 0$

The algebra may take some work . . .

We have: .$\displaystyle 2x^3 + 4x^2 - 2x - 4 \:=\:0$

Divide by 2: .$\displaystyle x^3 + 2x^2 - x - 2\:=\:0$

Factor by "grouping": .$\displaystyle x^2(x + 2) - (x + 2) \:=\:0$

. . and we have: .$\displaystyle (x +2)(x^2 - 1)\:=\:0\quad\Rightarrow\quad (x + 2)(x - 1)(x + 1)\:=\:0$

Therefore: .$\displaystyle x \:=\:-2,\:1,\:-1$

4. Okay, I was under the impression that zeros were basically x-intercepts. Am I completely off? So shouldn't the answer just be three y-values? Apparently I fail at math. Yay me. I mean, who WOULDN'T want to take Algebra 2 for a third time?

EDIT: Nevermind. I started posting before Soroban's post. Thank you.

5. Originally Posted by iGuess
Well, I have searched all over the internet for a simple explanation for how to

"find all real zeros of the function: f(x) = 2x^3 + 4x^2 - 2x - 4"

but to no avail. I knew how to do this at some point, and I don't remember it being that hard, but I think my mind erased it. Help please?
Factoring by grouping is probably the best method, and likely the one you were supposed to use, but here's another method anyway.

I agree with Soroban, go ahead and divide by 2 first.
$\displaystyle x^3 + 2x^2 - x - 2 = 0$

Now for something called the "Rational Root Theorem."
If any rational zeros of this polynomial exist, they will be in the form
$\displaystyle x = \pm \frac{\text{factor of 2}}{\text{factor of 1}}$
(The "2" is the constant term in the polynomial, in this case actually a -2. The "1" is the coefficient of the leading term, in this case the $\displaystyle x^3$ term.)

So all the possible rational roots of this polynomial are:
$\displaystyle x = \pm 1, \, \pm 2$

In this case running through the 4 possibilities gives you all three zeros. In the general case, say we found only one root: x = -2, for example. This means that x - (-2) = x + 2 is a factor of $\displaystyle x^3 + 2x^2 - x - 2$. You can divide the two polynomials to find that:
$\displaystyle x^3 + 2x^2 - x - 2 = (x + 2)(x^2 - 1) = 0$

So you know that the solutions to $\displaystyle x^2 - 1 = 0$ are also solutions, which you can solve by any means you like.

-Dan

6. Originally Posted by Soroban
Hello, iGuess!

The procedure is simple: .Solve $\displaystyle f(x) = 0$

The algebra may take some work . . .

We have: .$\displaystyle 2x^3 + 4x^2 - 2x - 4 \:=\:0$

Divide by 2: .$\displaystyle x^3 + 2x^2 - x - 2\:=\:0$

Factor by "grouping": .$\displaystyle x^2(x + 2) - (x + 2) \:=\:0$

. . and we have: .$\displaystyle (x +2)(x^2 - 1)\:=\:0\quad\Rightarrow\quad (x + 2)(x - 1)(x + 1)\:=\:0$

Therefore: .$\displaystyle x \:=\:-2,\:1,\:-1$

Hello Soroban, I'm having a test about this stuff in a few days, so I thought it would be better to ask you my questions on this thread rather than starting a new thread.

My problem starts right after factoring by grouping.

I understand until this part
x^2(x+2)-(x+2)=0
but then I'm totally lost in the next part when you say
"and we have (x+2)(x^2-1)"

I don't understand, when factoring by group you have two times (x+2) but then you only put it only once, I put in bold the part which confuses me, also where did you get from that
(x^2-1)?

Thank you so much for your help Soroban.

7. Originally Posted by jhonwashington
Hello Soroban, I'm having a test about this stuff in a few days, so I thought it would be better to ask you my questions on this thread rather than starting a new thread.

My problem starts right after factoring by grouping.

I understand until this part
x^2(x+2)-(x+2)=0
but then I'm totally lost in the next part when you say
"and we have (x+2)(x^2-1)"

I don't understand, when factoring by group you have two times (x+2) but then you only put it only once, I put in bold the part which confuses me, also where did you get from that
(x^2-1)?

Thank you so much for your help Soroban.
he pulled the common term out.

let's say you wanted to factor a out of ab - a, how would you do it?

ab - a = a(b - 1)

you pull the common term out and put in brackets beside it what is left after you take it out. Soroban did the same thing.

for simplicity, what if we replaced the (x + 2) with a:

we have: $\displaystyle ax^2 - a$ and we want to factor $\displaystyle a$ out.

we get: $\displaystyle a \left( x^2 - 1 \right)$ as shown above.

do you get it?

8. ## thanks

Thank you so much Jhevon for your great explanation, I think I now understand