x^3 - 12x - 16 = 0 how can i solve it?

Answer 1

Let p(x) = x^3 - 12x - 16

What you want to do now is test factors of 16 (both positive and negative). As soon as you get a factor of 16 that works, (let's call this r), we know that (x - r) will be a factor, and we use long division.

The values we want to test: 1, -1, 2, -2, 4, -4, 8, -8, 16, -16.

p(1) = 1 - 12 - 16 = -27, which is NOT zero.
p(-1) = -1 - 12(-1) - 16 = -1 + 12 - 16 = -5 (non-zero).
p(2) = 2^3 - 12(2) - 16 = 8 - 24 - 16 = 8 - 40 = -32 (nonzero)
p(-2) = (-2)^3 - 12(-2) - 16 = -8 + 24 - 16 = 0

Since -2 is a root, (x - (-2), or (x + 2) is a factor.

Now we do long division; x + 2 into x^3 - 12x - 16. Without showing you the details of the synthetic long division, you should get a quotient of (x^2 - 2x - 8). Therefore

x^3 - 12x - 16 = 0
(x + 2) (x^2 - 2x - 8) = 0
(x + 2) (x - 4) (x + 2) = 0
(x + 2)^2 (x - 4) = 0

Therefore, we equate each of these to 0, and end up with
x = {-2, 4}

Answer 2

Any root of this equation must be a factor of
16. So try all the possible divisors of 16.
You will find that 4 is a root, so x -4 is a
factor.
Dividing x³-12x-16 by x-4 gives x²+4x+4.
Setting x²+4x+4 = 0 and solving yields x = -2 as a
double root.
So the roots are x = -2(double root) and x = 4.

Answer 3

(x - 4)(x+2)(x+2) = 0

what I do when I have to solve for x^3
is that I replace x in the equation by the numbers like 1, -1, 2, -2 etc
when the number I use turns the equation = to 0 then I put this number in factor
i.e.
let's say for the above equation if I use x = -2
(-2)^3 - 12(-3) -16 is = 0 so (x+2) is a factor
etc

Answer 4

One way is just use a table of values to plot points. The given is not that complex so this is the easiest method to use. This is cubic equation so you expect to get a maximum of three roots. Easily using values of x from -3 to +5 or just -2,0, 1, 4, you will find that at x= -2 and x=+4 are solutions or roots. The choice of table values is further limited by the knowlege of using only factors of 16.

By the way, the -2 occurs 2x ( see factoring and actual graph which shows that at -2 the graph DOES NOT cross the x-axis or the graph just touches the x-axis)

Another way is using a scientific calculator. Press the Y= button to enter your equation and then Graph. Trace for x zeroes and you should get the same values of x.

The third way is using the concept of factoring and factors. This is discussed by 2 previous responders.

Answer 5

Factor.

x^3-12x-16=0

x^3-12x-16 = (x-4)(x+2)^2

x-4 =0
x+2 = 0

Solve for x.

Answer 6

x^4 + sixteen = 12x^2 enable u = x^2 u^2 + sixteen = 12u u^2 - 12u + sixteen = 0 u^2 - 12u + 36 + sixteen - 36 = 0 (u - 6)^2 - 20 = 0 (u - 6 + sqrt(20)) (u - 6 - sqrt(20)) = 0 so u = x^2 = 6 - sqrt(20) or u = x^2 = 6 + sqrt(20) as a result x = + sqrt( 6 - sqrt(20)), - sqrt( 6 - sqrt(20)) or + sqrt( 6 + sqrt(20)), - sqrt( 6 + sqrt(20))

Answer 7

Use polynomial division.

I get (x-4) (x2+4x+4)

Now you have one integer root, and the other term can be solved with the quadratic formula.

I get a double root of -2, and another root of 4.

Answer 8

But 16 on the other side of the equation and take the cubed root of it. It should be simple from there.

Answer 9

x<3-12x=16
its easy from here

Answer 10

you have to factor from (x+2)
then you have (x+2)(x-4)=0
so your answer is x=-2 & 4