How do I algebraically find the x-intercepts for this equation y = x^3 - 3x^2 + 3.
I know I need to plug in y = 0 to solve for x.
x^3 - 3x^2 + 3 = 0
But where do I go from there? I don't think I can factor, and I don't think I can use the quadratic formula.
It is actually very simple. To solve an equation of the form:
x^3 + a x^2 + b x + c = 0
you follow the following steps.
First, get rid of the quadratic term
a x^2 using the substitution:
x = y - a/3
This step is analogous to how you solve the quadratic equation when you write it as a perfect square.
In this case we aren't finshed yet as the equation now is of the form:
y^3 + p y + q = 0
How do we solve this equation?
Consider the identity:
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
We can rewrite the right hand side as follows:
a^3 + 3a^2b + 3ab^2 + b^3 =
a^3 + b^3 + 3ab(a+b)
So, we have:
(a+b)^3 = 3ab(a+b)+ a^3 + b^3 ---->
(a+b)^3 - 3ab(a+b) - [a^3+b^3] = 0
Of course, this equation is always satisfied whatever values for a and b you substitute in it. Now, if we choose a and b such that:
3ab = -p (1)
a^3 + b^3 = -q (2)
Then:
(a+b)^3 +p(a+b) + q = 0
so y = a + b is then the solution we are looking for.
So, it all boils down to solving equations (1) and (2) for a and b.
If you thake the third power of (1):
27 a^3b^3 = -p^3
So, if we put A = a^3 and B = B^3, we have:
A*B = -p^3/27
A + B = -q
If you eliminate, say, B , you get a quadratic equation for A which you know how to solve. You then only need to pay attention when you extract the cube roots to find a and b. Equation (1) has to be satisfied, which means that if you choose one of the three possible complex cube roots for A, the root for B is fixed.
So, let's see how this works out for in your case:
x^3 - 3x^2 + 3 = 0
We put x = y + 1, which gives after some simple algebra:
y^3 - 3y + 1 = 0
So, p = -3 and q = 1
We need to solve:
A*B = -p^3/27 = 1
A + B = -q = -1
The second equation says that
B = -(1+A)
Substituting in the first gives:
-A^2 - A = 1 --->
A^2 + A + 1 = 0 ----->
A = -1/2 ± i sqrt[3]/2
We have two solutions because interchanging A and B will give you another solution. But if you do that the other solution for A will be the first solution for B. So we can take:
A = -1/2 + i sqrt[3]/2
B = -1/2 - i sqrt[3]/2
We can rewrite this as:
A = cos(4 pi/3) - i sin(4 pi/3)
B = cos(4 pi/3) + i sin(4 pi/3)
Which we can also write as:
A = exp(-4 pi i/3)
B = exp(4 pi i/3)
Which means that:
a = exp(-4 pi i/9 + 2 pi i n/3)
b = exp(4 pi i/9 - 2 pi i n/3)
Note that the product of a and b must be 1, the integer n which selects which of the three complex cube roots we choose for A fixes the cube root for B.
The solution is thus:
y = a + b =
exp(-4 pi i/9 + 2 pi i n/3)
+ exp(4 pi i/9 - 2 pi i n/3) =
2 cos(4 pi/9 + 2 pi n/3)
And
x = y + 1 = 2 cos(4 pi/9 + 2 pi n/3)
The three solutions are thus:
x = 2 cos(4 pi/9) + 1
x = 2 cos(10 pi/9) + 1
x = 2 cos(2 pi/9) + 1
I guess we will have to find one of the roots by brute force.
We are looking for zeros of f(x) for some x
Make a table to try to find the zeros
for example
x f(x)
0 +3
1 +1
2 -1
3 +3
That is interesting, f(x) goes through zero at least once between x = +1 and x +2. Try x = 1.3
1.3 +1.27
1.4 -.136
Oh, so between 1.3 and 1.4
Try 1.33
1.33 +.045937
1.34 +.019304
1.35 -.007125
Humm, I am not going do go any further. That is close enough for now. The function crosses the x axis at x is about 1.35
Now what. I need to factor that root out so I am left with a quadratic or else continue graphing and calculating to find the other two roots.
To factor it out, note that (x-1.35) is a factor (to our level of approximation) so
(x-1.35)(a x^2 + b x +c) = x^3-3x^2+3
so we need to divide the right by (x-1.35)
I get
x^2 - 1.65 x - 2.2275
solve that quadratic to find the other two roots.
However I bet the problem is a typo unless you have covered this stuff in class.
Wow... that's a long process. Thanks for the help!
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To algebraically find the x-intercepts for the equation y = x^3 - 3x^2 + 3, you're on the right track by setting y = 0 and solving for x.
So you have x^3 - 3x^2 + 3 = 0
One possible way to solve this equation is by using synthetic division or long division to find one root, and then factorizing the remaining quadratic equation to find the other two roots.
Let's start by using synthetic division to find one root. Synthetic division is a shortcut method for dividing a polynomial by a binomial of the form (x - r), where r is a root of the polynomial.
First, to make things easier, let's rewrite the equation as 0 = x^3 - 3x^2 + 3:
1 | 3 -3 0
|_________
| 3 0
|_________
3 0
Using synthetic division, we divide (x^3 - 3x^2 + 3) by (x - 1) since we are looking for the value of x that makes y equal to zero.
We have obtained two terms, 3 and 0. The last term, 0, represents the remainder when dividing (x^3 - 3x^2 + 3) by (x - 1). This means that (x - 1) is a factor of the equation and x = 1 is a root.
Now that we have found one root, x = 1, we can divide the original cubic equation by (x - 1) using long division or synthetic division to obtain a quadratic equation.
Performing long division or synthetic division, you will find that:
(x^3 - 3x^2 + 3) / (x - 1) = x^2 - 2x - 3
So we have reduced the original equation to the quadratic equation x^2 - 2x - 3 = 0.
Now it's easier to solve this quadratic equation by factoring or using the quadratic formula. In this case, you can factor the quadratic equation as:
(x - 3)(x + 1) = 0
Setting each factor equal to zero, we get:
x - 3 = 0 or x + 1 = 0
Solving for x, we find:
x = 3 or x = -1
Therefore, the x-intercepts of the equation y = x^3 - 3x^2 + 3 are x = 1, x = 3, and x = -1.
Remember, synthetic division or long division is used to find one root, and then you factorize the resulting quadratic equation to find the remaining roots.