Find all zeros of p(x). Include any multiplicities greater than one.

Question

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

 

 

I solved the problem but I'm almost positive I did it wrong. Can someone please explain how to solve it? Thanks. :)

Answer 1

Hello, Michelle,

 

My approach to this problem is through use of the factor theorem.

 

p(x) = x^3 – 3x^2 + 4x – 12

Finding the zeroes of the function p = 0 means that p(x) = 0, and therefore, x^3 – 3x^2 + 4x – 12 = 0
We would have a line of the coefficients 1, -3, 4, and -12;
Next, place the 3, representing the factor (x-3):

 

3 1 -3   4 -12 for x^3 – 3x^2 + 4x – 12
       3   0  12
   1  0   4    0 for x^2 + 0 x^1 + 4, or x^2 + 4

From here, we established that the factor (x - 3) is a factor and that (x^2 + 4) is a factor.

 

x - 3 = 0          x^2 + 4 = 0

x = 3

From this latter part, we have something similar to the special polynomial or quadratic product x^2 - 4. However, this product is not one of the special products we know, nor do we have this factor as a part of our solution.

From this, we subtract 4 on both sides, which implies that

x^2 = -4; x = √(-4); x = ±2i

([-2]i)^2 = ([-2]^2)(i^2) = (4)(-1) = -4
(2i)^2 =         (2^2)(i^2) = (4)(-1) = -4

 

Therefore, we have two imaginary solutions ±2i

Remember: Every imaginary or complex solutions come in pairs.  The imaginary part is in the form x±c, where c , in our case, is ±2i, where the real part is 0: 0±2i

⇒ (implies [that]) x = {0-2i, 0+2i, 3}, or x = {-2i, 2i, 3}.

 

Remember also that IF x is only in the REAL numbers, then ±2i would not be a part of the solution.  That is, if x is in only the set of real numbers, and x = {3}.

Answer 2

To find the zeros you need to find all values of x for which y=0

    Do this by completely factoring the function,

    then setting each of the factors equal to 0

    solve for the x's and these are the zeros of the function.

 

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

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

      = x^2 (x-3) + 4(x-3)  factor out x-3

      = (x-3)(x^2 + 4)

 

SO either (x-3)=0  or  (x^2+4)=0

giving x=3 for a real root,

and x^2= –4   giving  x = √(–4) for imaginary roots

 

Answer 3

Michelle, it looks like you could factor out x^2 from the first two terms and 4 from the last two terms in which the function becomes:

 

p(x)=x^2(x-3)+4(x-3) This implies

p(x)=(x^2+4)(x-3)

 

Thus, there is only one real zero that occurs at x=3 and there is only one multiplicity at this value.

 

Hope this helps.