Tuesday, August 13, 2013

Find the complex zeros of each polynomial function. write f in factored form. f(x)= x^3- 8x^2+ 25x-26

f(x)= x^3- 8x^2+ 25x-26 

working with the expression... 

x^3- 8x^2+ 25x-26=
x^3- 8x^2+12x +13x-26=
x(x^2-8x+12)+13(x-2)=
x(x-2)(x-6)+13(x-2)
(x-2)(x^2-6x+13) 
now a=1,b=-6, c=13 

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-6x%2B13+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4acb%5E2-4ac=%28-6%29%5E2-4%2A1%2A13=-16.

The discriminant -16 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -16 is + or - sqrt%28+16%29+=+4.

The solution is x%5B12%5D+=+%28--6%2B-+i%2Asqrt%28+-16+%29%29%2F2%5C1+=++%28--6%2B-+i%2A4%29%2F2%5C1+

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B13+%29

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