Integral Zero Theorm and Factoring High-Degree Polynomials

Hi everyone,it's Ashriti
On Wednesday,we were taught the Integral Zero Theorem.
This theorem basically helps you know which integer values of a to try when determining if p(a)=0.Consider the polynomial x^3+3x^2-6x-8.If x=a satisfies p(a)=0,then x^3+3x^2-6x-8=0 or x^3+3x^2-6x=8.Factoring out the common factor in the left side of the equation gives the product x(x^2+3x-6)=8
Then,the possible integer values for the factors in the the product on the left side are factors of 8.They are +_1,+_2,+_4 and +_8.
So,if x-a is a factor of a polynomial function p(x) with integral coefficients,then  a is factor of the constant term p(x).
Example:
Factor   f(x)=x^3+3x^2-6x-8
f(x)=x^3+3x^2-6x=8
The possible integral zeros of the polynomial are:+-1,+-2,+-4,+-8
Now,first try it with the value of -1
f(-1)= (-1)(-1)(-1)+3(-1)(-1)-6(-1)-8
f(-1)= -1+3+6-8
f(-1)=0
So, (x+1) is one of the factors of f(x).
Now divide (x+1) by f(x)
On dividing it we will get x^2+2x-8
And then factor this again 
On factoring it we get:(x-2)(x+4)

So,f(x)=(x+1)(x-2)(x+4)

Factoring High-Degree Polynomials

When factoring high degree Polynomials we have to take the x-value common and move the constant over to the other side of the equation(on the right side of=).After this check for the possible values of the factor.It basically reduces the number of options for the factor.

Example:

Factor:
P(k)=k^5+3k^4-5k^3-15k^2+4k+12
P(k)=k^5 +3k^4-5k^3-15k^2+4k= -12
The possible factors are:+- 1,2,3,4,6,12
Now,P(1)=(1)(1)(1)(1)(1)+3(1)(1)(1)(1)-5(1)(1)(1)-15(1)(1)+4(1)+12
P(1)= 1+3-5-15+4+12
P(1)=0
So, (k-1) is a factor of f(k).
Now, divide f(k) by (k-1)
After dividing we will get k^4+4k^3-k^2-16k-12
Now,factor this again
(k+1) is one of the factors of above equation.
After factoring this we get (k+1)(k-2)(k+2)(k+3)
So, the given equation:
f(k)= (k-1)(k+1)(k-2)(k+2)(k+3)

I choose number 11 to blog next.