SOLUTION: Let f(x)=x^4+2x^3-7x^2-20x-12 If -2 is a zero, factor f(x) completely - please show how to factor it as well.

Algebra ->  Equations -> SOLUTION: Let f(x)=x^4+2x^3-7x^2-20x-12 If -2 is a zero, factor f(x) completely - please show how to factor it as well.       Log On


   



Question 1043460: Let f(x)=x^4+2x^3-7x^2-20x-12
If -2 is a zero, factor f(x) completely
- please show how to factor it as well.

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Expect the remainder for checking the given root to be 0.
-2   |   1   2   -7   -20   -12
     |
     |      -2    0    14    12
     |________________________________
        1   0    -7    -6    0

This factorization shows f(x)=(x+2)(x^3+0x^2-7x-6) and other factors to check would be -1,-2,-3,1,2,3.

Carry on with synthetic divisions, and you will find the roots to be, for that cubic factor, -2,-1,3.
f%28x%29=%28x%2B2%29%5E2%28x%2B1%29%28x-3%29

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let f(x)=x^4+2x^3-7x^2-20x-12
If -2 is a zero, factor f(x) completely
- please show how to factor it as well.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~




Figure. Plot f(x) = x%5E4%2B2x%5E3-7x%5E2-20x-12

Let's apply grouping method.

f(x) = x%5E4%2B2x%5E3-7x%5E2-20x-12 = %28x%5E4%2B2x%5E3%29+-+%287x%5E2+%2B20x+%2B12%29 = 

= %28x%5E3%28x%2B2%29%29 - %287x%5E2+%2B+14x%29+%2B+%286x%2B12%29 = 

= %28x%5E3%28x%2B2%29%29 - %287x%28x+%2B+2%29%29+%2B+6%28x%2B2%29 = 

= %28x%2B2%29%2A%28x%5E3+-+7x+-6%29 =       (now check the polynomial x%5E3+-7x+-6 for the roots. Check for x =1. 
                             See how I will extract this root by applying grouping again)


= %28x%2B2%29%2A%28%28x%5E3+-x%29+-+%286x%2B6%29%29 = %28x%2B2%29%2A%28x%28x%5E2-1%29+-+6%28x%2B1%29%29 = %28x%2B2%29%2A%28x%28x-1%29%2A%28x%2B1%29+-+6%28x%2B1%29%29 =

= %28x%2B2%29%2A%28x%2B1%29%2A%28x%2A%28x-1%29-6%29 = = %28x%2B2%29%2A%28x-1%29%2A%28x%5E2+-+x-6%29 =

     Now you have to find the roots of the quadratic polynomial in parentheses.
     You can do it by many ways. Its roots are 2 and -3. Therefore

=  %28x%2B2%29%2A%28x-1%29%2A%28x%2B2%29%2A%28x-3%29 =    (so you finally get)

=  %28x%2B2%29%5E2%28x-1%29%2A%28x-3%29.

Now identify the roots of the polynomial.
Check the correspondence between the roots and the plot in the figure.


What method did I apply?  - The grouping method.

What method did I apply?  - The grouping method.

See also the lesson
    - Solving polynomial equations of high degree by factoring
in this site.