SOLUTION: For g(x)=x^3-6x^2-13x+42: a. Prove x-2 is a factor (by using the remainder theorem). b. Find the other factors. Here is what I have done so far: x^3-6x^2-13x+42 x-(2)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: For g(x)=x^3-6x^2-13x+42: a. Prove x-2 is a factor (by using the remainder theorem). b. Find the other factors. Here is what I have done so far: x^3-6x^2-13x+42 x-(2)       Log On


   

Question 1129665: For g(x)=x^3-6x^2-13x+42:
a. Prove x-2 is a factor (by using the remainder theorem).
b. Find the other factors.
Here is what I have done so far:
x^3-6x^2-13x+42
x-(2)
2|1-6-13 42
|2 -8 -42
= 1 -4 -21 0
1x^2+4x-21
x^2-4x-21= 0 Which my professor said was correct for part A.
So for B I put: There are no factors because part A is 0 and got that wrong. What did I do wrong?
Found 2 solutions by MathLover1, solver91311:
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!

For g%28x%29=x%5E3-6x%5E2-13x%2B42:
a. Prove+x-2 is a factor (by using the remainder theorem).
b. Find the other factors.
you did:
x^3-6x^2-13x+42
x-(2)
2|1-6-13 42
|2 -8 -42
= 1 -4 -21 0
1x^2+4x-21
x%5E2-4x-21 and reminder is +0+
so you did part A
and given equation can be written as: %28x%5E2-4x-21%29%28x-2%29%2B0%2F%28x-2%29
or just %28x%5E2-4x-21%29%28x-2%29
for part b. you need to find the other factors, means you already have one factor which is x-2, and you need to find factors of part a result which is x%5E2-4x-21

to factor x%5E2-4x-21, first write -4x as 3x-7x
x%5E2%2B3x-7x-21..............group
%28x%5E2%2B3x%29-%287x%2B21%29
x%28x%2B3%29-7%28x%2B3%29
%28x-7%29%28x%2B3%29-> solution for part b

now use given factor %28x-2%29 and factors from b., and write given polinomial as a product of these factors
g%28x%29=x%5E3-6x%5E2-13x%2B42=%28x-2%29%28x-7%29%28x%2B3%29


Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Your synthetic division process demonstrated the following fact:



I don't know where you got the idea that "Part A is 0". The remainder resulting from the synthetic division in Part A is, indeed, zero, but that is simply the indicator that the divisor is a zero of the dividend function and that the quotient is the set of coefficients for the other factor. So:



is a factor of the original function. However, you aren't done yet because the quadratic is factorable into, you guessed it, two factors. I'll leave that part up to you.


John

My calculator said it, I believe it, that settles it