Solve the following equation by factoring using rational zeros to find the factors.?

Question

(Hint: Use synthetic Division to simplifiy the process.)
x^3 + x^2 - 4x -4 = 0

Answer 1

No need to use synthetic division really, if you notice that there are two pairs of similar coefficients:

x^2(x + 1) - 4(x + 1) = 0
(x^2 - 4)(x + 1) = 0
(x + 2)(x - 2)(x + 1) = 0

x = 2, -2, -1

Answer 2

OK, the procedure consists of these steps:

1. Sort your equation from higher to lower degree: You already have it: x^3 + x^2 - 4x -4 = 0

2. Find positive and negative integer divisors for the first coefficient (the one of the variable with highest degree) and last coefficient (the numeric value). In this case, first coefficient is 1 and last is -4.

Its factors are:
1: 1, -1
4: 1,-1, 2,-2, 4,-4
Notice you find all the divisors and you set them positive AND negative. You should find all possible integer divisors.

3.If you had this equation:
(x+a/b)(x+c/d)(x+e/f)=0... and you expanded it, you would get something like this: x^3+...something... +ace/bdf=0
This shows that information for rational roots is contained in the first and last coefficient. And for finding them, we have to try all possible combinations of the divisors we found above:
make all possible combinations of the divisors of the last number divided by the divisors of the first number: (you can make a table)

1 -1 2 -2, 4 -4
----+---------------------------
1 | 1 -1 2 -2 4 -4
-1 | -1 -1 -2 2 -4 4

Note the cells of the table are the divisions of each divisor of 4 by each divisor by 1.

4.You can get many repeated numbers with this method but doesn't matter: it's safer than taking the risk to forget one! So now we only keep unrepeated numbers:
1,-1,2,-2,4,-4 . These are the original divisors of 4, but don't get confused. That is because the first coefficient (from x^3) was 1. If it were different you could get something else.

5. Finally, you test the numbers above with synthetic division. When you get a 0 in the remaining, that means you got one root!. Remember you can get a maximum of 3 roots because your equation is 3th-degree.

When trying the numbers we can find -1, 2, and -2 are the rational roots.

Answer 3

(x-2)(x+1)(x+2)=0
x-2=0 or x+1=0 or x+2=0
x=2 or x=-1 or x=-2

Answer 4

(x - 2)(x^2 + 3x + 2) = 0
(x - 2)(x + 1)(x + 2) = 0

Answer 5

i think u have to use the quadratic equation since the numbers are in y=aX^2+bX+c the quadratic formula= X=-b+/-square root of b square- 4ac over(divided) by 2a or u could factor them

Answer 6

(-2 + x) (1 + x) (2 + x)=0
x=-2,-1,2