SOLUTION: Factor the polynomial completely, given that hte binomial following it is a factorof the polynomial. x^3+2x^2-5x-6, x+3

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor the polynomial completely, given that hte binomial following it is a factorof the polynomial. x^3+2x^2-5x-6, x+3      Log On


   



Question 120066: Factor the polynomial completely, given that hte binomial following it is a factorof the polynomial. x^3+2x^2-5x-6, x+3
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Since x%2B3 is a factor of x%5E3+%2B+2x%5E2+-+5x+-+6, x=-3 is a zero of x%5E3+%2B+2x%5E2+-+5x+-+6

So our test zero is -3


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
-3|12-5-6
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-3|12-5-6
|
1

Multiply -3 by 1 and place the product (which is -3) right underneath the second coefficient (which is 2)
-3|12-5-6
|-3
1

Add -3 and 2 to get -1. Place the sum right underneath -3.
-3|12-5-6
|-3
1-1

Multiply -3 by -1 and place the product (which is 3) right underneath the third coefficient (which is -5)
-3|12-5-6
|-33
1-1

Add 3 and -5 to get -2. Place the sum right underneath 3.
-3|12-5-6
|-33
1-1-2

Multiply -3 by -2 and place the product (which is 6) right underneath the fourth coefficient (which is -6)
-3|12-5-6
|-336
1-1-2

Add 6 and -6 to get 0. Place the sum right underneath 6.
-3|12-5-6
|-336
1-1-20

Since the last column adds to zero, we have a remainder of zero. This means x%2B3 is a factor of x%5E3+%2B+2x%5E2+-+5x+-+6

Now lets look at the bottom row of coefficients:

The first 3 coefficients (1,-1,-2) form the quotient

x%5E2+-+x+-+2


So %28x%5E3+%2B+2x%5E2+-+5x+-+6%29%2F%28x%2B3%29=x%5E2+-+x+-+2

You can use this online polynomial division calculator to check your work

Basically x%5E3+%2B+2x%5E2+-+5x+-+6 factors to %28x%2B3%29%28x%5E2+-+x+-+2%29

Now lets break x%5E2+-+x+-+2 down further


Looking at x%5E2-x-2 we can see that the first term is x%5E2 and the last term is -2 where the coefficients are 1 and -2 respectively.

Now multiply the first coefficient 1 and the last coefficient -2 to get -2. Now what two numbers multiply to -2 and add to the middle coefficient -1? Let's list all of the factors of -2:



Factors of -2:
1,2

-1,-2 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to -2
(1)*(-2)
(-1)*(2)

note: remember, the product of a negative and a positive number is a negative number


Now which of these pairs add to -1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -1

First NumberSecond NumberSum
1-21+(-2)=-1
-12-1+2=1



From this list we can see that 1 and -2 add up to -1 and multiply to -2


Now looking at the expression x%5E2-x-2, replace -x with x%2B-2x (notice x%2B-2x adds up to -x. So it is equivalent to -x)

x%5E2%2Bhighlight%28x%2B-2x%29%2B-2


Now let's factor x%5E2%2Bx-2x-2 by grouping:


%28x%5E2%2Bx%29%2B%28-2x-2%29 Group like terms


x%28x%2B1%29-2%28x%2B1%29 Factor out the GCF of x out of the first group. Factor out the GCF of -2 out of the second group


%28x-2%29%28x%2B1%29 Since we have a common term of x%2B1, we can combine like terms

So x%5E2%2Bx-2x-2 factors to %28x-2%29%28x%2B1%29


So this also means that x%5E2-x-2 factors to %28x-2%29%28x%2B1%29 (since x%5E2-x-2 is equivalent to x%5E2%2Bx-2x-2)

So x%5E2-x-2 factors to %28x-2%29%28x%2B1%29

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Answer:


So x%5E3+%2B+2x%5E2+-+5x+-+6 factors to %28x%2B3%29%28x-2%29%28x%2B1%29