SOLUTION: use synthetic division to show that the number given to the right of the equation is a solution of the equation, then solve the polynomial equation. x^3+2x^2-11x-12=0;-4 a){3,1

Algebra ->  Equations -> SOLUTION: use synthetic division to show that the number given to the right of the equation is a solution of the equation, then solve the polynomial equation. x^3+2x^2-11x-12=0;-4 a){3,1      Log On


   

Question 147748: use synthetic division to show that the number given to the right of the equation is a solution of the equation, then solve the polynomial equation.
x^3+2x^2-11x-12=0;-4
a){3,1,-4}
b){3.-1.-4}
c){-3,1,-4)
d){-3,-1,-4}
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

First set up the synthetic division table by placing the test zero (which in this case is -4) in the upper left corner and placing the coefficients of the function to the right of the test zero.
-4|12-11-12
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-4|12-11-12
|
1

Multiply -4 by 1 and place the product (which is -4) right underneath the second coefficient (which is 2)
-4|12-11-12
|-4
1

Add -4 and 2 to get -2. Place the sum right underneath -4.
-4|12-11-12
|-4
1-2

Multiply -4 by -2 and place the product (which is 8) right underneath the third coefficient (which is -11)
-4|12-11-12
|-48
1-2

Add 8 and -11 to get -3. Place the sum right underneath 8.
-4|12-11-12
|-48
1-2-3

Multiply -4 by -3 and place the product (which is 12) right underneath the fourth coefficient (which is -12)
-4|12-11-12
|-4812
1-2-3

Add 12 and -12 to get 0. Place the sum right underneath 12.
-4|12-11-12
|-4812
1-2-30


Since the last column adds to zero, we have a remainder of zero. This means that -4 is a solution of the equation


Now lets look at the bottom row of coefficients:

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

x%5E2+-+2x+-+3


So %28x%5E3+%2B+2x%5E2+-+11x+-+12%29%2F%28x%2B4%29=x%5E2+-+2x+-+3

Basically x%5E3+%2B+2x%5E2+-+11x+-+12 factors to %28x%2B4%29%28x%5E2+-+2x+-+3%29

Now lets solve x%5E2+-+2x+-+3=0:


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%281%29%28-3%29+%29%29%2F%282%281%29%29 Plug in a=1, b=-2, and c=-3


x+=+%282+%2B-+sqrt%28+%28-2%29%5E2-4%281%29%28-3%29+%29%29%2F%282%281%29%29 Negate -2 to get 2.


x+=+%282+%2B-+sqrt%28+4-4%281%29%28-3%29+%29%29%2F%282%281%29%29 Square -2 to get 4.


x+=+%282+%2B-+sqrt%28+4--12+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-3%29 to get -12


x+=+%282+%2B-+sqrt%28+4%2B12+%29%29%2F%282%281%29%29 Rewrite sqrt%284--12%29 as sqrt%284%2B12%29


x+=+%282+%2B-+sqrt%28+16+%29%29%2F%282%281%29%29 Add 4 to 12 to get 16


x+=+%282+%2B-+sqrt%28+16+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%282+%2B-+4%29%2F%282%29 Take the square root of 16 to get 4.


x+=+%282+%2B+4%29%2F%282%29 or x+=+%282+-+4%29%2F%282%29 Break up the expression.


x+=+%286%29%2F%282%29 or x+=++%28-2%29%2F%282%29 Combine like terms.


x+=+3 or x+=+-1 Simplify.


So the answers are:

x=-4, x+=+3 or x+=+-1