SOLUTION: use the p over q method and synthetic division to factor the polynomial P(x). then solve P(x)=0 P(x)=x^3+4x^2+x-6

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Question 1090305: use the p over q method and synthetic division to factor the polynomial P(x). then solve P(x)=0
P(x)=x^3+4x^2+x-6
Found 2 solutions by Boreal, MathLover1:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
P(x)=x^3+4x^2+x-6
p is 6 and q is 1, the coefficient of x^3
therefore the possible integer roots are +/-1,2,3,6
synthetic division with
1/1----4----1---minus 6
==1==5----6----0
1 is a root, so (x-1) is a factor
the other factor may be read off the division as x^2+5x+6, which is (x+3)(x+2)
Those are the three factors, and roots are -3, -2, and 1
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E3%2B4x%5E2%2Bx-6%29

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
P(x)=x^3+4x^2+x-6

1. Arrange the polynomial in descending order:
P%28x%29=x%5E3%2B4x%5E2%2Bx-6+
2. Write down all the factors of the constant term. These are all the possible values of p .
p=6, all the factors are
p= 1,± 2,3,6
3. Write down all the factors of the leading coefficient. These are all the possible values of q .
q=1

4.Write down all the possible values of p%2Fq. Remember that since factors can be negative, p%2Fq+and -+p%2Fq must both be included. Simplify each value and cross out any duplicates.
p%2Fq=±1%2F1, ±2%2F1,±3%2F1, ±6%2F1,... (eliminate 6%2F1, it is product of 2%2F1 and 3%2F1)
p%2Fq=±1, ±2,±3,
5. Use synthetic division to determine the values of +p%2Fq for which P%28p%2Fq%29+=+0 . These are all the rational roots of P%28x%29+.
The synthetic division table is: if p%2Fq=+1
1|.... 1.... +4.... ..1.... +-6
...................1.... ..5.... ..6
-----------------------------------------
.........1........5.......6........0-> reminder=> 0,=> highlight%281%29 is a root

The synthetic division table is: if p%2Fq=+-1
-1|.... 1.... 4.... ..1.... +-6
...................-1.... -3.... ....+2
-----------------------------------------
.........1........3......-2....-4-> reminder -4=> -1 is not root
The synthetic division table is: if p%2Fq=+2
2|.... 1.... 4.... ..1.... +-6
...................2.... ..12....+26
-----------------------------------------
.........1........6.......13........20-> reminder=> 20,=> 2 is not a root

The synthetic division table is: if p%2Fq=+-2
-2|.... 1.... +4.... .. 1.... +-6
................... -2.... .. -4.... 6
-----------------------------------------
......... 1........ 2....... -3....... 0-> reminder=> 0,=> highlight%28-2%29 is a root

The synthetic division table is: if p%2Fq=+3
3|.... 1.... +4.... .. 1.... +-6
................... 3.... .. 21.... 66
-----------------------------------------
......... 1........ 7....... 22........ 60-> reminder=> 60,=> 3 is not a root

The synthetic division table is: if p%2Fq=+-3
-3|.... 1.... 4.... ..1.... +-6
...................-3.... ..-3.... 6
-----------------------------------------
.........1........1.......-2........0-> reminder=> 0,=> highlight%28-3%29 is a root
P%28x%29=%28x+-+1%29+%28x+%2B+2%29+%28x+%2B+3%29
then solve+P%28x%29=0+
%28x+-+1%29+%28x+%2B+2%29+%28x+%2B+3%29=0
if x+-+1=0=>x=1
if x+%2B+2=0=>x=-2
if x+%2B+3=0=>x=-3