How do you use synthetic division to find factors for #x^3 + 2x^2 - 5x - 6#?

1 Answer
Sep 26, 2015

Answer:

Noting that #(x-2)# is a factor [see explanation], we can use synthetic division to reduce #x^3+2x-5x-6# to a more easily factored expression.
Complete factors: #(x-2)(x+3)(x+1)#

Explanation:

Part 1: The initial Factor
If #x^3+2x-5x-6 = 0#
then setting the negative terms on one side and the positive terms on the other:
#color(white)("XX")x^3+2x^2=5x+6#
then testing a few values:

#{: (x,color(white)("XX"),x^3+2x^2,color(white)("XX"),5x+6), (0,color(white)("XX"),0,color(white)("XX"),6), (1,color(white)("XX"),3,color(white)("XX"),11), (2,color(white)("XX"),16,color(white)("XX"),16) :}#
So if #x=2# then #x^3+2x^2-5x-6 =0#
#rarr (x-2)# is a factor of #x^3+2x^2-5x-6#

Part 2: Use of synthetic division
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So
#color(white)("XX")x^2+2x^2-5x-6 = (x-2)(x^2+4x+3)#

Part 3: Factoring the remaining quadratic
By observation or using the quadratic formula we can factor:
#color(white)("XX")x^2+4x+3=(x+3)(x+1)#

Part 4: Summarize results
#x^3+2x^2-5x-6 = (x-2)(x+3)(x+1)#