Question 53491
{{{(x^3+2x^2-5x-6)}}}; factor: x-2 [Use synthetic division or regular division to solve]

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Since (x-2) is one of the factors, divide the polynomial by (x-2):
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x-2=0
x=2 [Divisor]

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1+2-5-6 [Use only the coefficients for the dividend]
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1 [Bring down the very first coefficient of "1"]

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1+2-5-6 [Dividend]
0+2
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1+4 [Begin to "divide" by 2 by multiplying [(1*2) + 2)=4]

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1+2-5-6 [Dividend]
0+2+8
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1+4+3 [Continue to "divide" by 2 by multiplying [(4*2)+(-5)=3]
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1+2-5-6 [Dividend]
0+2+8+6
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1+4+3+0 [Continue to "divide" by 2 by multipling [(3*2) + (-6)=0][Quotient]


[Re-apply the coefficents to the "x"terms minus one.  For example you started with x^3, so the quotient will start with x^2]
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So, the factors are {{{(x-2)(x^2+4x+3)}}}

Check by multiplying the factors to get the original polynomial.