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x-2 is to be a factor of 2x³-6x²+5x+a
We divide either by long division or by synthetic division:
and set the remainder = 0
By long division: By synthetic division
2x²-2x+1 2|2 -6 5 a
x-2)2x³-6x²+5x+a | 4 -4 2
2x³-4x² 2 -2 1 a+2
-2x²+5x
-2x²+4x
x+a
x-2
a+2
Either way, the remainder is a+2.
We want the remainder to be 0.
So we set a+2 = 0
a = -2
Edwin