When ax^3-2x^2+3x-5 is divided by 2x-1, the remainder is -8
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2x - 1)ax³ - 2x² - 2x² + 3x - 5
We won't divide that out, but suppose we did.
We would get some quotient Q(x) and a remainder of -8
Q(x)
2x - 1)ax³ - 2x² - 2x² + 3x - 5
.....
......
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-8
Then use
DIVIDEND = QUOTIENT*DIVISOR + REMAINDER
ax³ - 2x² + 3x - 5 = Q(x)*(2x - 1) + (-8)
We want to make the factor (2x - 1) equal to zero so that
term will be eliminated.
2x - 1 = 0
2x = 1
x = ½
Now substitute x = ½ in
ax³ - 2x² + 3x - 5 = Q(x)*(2x - 1) + (-8)
a(½)³ - 2(½)² + 3(½) - 5 = Q(½)*(2*½ - 1) + (-8)
a(⅛) - 2(¼) + - 5 = Q(½)*(1 - 1) - 8
- ½ + - 5 = Q(½)*(0) - 8
- ½ + - 5 = 0 - 8
- ½ + - 5 = -8
Clear of fractions by multiplying through by 8
a - 4 + 12 - 40 = -64
a - 32 = -64
a = -32
Checking:
-16x² - 9x - 3
2x - 1)-32x³ - 2x² + 3x - 5
-32x³ + 16x²
-18x² + 3x
-18x² + 9x
-6x - 5
-6x + 3
-8
It leaves the remainder -8, so we are correct!
Edwin