SOLUTION: Please use the remainder theorem to evaluate {{{4x^3 - 3x^2 + 2x - 3 when x=2}}}

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Question 981966: Please use the remainder theorem to evaluate 4x%5E3+-+3x%5E2+%2B+2x+-+3+when+x=2
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
This means, use synthetic division with "root to check" being 2 and dividend 4x%5E3-3x%5E2%2B2x-3. The remainder will be the given expression (your dividend) evaluated for x=2.

Do you need to see the work?
response: YES.

Root to check on the left, dividend coefficients on the other side of the "vertical wall". This shows a start setup of the synthetic division:
___________________________________________
___________|
______2____|_____4______-3______2_______-3
___________|
___________|_________________________________


___________________________________________
___________|
______2____|_____4______-3______2_______-3
___________|____________8_______10______24
___________|_________________________________
_________________4_______5______12_______21

The remainder is 21;
this means that 4x^3 - 3x^2 + 2x - 3 will be 21 for x=2.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Please use the remainder theorem to evaluate 4x%5E3+-+3x%5E2+%2B+2x+-+3+when+x=2
Since f%28x%29+=+4x%5E3+-+3x%5E2+%2B+2x+-+3, then remainder when x = 2 is:
f%282%29+=+4%282%29%5E3+-+3%282%29%5E2+%2B+2%282%29+-+3
f(2) = 4(8) - 3(4) + 4 - 3
f(x) = 32 - 12 + 4 - 3
highlight_green%28f%28x%29+=+21%29 (Remainder)