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Use the remainder theorem to evaluate 2x^3 - 5x^2 + 2x + 5 when x = -2
ok i only got up to here i dont know what else to do
p(-2)
-2 | 2 -5 2 5
___ -5
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For this question you need to divide (2x³ - 5x² + 2x + 5) by (x+2), using algebraic division.
(2x³ - 5x² + 2x + 5)/(x+2)
Split the x² term: (2x³ + 4x² - 9x² + 2x + 5)/(x+2)
Factorise partially: 2x² + (-9x² + 2x + 5)/(x+2)
Split the x term: 2x² + (-9x² - 18x + 20x + 5)/(x+2)
Factorise partially: 2x² - 9x + (20x+5)/(x+2)
Split the constant: 2x² - 9x + (20x + 40 - 35)/(x+2)
Factorise partially: 2x² - 9x + 20 - 35/(x+2)
And now, by the remainder theorem, we know that the function has a value of -35, when x = -2.
Why did the vector cross the road?
It wanted to be normal.
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mathsy's right.
With horner's scheme (i don't know exactly how's in English): ( changing the sign:)
| 2 -5 2 5
2 | 2 -9 20 -35 ->remainder;
IPBLE: Increasing Performance By Lowering Expectations.
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