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3 years ago

What should be added in the polynomial x3-6x2+11x+8 so that it is completely divisible by 1-3x +x2

Mathematics

1 answer:

yanalaym [24]3 years ago

5 0

Answer:

The value to be added to the polynomial x³ - 6·x² + 11·x + 8 so that it is completely divisible by 1 - 3·x + x² is -(x + 11)

Step-by-step explanation:

By long division, we have;

${\left ({x^{3}-6x^{2}+11x+8} \right )\div 1 - 3x + x^{2}}$ = x - 3

-(x³ - 3·x² + x)

-3·x² + 10·x + 8

-(-3·x² + 9·x -3)

x + 11

Therefore, -(x + 11) should be added to the polynomial x³ - 6·x² + 11·x + 8 so that it is completely divisible by 1 - 3·x + x².

That is (x³ - 6·x² + 11·x + 8 - x - 11) ÷ (1 - 3·x + x²) = x - 3.

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<h2>First Option: {6, 10, 12}</h2>

Let A = 6, B = 10, and C = 12:

$6+10>12\\16>12\ \textsf{true}\\\\6+12>10\\18>10\ \textsf{true}\\\\12+10>6\\22>6\ \textsf{true}$

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<h2>Third Option: {4, 4, 9}</h2>

Let A = 4, B = 4, and C = 9

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Since the third option had an inequality that was false, <u>the third option cannot be a triangle.</u>

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