Question

7. Two zeroes of cubic polynomial ax^3+3x^2-bx-6 are -1 and -2. Find the 3rd zero and value of a and b.

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Answer 1

Hey!!!

As promised I am here to help you

Difficulty Level : Above Average

Chances of being asked in Board : 70%

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let p(x) = ax³ + 3x² - bx - 6

zeros = (-1) and (-2)

Thus

=> p(-1) = 0

=> a(-1)³ + 3(-1)² - b(-1) - 6 = 0

=> - a + 3 + b - 6 = 0

=> a - b = - 3 ------(1)

Similarly p(-2) = 0

=> a(-2)³ + 3(-2)² - b(-2) - 6 = 0

=> - 8a + 12 + 2b - 6 = 0

=> 8a - 2b = 6

=> 4a - b = 3

=> b = 4a - 3 ---(2)

Substituting (2) in (1)

=> a - (4a - 3) = - 3

=> a - 4a + 3 = -3

=> - 3a = -6

=> a = 2 <<<<<< Answer

Using a in (2)

=> b = 4(2) - 3

=> b = 5 <<<<<< Answer

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Hope this helps ✌️

Have a Marvelous Monday ahead

Answer 2

Heya !!!

Good evening :))

here is your answer,

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Given: Let f (x) = ax3 + 3x2 – bx – 6

also –1 and –2 are the zeroes of f (x)

=> f (-1) = 0

=> -a +3 +b -6 = 0

=> b - a = 3 - - - - - - - (1)

and f (-2) = 0

=> -8a +12 +2b -6 = 0

=> 4a -b = 3 - - - - - - - - (2)

Adding (1) and (2) we get

3a = 6

$= > a = \frac{6}{3} = 2$

Now, substitute the value of a in equation (1)
we get ,

=> b - a = 3

=> b - 2 = 3

=> b = 3 + 2 = 5

=> b = 5


So, f (x) = 2x3 + 3x2 – 5x – 6

 = (2x – 3) (x + 1) (x + 2)

Thus the required zero is 3/2.

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Hope it satisfied your answer !!!

Cheers :))

# Nikky