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siddhartharao77:
Option (B)
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option C is the right maybe
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report nonsense answers
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Given Equation is (5x^2 + 14x + 2)^2 - (4x^2 - 5x + 7)^2
We know that (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.
We know that (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca
= [25x^4 + 196x^2 + 4 + 2(5x^2 * 14x) + 2(14x * 2) + 2(2 * 5x^2)] - [16x^4 + 25x^2 + 49 - 2(4x^2 * 5x) - 2(5x * 7) + 2(7 * 4x^2)
= 25x^4 + 196x^2 + 4 + 140x^3 + 56x + 20x^2 - 16x^4 - 25x^2 - 49 + 40x^3 + 70x - 56x^2
= 9x^4 + 180x^3 + 135x^2 + 126x - 45.
Now,
We have to divide this by (x^2 + x + 1).
9x^2 + 171x - 45
--------------------------------------------------
x^2 + x + 1) 9x^4 + 180x^3 + 135x^2 + 126x - 45
9x^4 + 9x^3 + 9x^2
------------------------------------------------------
171x^3 + 126x^2 + 126x
171x^3 + 171x^2 + 171x
-----------------------------------------------------------
- 45x^2 - 45x - 45
- 45x^2 - 45x - 45
----------------------------------------------------------
0.
Therefore the quotient is 9x^2 + 171x - 45 (or) 9(x^2 + 19x - 5) with a remainder 0.
Hope this helps! --------------------- Good Luck!
We know that (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.
We know that (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca
= [25x^4 + 196x^2 + 4 + 2(5x^2 * 14x) + 2(14x * 2) + 2(2 * 5x^2)] - [16x^4 + 25x^2 + 49 - 2(4x^2 * 5x) - 2(5x * 7) + 2(7 * 4x^2)
= 25x^4 + 196x^2 + 4 + 140x^3 + 56x + 20x^2 - 16x^4 - 25x^2 - 49 + 40x^3 + 70x - 56x^2
= 9x^4 + 180x^3 + 135x^2 + 126x - 45.
Now,
We have to divide this by (x^2 + x + 1).
9x^2 + 171x - 45
--------------------------------------------------
x^2 + x + 1) 9x^4 + 180x^3 + 135x^2 + 126x - 45
9x^4 + 9x^3 + 9x^2
------------------------------------------------------
171x^3 + 126x^2 + 126x
171x^3 + 171x^2 + 171x
-----------------------------------------------------------
- 45x^2 - 45x - 45
- 45x^2 - 45x - 45
----------------------------------------------------------
0.
Therefore the quotient is 9x^2 + 171x - 45 (or) 9(x^2 + 19x - 5) with a remainder 0.
Hope this helps! --------------------- Good Luck!
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