Consider the following three polynomials : f(x) = x2 -8x + 15 g(x) = 2x2 - 11x + 5 h(x) = 2x2 - 7x + 3 Consider the polynomial p(x) = f(x)g(x)h(x). It turns out that p(x) is a perfect square. If the square - root of p(x) is written as p(x) = ax + bx2 + x + d, where a is positive, the value of 6a + 2b + C + d is =
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Answers
Answer:
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Explanation:
Alipore Sadar subdivision is an administrative subdivision of the South 24 Parganas district in the Indian state of West Bengal.
Given :
f(x) = x² - 8x + 15
g(x) = 2x² - 11x + 5
h(x) = 2x² - 7x + 3
p(x) = f(x)g(x)h(x) is a perfect square
√p(x) = ax³ + bx²+ cx + d , a is positive
To Find : 6a + 2b + C + d
Solution:
f(x) = x² - 8x + 15
= (x - 3)(x - 5)
g(x) = 2x² - 11x + 5
=2x² - 10x - x + 5
= 2x(x - 5) - 1(x - 5)
= (2x - 1)(x - 5)
h(x) = 2x² - 7x + 3
=2x² - 6x - x + 3
= 2x(x - 3)-1(x - 3)
= (2x-1)(x - 3)
p(x) = f(x)g(x)(hx)
= (x - 3)(x - 5)(2x - 1)(x - 5)(2x-1)(x - 3)
={ (x - 3)(2x - 1)(x - 5) }²
√p(x) = | (x - 3)(2x - 1)(x - 5) |
= | (x² - 8x + 15)(2x - 1) |
= | 2x³ - 16x² + 30x - x² + 8x - 15 |
= | 2x³ - 17x² + 38x - 15 |
as 2 is positive
Hence
2x³ - 17x² + 38x - 15
Comparing with
ax³ + bx²+ cx + d
a = 2 , b = -17 , c = 38 , d = -15
6a + 2b + c + d
= 6(2) + 2(-17) + 38 - 15
= 12 - 34 + 38 - 15
= 50 - 49
= 1
6a + 2b + c + d = 1
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