Please solve these questions....
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solution of question is in above pic
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Hello Mate!
(i) f(x) = 2x³ + 7x² - 3
[A] f(1), this means that value of x is when 1 then
f(1) = 2(1)³ + 7(1)² - 3
= 2 + 7 - 3 = 6
Hence then value is 6
[B] f(0), this means that value of x is when 0 then
f(0) = 2(0)³ + 7(0)² - 3
= - 3
Hence then value is - 3.
(ii) [A] In order to find roots, we must find such value of x where final value of p(x) comes "0".
f(x) = x³ + 8
0 = x³ + 8
- 8 = x³
- 2 = x.
Hence root of x³ + 8 is - 2.
[B] f(x) = 4x
0 = 4x => x = 0.
Hence root of 4x is 0.
(iii) In order to identity polynomial, remember that,
1. The variable x should not be in denominator.
2. The exponent of x should not be in farction i.e should not be under any root.
[A] x² - 5x + 2/x ( Wrong, violates 1st point )
[B] x² + x ( Right, it is! )
[C] x - 7√x ( Wrong, violates 2nd point )
Have great future ahead!
(i) f(x) = 2x³ + 7x² - 3
[A] f(1), this means that value of x is when 1 then
f(1) = 2(1)³ + 7(1)² - 3
= 2 + 7 - 3 = 6
Hence then value is 6
[B] f(0), this means that value of x is when 0 then
f(0) = 2(0)³ + 7(0)² - 3
= - 3
Hence then value is - 3.
(ii) [A] In order to find roots, we must find such value of x where final value of p(x) comes "0".
f(x) = x³ + 8
0 = x³ + 8
- 8 = x³
- 2 = x.
Hence root of x³ + 8 is - 2.
[B] f(x) = 4x
0 = 4x => x = 0.
Hence root of 4x is 0.
(iii) In order to identity polynomial, remember that,
1. The variable x should not be in denominator.
2. The exponent of x should not be in farction i.e should not be under any root.
[A] x² - 5x + 2/x ( Wrong, violates 1st point )
[B] x² + x ( Right, it is! )
[C] x - 7√x ( Wrong, violates 2nd point )
Have great future ahead!
ShuchiRecites:
Thanks for attempting brainliest
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