If x -1 is a factor of f x - a 0 x n - a 1 x n -1- a 2 x n -2- a n -0- thenA- a 0- a 2- a 4- a n -0B- a 1- a 3- a 5- a n -0C- a 0- a 1- a 2- a n -0D- a 0- a 2- a 4- a n -a 1- a 3- a 5- a n

The correct option is D a_0+a_2+a_4+⋯+a_n = a_1+a_3+a_5+⋯ +a_nSince, x+1 is a factor. x+1 = 0 ⇒ x = 1 1 is zero. ∴ f( 1)=0 nbsp;f( 1)=a_0( 1)^n+a_1( 1)^n ...

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Standard X

Mathematics

Zeros of a Polynomial

If x +1 is a ...

Question

If $x + 1$ is a factor of $f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0$ , then

a_{0} + a_{1} + a_{2} + \dots + a_{n} = 0

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a_{0} + a_{2} + a_{4} + \dots + a_{n} = 0

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a_{1} + a_{3} + a_{5} + \dots + a_{n} = 0

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a_{0} + a_{2} + a_{4} + \dots + a_{n} =

a_{1} + a_{3} + a_{5} + \dots + a_{n}

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Solution

The correct option is D $a_{0} + a_{2} + a_{4} + \dots + a_{n} =$
$a_{1} + a_{3} + a_{5} + \dots + a_{n}$
Since, $x + 1$ is a factor.
$x + 1 = 0 \Rightarrow x = - 1$
$- 1$ is zero.
$∴ f (- 1) = 0$

$f (- 1) = a_{0} (- 1)^{n} + a_{1} (- 1)^{n - 1} + a_{2} (- 1)^{n - 2} + \dots + a_{n}$
$0 = a_{0} - a_{1} + a_{2} - a_{3} + \dots - a_{n - 1} + a_{n}$
$∴ a_{0} + a_{2} + a_{4} + \dots + a_{n} = a_{1} + a_{3} + a_{5} + \dots + a_{n}$

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Similar questions

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

then

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

Q. Let the sequence

a_{1}, a_{2},

________

a_{n}

form an A.P. and let

a_{1} = 0

, prove that

\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \frac{a_{5}}{a_{4}} +

________

+ \frac{a_{n}}{a_{n - 1}} - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{3}} + . . . + \frac{1}{a_{n - 2}}) = \frac{a_{n - 1}}{a_{2}} + \frac{a_{2}}{a_{n - 1}}

Q. Let

A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}

be a regular hexagon inscribed in a circle of unit radius.Then the product of the length of

A_{0} A_{1} . A_{0} A_{2} . A_{0} A_{4}

Q. If

a_{1}, a_{2}, a_{3}, . . . . . . . a_{n}

are first n terms of an A.P. with first term

a_{1} = 0

and common difference

d \neq 0

Then value of

\frac{a_{3} - a_{2}}{a_{2}} + \frac{a_{4} - a_{2}}{a_{3}} + \frac{a_{5} - a_{2}}{a_{4}} . . . . . . . . . . . . \frac{a_{n} - a_{2}}{a_{n - 1}}

Q. If for an

A . P .

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If x+1 is a factor of f(x)=a0xn+a1xn−1+a2xn−2+⋯+an=0, then

The correct option is D a0+a2+a4+⋯+an= a1+a3+a5+⋯+anSince, x+1 is a factor. x+1=0⇒x=−1 −1 is zero. ∴f(−1)=0 f(−1)=a0(−1)n+a1(−1)n−1+a2(−1)n−2+⋯+an 0=a0−a1+a2−a3+⋯−an−1+an ∴a0+a2+a4+⋯+an=a1+a3+a5+⋯+an

If $x + 1$ is a factor of $f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0$ , then