If x-1 is a factor of fx-a0xn-a1xn-1-a2xn-2- -an -0- then

Since, 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 1+a_2( 1)^n 2+⋯ +a_n 0 = a_0 a_1 + a_2 a_3 + ⋯ a_n 1 + a ...

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Byju's Answer

Standard X

Mathematics

Zeros of a Polynomial

If x+1 is a f...

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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has in (0,1)

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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