If x- -1 is one of the zeroes of polynomial fx-a0xn-a1xn-1-a2xn-2- -an -0- then prove that a0-a2-a4-an - a1-a3-a5- -an

Since, x= 1 is a 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_n ∴ a_0+a_2+a_4+⋯ +a_n =a_1+a ...

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If x= -1 is o...

Question

If $x = - 1$ is one of the zeroes of polynomial $f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0$ , then prove that
$a_{0} + a_{2} + a_{4} + \dots + a_{n} =$ $a_{1} + a_{3} + a_{5} + \dots + a_{n}$

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Solution

Since, $x = - 1$ is a 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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If x=−1 is one of the zeroes of polynomial f(x)=a0xn+a1xn−1+a2xn−2+⋯+an=0, then prove that a0+a2+a4+⋯+an= a1+a3+a5+⋯+an

Since, x=−1 is a 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 one of the zeroes of polynomial $f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0$ , then prove that
$a_{0} + a_{2} + a_{4} + \dots + a_{n} =$ $a_{1} + a_{3} + a_{5} + \dots + a_{n}$

Since, $x = - 1$ is a zero.
$∴ f (- 1) = 0$

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