Question

Find the value of A and B so that (X + 1) and (x-1) are the factor of X^4 + ax^3 - 3x^2 + 2x + b

Answer 1

EXPLANATION.

(x + 1) and (x - 1) are the factors of equation,

⇒ F(x) = x⁴ + ax³ - 3x² + 2x + b.

As we know that,

Zeroes of the polynomial,

⇒ (x + 1) = 0.

⇒x = -1.

Put the value of x = -1 in equation, we get.

⇒ F(-1) = (-1)⁴ + a(-1)³ - 3(-1)² + 2(-1) + b.

⇒ 1 - a - 3 - 2 + b = 0.

⇒ - a - 4 + b = 0.

⇒ b = a + 4 ⇒ (1).

⇒ (x - 1) = 0.

⇒ x = 1.

Put the value of x = 1 in equation, we get.

⇒ F(1) = (1)⁴ + a(1)³ - 3(1)² + 2(1) + b.

⇒ 1 + a - 3 + 2 + b = 0.

⇒ a + b = 0.

⇒ a = -b ⇒ (2).

From equation, (1) and (2) we get,

Put the value of equation (2) in (1) we get,

⇒ b = a + 4.

⇒ b = -b + 4.

⇒ b + b = 4.

⇒ 2b = 4.

⇒ b = 2.

Put the value of b = 2 in equation (2), we get.

⇒ a = -b.

⇒ a = -2.

Value of A = -2 & B = 2.

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

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$\tt \ \: :  ⟼ Let \: f(x) = {x}^{4} + a {x}^{3} - {3x}^{2} + 2x + b$

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$\large\underline{\bold{❥︎Step :- 1 }}$

$\tt \ \: :  ⟼ Now, \: x + 1 \: is \: a \: factor \: of \: f(x) \:$

$\tt\implies \:f( - 1) = 0$

$\tt\implies \: {( - 1)}^{4} + a {( - 1)}^{3} - 3 {( - 1)}^{2} + 2( - 1) + b = 0$

$\tt\implies \:1 - a - 3 - 2 + b = 0$

$\tt\implies \:b = 4 + a \: - - - (1)$

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$\large\underline{\bold{❥︎Step :- 2 }}$

$\tt \ \: :  ⟼ Now, \: again \: (x - 1) \: is \: factor \: of \: f(x)$

$\tt\implies \:f(1) = 0$

$\tt\implies \: {( 1)}^{4} + a {( 1)}^{3} - 3 {( 1)}^{2} + 2( 1) + b = 0$

$\tt\implies \:1 + a - 3 + 2 + b = 0$

$\tt\implies \:a + b = 0$

$\tt\implies \:a + 4 + a = 0 \: \: \: \: \: \: \: \bigg(using \: (1) \bigg)$

$\tt\implies \:2a + 4 = 0$

$\tt\implies \:2a = - \: 4$

$\tt\implies \:a \: = \: - \: 2$

$\tt \:  ⟼ So \: b \: = 4 - 2 = 2$

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$\begin{gathered}\begin{gathered}\bf So - \: \begin{cases} &\sf{a = - \: 2} \\ &\sf{b \: = \: 2} \end{cases}\end{gathered}\end{gathered}$

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