Question

. If x +1 and x -1 are factors of f(x) = x³ +2 ax +b, calculate the values of a and b. Using these values of a and b, factorise f(x) completely. 

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Answer 1

$\large\underline{\sf{Solution-}}$

Given polynomial is

$\rm :\longmapsto\:f(x) = {x}^{3} + 2ax + b - - - (1)$

Further Given that,

$\red{\rm :\longmapsto\:x + 1 \: is \: a \: factor \: of \: f(x)}$

We know,

Factor Theorem states that if a polynomial p(x) is divided by linear polynomial x - a, then remainder is 0.

So, using this concept of Factor Theorem, we have

$\rm :\longmapsto\:f( - 1) = 0$

$\rm :\longmapsto\:{( - 1)}^{3} + 2a( - 1) + b = 0$

$\bf :\longmapsto\: - 1 - 2a + b = 0 - - - (2)$

Again, given that

$\red{\rm :\longmapsto\:x - 1 \: is \: a \: factor \: of \: f(x)}$

So, by using concept of Factor Theorem, we have

$\rm :\longmapsto\:f( 1) = 0$

$\rm :\longmapsto\:{(1)}^{3} + 2a(1) + b = 0$

$\bf :\longmapsto\:1 + 2a+ b = 0 - - - - (3)$

On adding equation (2) and (3), we get

$\rm :\longmapsto\:2b = 0$

$\rm \implies\:\boxed{ \tt{ \: b \: = \: 0 \: }}$

On substituting the value of b, in equation (3), we get

$\rm :\longmapsto\:1 + 2a = 0$

$\rm :\longmapsto\:2a = - 1$

$\rm \implies\:\boxed{ \tt{ \: a \: = \: - \: \frac{1}{2} \: }}$

Now, given polynomial can be rewritten as on substituting the values of a and b, we have

$\rm :\longmapsto\:f(x) = {x}^{3} - x$

$\rm :\longmapsto\:f(x) = x( {x}^{2} - 1)$

$\rm :\longmapsto\:f(x) = x( {x}^{2} - {1}^{2} )$

$\rm :\longmapsto\:f(x) = x( x - 1)(x + 1)$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: f(x) = x( x - 1)(x + 1) \: }}}$

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More Identities to know :-

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)