Question

2x⁴+x³ 14x² - 19x- 6

Answer 1

Hey friend , Harish here.

Here is your answer:

$2x^4 +x^3 - 14x^2 - 19x -6$

⇒  $\bigl ( 2x^4 + x^3 \bigr ) - \bigl ( 14x^2 + 19x + 6 \bigr )$

⇒  $x^3 \bigl( 2x +1 \bigr ) - \bigl ( 14x^2 + 12x + 7x + 6 \bigr )$

⇒  $x^3 \bigl( 2x +1 \bigr ) - \bigl [ 2x \bigl (7x+6 \bigr ) + 1 \bigl (7x+6 \bigr ) \ \bigr ]$

⇒  $x^3 \bigl( 2x +1 \bigr ) - \bigl [ \bigl (2x+1 \bigr ) \bigl (7x+6 \bigr ) \ \bigr ]$

(Taking 2x + 1 as common ).

⇒ $\bigl (2x+1 \bigr ) \bigl [ x^3 - 7x - 6 \bigr ]$

⇒ $\bigl (2x+1 \bigr ) \bigl [ x^3 + x^2 -x^2 - 7x - 6 \bigr ]$

( Here we are adding and subtracting x². )

⇒ $\bigl (2x+1 \bigr ) \bigl [ (x^3 + x^2 ) - \bigl( x^2 + 7x +6 \bigr ) \bigr ]$

⇒ $\bigl (2x+1 \bigr ) \bigl [ x^2(x + 1 ) - \bigl( x^2 + 1x+ 6x + 6 \bigr ) \bigr ]$

⇒ $\bigl (2x+1 \bigr ) \bigl [ x^2(x + 1 ) - \bigl( x+6 \bigr )(x+1) \bigr ]$

( Now inside the square brackets, take x+1 as common )

⇒ $\bigl (2x+1 \bigr ) \bigl [ (x+1) \bigl(x^2 -x - 6 \bigr ) \bigr ]$

⇒ $(2x-1)(x+1) \bigl ( x^2 + 2x - 3x - 6 \bigr )$

⇒ $\boxed{\bigl ( 2x-1 \bigr ) \bigl (x+1 \bigr ) \bigl ( x+2 \bigr ) \bigl( x-3 \bigr )}$
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Hope my answer is helpful to you.