Question

Factorize
x^4 - 2x^3 - 7x^2 + 8x +12
plzzz

Answer 1

HERE IS YOUR ANSWER...⬇⬇

$\boxed{SOLUTION}$ ➡

$x {}^{4} - 2x {}^{3} - 7x {}^{2} + 8x + 12 \\ = x {}^{4} - 2x {}^{3} + x {}^{2} - x {}^{2} - 7x {}^{2} + 8x + 12 \\ = (x {}^{4} - 2x {}^{3} + x {}^{2} ) - 8x {}^{2} + 8x + 12 \\ =( x {}^{2} - x) {}^{2} - 8(x {}^{2} - x) + 12 \: \: \: \: - - - - - - - - - - - - > (1)$

Let,

$x {}^{2} - x = y$

So,

$(1) = y {}^{2} - 8y + 12 \\ = y {}^{2} - (6 + 2)y + 12 \\ = y {}^{2} - 6y - 2y + 12 \\ = y(y - 6) - 2(y - 6) \\ = (y - 2)(y - 6) \: \: - - - - - - - - - - - - - - - > (2)$

By putting the value of "y" in (2), we get

(2)= { ( x^2 - x ) - 2 } { ( x^2 - x ) - 6 }

= ( x^2 - x -2 ) ( x^2 - x - 6 )

= ( x^2 - 2x + x - 2 )( x^2 - 3x + 2x - 6 )

= { x ( x - 2 ) + 1 ( x - 2 ) } { x ( x - 3) + 2 ( x - 3) }

= ( x + 1) ( x - 2) ( x + 2) ( x -3 )



Therefore,

$x {}^{4} - 2x {}^{3} - 7x {}^{2} + 8x + 12 = (x + 1)(x - 2)(x + 2)(x -3 )$

✝✝...HOPE IT HELPS YOU...✝✝

Answer 2

Answer:

x^4-2x^3-7x^2+8x+12

=x^4-3x^3+x^3-3x^2-4x^2+12x-4x+12

=x^3 (x-3)+x^2 (x-3)-4x (x-3)-4 (x-3)

=(x-3)(x^3+x^2-4x-4)

=(x-3)(x^3+2x^2-x^2-2x-2x-4)

=(x-3)[x^2 (x+2)-x (x+2)-2 (x+2)]

=(x-3)(x+2)(x^2-x-2)

=(x-3)(x+2)(x^2+x-2x-2)

=(x-3)(x+2)[x (x+1)-2 (x+1)]

=(x-3)(x+2)(x+1)(x-2)

This is your required answer

Step-by-step explanation: