Question

find the roots of the polynomial equation(x-1)^3(x+1)^2(x+5)=0

Answer 1

$(x-1)^3(x+1)^2(x+5) = 0$

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

$⟹ x(3x+2)-1(3x+2) = 0$

$⟹ 3x²+2x-3x-2 = 0$

$⟹ x(3x+2)-1(3x+2) = 0$

$⟹ (3x+2)(x-1) = 0$

$⟹ (3x+2) = 0\: and (x-1) = 0$

$⟹ 3x = -2\: and\: x = 1$

$⟹ x = -2/3\: and\: x = 1$

Answer 2

we have ....

( x-1)^3(x+1)^2(x+5)=0

WKT ...

( x-1)^3 = x^3- 1^3- 3x×1 ( x-1)

= x^3 - 1 - 3x (x-1)

then...

( x-1)^3(x+1)^2(x+5)=0

= [x^3 -1 - 3x (x - 1 ) ]. [ x^2 + 2x +1 ] [x+5]

= [ x^3 - 1- 3x^2 +1 ] [ x^2 + 2x +1 ] [x+5]

=[ x^3 -3 x^2 ] [x^2 + 2x +1 ] [x+5]

= [ x^3 × x^2 + x^3 ×2x + x^3 - 3 x^2 × x^2- 3 x^2 × 2x - 3 x^2 ] [ x-5]

=[ x^5 + 2x^4 + x^3- 3 x^5 - 3 x^4 - 6 x ^3 - 3 x^2] [ x-5]

= [ -2 x^5 -x^4 - 5^3 - 3 x^2] [ x-5]

= [ -2 x^6 -x^5 - 5^4 - 3 x^3 + 10 x^6 +5x^5 +25^4 +15x^3 ]

= 8 x^6 +4x^5 +20 x ^4 +12 x^3 ]

divided by x^3 in the all term ...

= 8 x^3 +4x^2 +20 x +12 ]