Question

Factor the polynomial as a product of linear factors: x⁴-5x³+12x²-2x-20

Answer 1

Given -----> f( x ) = x^4 - 5*x^3 + 12*x^2 - 2*x - 20

First we have to check for what value of x the expression becomes zero

Put x = 2

f( 2 ) = 2^4 - 5*2^3 + 12*2^2 - 2*2 - 20

= 16 - 40 + 48 - 4 - 2

= 0

So ( x - 2 ) is a factor of f( x )

f( x ) = x^3*( x - 2 ) - 3*x^2*( x - 2 ) + 6*x*( x - 2 ) + 10*( x - 2 )

=   ( x - 2 )*( x^3 - 3*x^2 + 6*x + 10 )

Now considering------> g( x ) = x^3 - 3*x^2 + 6*x + 10

Again we have to check for what value of x the expression becomes zero

Put x = - 1

g( - 1 ) = -1^3 - 3*( - 1 )^2 + 6*( - 1 ) + 10

= - 1 - 3 - 6 + 10 = 0

So ( x + 1 ) is a factor of g( x )

g( x ) = x^2* ( x + 1 ) - 4*x*( x + 1 ) + 10*( x + 1 )

= ( x + 1 )*( x^2 - 4*x + 10 )

Now check for quadratic equation roots

x = [ - b + / - sqrt( b^2 - 4*a*c ) ] / 2*a

x = [ 4 + / - sqrt( ( - 4 )^2 - 4*1*10 ) ] / 2*1

So , x =  [ 4 + sqrt( 16 - 40 ) ] / 2

= [ 4 + sqrt( - 24 ) ] / 2  

= [ 4 + √ ( - 4*6 ) ] / 2

= [ 4 + 2* √ ( - 1*6 ) ] / 2

= 2 + √ ( - 1*6 )

Since  √ ( -1 ) = i

x = 2 + [ √ ( 6 ) ]*i

Similarly x = 2 - [ √ ( 6 ) ]*i   

Therefore ( x - { 2 + [ √ ( 6 ) ]*i } ) is a factor of f( x )

Also ( x - { 2 - [ √ ( 6 ) ]*i } ) is a factor of f( x )

So f( x ) can be written as

( x + 1 )* ( x - 2 )* ( x - { 2 + [ √ ( 6 ) ]*i } )* ( x - { 2 - [ √ ( 6 ) ]*i } )

=> ( x + 1 )* ( x - 2 )* ( x - 2 - i √ ( 6 ) )* ( x - 2 + i √ ( 6 ) )

So f( x ) = ( x + 1 )* ( x - 2 )* ( x - 2 - i √ ( 6 ) )* ( x - 2 + i √ ( 6 ) )