Question #3772b

Given
#f(x)=kx^3-6x^2+bx-6#
Given also that #(x+1) and (x-3)# are factors of #f(x)#

To find out #k and b# we need two equations.

We know, if #(x+a)# is a factor of #f(x)# then #f(-a)=0#

#:.# from first factor we obtain #f(-1)=0# and from the second factor we have #f(3)=0#
First equation is
#f(-1)=0=k(-1)^3-6(-1)^2+b(-1)-6#
or #-k-6-b-6=0#, rearranging we obtain

#k+b=-12#.......(1)

Similarly second equation is

#f(3)=0=kxx3^3-6xx3^2+bxx3-6#

or #27k+3b=60#, dividing both sides with 3

#9k+b=20#......(2)

Inserting the value of #b=-12-k# from (1) in (2), we obtain

#9k-12-k=20#
or #8k=32#, gives us the value of #k=4#,

and from (1) we get the value of #b=-16#

#:. f(x)=4x^3-6x^2-16x-6#
First dividing #f(x)# by #(x+1)# using long division, we obtain the quotient as

#4x^2-10x-6#

To factorize this quadratic we see that the middle term #-10x# can be written as #-12x+2x#,

[or we can use long division once more to find the new quotient after dividing the quadratic with #(x-3)#]

Rewriting the quadratic
#4x^2-12x+2x-6#
or #4x(x-3)+2(x-3)#
or #(x-3)(4x+2)#

Which gives us the third factor as #(4x+2)#

Given:#" "kx^3-6x^2+bx-6 = (x+1)(x-3)(mx+n)#

Write as#" "kx^3-6x^2+bx-6= (x^2-2x-3)(mx+n)#

First consider the final constant of -6 from the LHS

Comparing LHS to RHS

#-6=-3n#

#color(red)(n=+2)" "# giving

#kx^3-6x^2+bx-6= (x^2-2x-3)(mx+2)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Multiplying out we have:

#kx^3-6x^2+bx-6= mx(x^2-2x-3)+2(x^2-2x-3)#

#kx^3-6x^2+bx-6= mx^3-2mx^2-3mx+2x^2-4x-6#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Comparing coefficients of "x^3: m=k" giving")#

#kx^3-6x^2+bx-6= kx^3-2kx^2-3kx+2x^2-4x-6#

Collecting like terms

#kx^3-6x^2+bx-6= kx^3-2kx^2+2x^2 -3kx-4x-6#

#kx^3-6x^2+bx-6= kx^3-2x^2(k-1)-x(3k+4)-6#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Comparing coefficients of "x^2:" " -2(k-1)=-6" giving")#

#k-1=+6/2#

#color(red)(k=+8/2=+4)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Comparing coefficients of "x " given that "k=+4)#

#+bx=-(3k+4)x#

#=>color(red)(b=-(12+4)=-16)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Factors are:#" "(x+1)(x-3)(4x+2) =0#

#color(magenta)(=>" "x=-1"; "x=+3"; "x=-1/2)#