Question #3772b

2 Answers
Feb 23, 2016

k=4, b=-16,
three roots are given by factors: (x+1), (x-3) and (4x+2)
x=-1, 3 and -1/2

Explanation:

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)

Feb 23, 2016

Whilst I have done my best not to have a slip you should check my answer.
color(magenta)(=>" "x=-1"; "x=+3"; "x=-1/2)

Explanation:

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)