Question

Question #3772b

Answer 1

`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)`

Answer 2

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)`