# Question #3772b

Feb 23, 2016

$k = 4$, $b = - 16$,
three roots are given by factors: $\left(x + 1\right) , \left(x - 3\right) \mathmr{and} \left(4 x + 2\right)$
$x = - 1 , 3 \mathmr{and} - \frac{1}{2}$

#### Explanation:

Given
$f \left(x\right) = k {x}^{3} - 6 {x}^{2} + b x - 6$
Given also that $\left(x + 1\right) \mathmr{and} \left(x - 3\right)$ are factors of $f \left(x\right)$

To find out $k \mathmr{and} b$ we need two equations.

We know, if $\left(x + a\right)$ is a factor of $f \left(x\right)$ then $f \left(- a\right) = 0$

$\therefore$ from first factor we obtain $f \left(- 1\right) = 0$ and from the second factor we have $f \left(3\right) = 0$
First equation is
$f \left(- 1\right) = 0 = k {\left(- 1\right)}^{3} - 6 {\left(- 1\right)}^{2} + b \left(- 1\right) - 6$
or $- k - 6 - b - 6 = 0$, rearranging we obtain

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

Similarly second equation is

$f \left(3\right) = 0 = k \times {3}^{3} - 6 \times {3}^{2} + b \times 3 - 6$

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

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

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

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

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

$\therefore f \left(x\right) = 4 {x}^{3} - 6 {x}^{2} - 16 x - 6$
First dividing $f \left(x\right)$ by $\left(x + 1\right)$ using long division, we obtain the quotient as

$4 {x}^{2} - 10 x - 6$

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

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

$4 {x}^{2} - 12 x + 2 x - 6$
or $4 x \left(x - 3\right) + 2 \left(x - 3\right)$
or $\left(x - 3\right) \left(4 x + 2\right)$

Which gives us the third factor as $\left(4 x + 2\right)$

Feb 23, 2016

Whilst I have done my best not to have a slip you should check my answer.
$\textcolor{m a \ge n t a}{\implies \text{ "x=-1"; "x=+3"; } x = - \frac{1}{2}}$

#### Explanation:

Given:$\text{ } k {x}^{3} - 6 {x}^{2} + b x - 6 = \left(x + 1\right) \left(x - 3\right) \left(m x + n\right)$

Write as$\text{ } k {x}^{3} - 6 {x}^{2} + b x - 6 = \left({x}^{2} - 2 x - 3\right) \left(m x + n\right)$

First consider the final constant of -6 from the LHS

Comparing LHS to RHS

$- 6 = - 3 n$

$\textcolor{red}{n = + 2} \text{ }$ giving

$k {x}^{3} - 6 {x}^{2} + b x - 6 = \left({x}^{2} - 2 x - 3\right) \left(m x + 2\right)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Multiplying out we have:

$k {x}^{3} - 6 {x}^{2} + b x - 6 = m x \left({x}^{2} - 2 x - 3\right) + 2 \left({x}^{2} - 2 x - 3\right)$

$k {x}^{3} - 6 {x}^{2} + b x - 6 = m {x}^{3} - 2 m {x}^{2} - 3 m x + 2 {x}^{2} - 4 x - 6$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Comparing coefficients of "x^3: m=k" giving}}$

$k {x}^{3} - 6 {x}^{2} + b x - 6 = k {x}^{3} - 2 k {x}^{2} - 3 k x + 2 {x}^{2} - 4 x - 6$

Collecting like terms

$k {x}^{3} - 6 {x}^{2} + b x - 6 = k {x}^{3} - 2 k {x}^{2} + 2 {x}^{2} - 3 k x - 4 x - 6$

$k {x}^{3} - 6 {x}^{2} + b x - 6 = k {x}^{3} - 2 {x}^{2} \left(k - 1\right) - x \left(3 k + 4\right) - 6$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Comparing coefficients of "x^2:" " -2(k-1)=-6" giving}}$

$k - 1 = + \frac{6}{2}$

$\textcolor{red}{k = + \frac{8}{2} = + 4}$

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

$\textcolor{b l u e}{\text{Comparing coefficients of "x " given that } k = + 4}$

$+ b x = - \left(3 k + 4\right) x$

$\implies \textcolor{red}{b = - \left(12 + 4\right) = - 16}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Factors are:$\text{ } \left(x + 1\right) \left(x - 3\right) \left(4 x + 2\right) = 0$

$\textcolor{m a \ge n t a}{\implies \text{ "x=-1"; "x=+3"; } x = - \frac{1}{2}}$