# When p is -4<p<=3, (x+2) is a factor of the function f(x) =x^2-(p+1)x +p+4. Moreover, g(x) = xf(2x) - f(x - 1/3) +2/3x. Determine the factors of g(x) without finding f(2x) and f(x-1/3)?

Jul 13, 2016

color(red)("Factors:- " x,(x+1),(12x-1)

#### Explanation:

$\textcolor{b l u e}{\left(x + 2\right) \text{ being a factor of "f(x)," the value of "f(-2) " should be zero.}}$

So
$f \left(- 2\right) = {\left(- 2\right)}^{2} - \left(p + 1\right) \left(- 2\right) + p + 4 = 0$

$\implies 4 + 2 p + 2 + 4 = 0$

$\implies p = - \frac{10}{3}$

So $f \left(x\right) = {x}^{2} - \left(- \frac{10}{3} + 1\right) - \frac{10}{3} + 4$

$\implies f \left(x\right) = {x}^{2} + \frac{7}{3} x + \frac{2}{3}$

$\implies f \left(x\right) = \frac{1}{3} \left(3 {x}^{2} + 7 x + 2\right)$

$\implies f \left(x\right) = \frac{1}{3} \left(3 {x}^{2} + 6 x + x + 2\right)$

=>f(x)=1/3(3x(x+2)+1(x+2)

$\implies f \left(x\right) = \frac{1}{3} \left(3 x + 1\right) \left(x + 2\right)$

$\implies f \left(x\right) = \left(x + \frac{1}{3}\right) \left(x + 2\right)$

This factorisation of f(x) reveals that (3x+1) and (x+2) are two fators of f(x)

So color(red)(f(-1/3)=0" and "f(-2)=0

Now we are to determine the factors of g(x) by trial,
where
$g \left(x\right) = x f \left(2 x\right) - f \left(x - \frac{1}{3}\right) + \frac{2}{3} x$

Trial-1

$\textcolor{g r e e n}{\text{Putting "x =-1->g(x)" we have}}$

$g \left(- 1\right) = - 1 \cdot f \left(- 2\right) - f \left(- 1 - \frac{1}{3}\right) - \frac{2}{3}$

$\implies g \left(- 1\right) = - 1 \cdot 0 - f \left(- \frac{4}{3}\right) - \frac{2}{3}$

$= g \left(- 1\right) = - f \left(- \frac{4}{3}\right) - \frac{2}{3} =$

Now
$\implies f \left(x\right) = \left(x + \frac{1}{3}\right) \left(x + 2\right)$

$\implies f \left(- \frac{4}{3}\right) = \left(- \frac{4}{3} + \frac{1}{3}\right) \left(- \frac{4}{3} + 2\right)$

$\implies f \left(- \frac{4}{3}\right) = \left(- 1\right) \left(\frac{2}{3}\right) = - \frac{2}{3}$

$\therefore g \left(- 1\right) = - f \left(- \frac{4}{3}\right) - \frac{2}{3}$

$= - \left(- \frac{2}{3}\right) - \frac{2}{3} = 0$

color(red)("This means " (x+1)" is a factor of " g(x)

Trial-2

$\textcolor{g r e e n}{\text{Putting "x =0->g(x)" we have}}$

$g \left(0\right) = 0 f \left(0\right) - f \left(0 - \frac{1}{3}\right) + \frac{2}{3} \times o$

$\implies g \left(x\right) = - f \left(- \frac{1}{3}\right) = 0$

color(red)("This means " (x)" is a factor of " g(x)

Trial-3

On inspection it is found that g(x) contains a ${x}^{3}$ as highest power of x. So another linear factor should exist.

$\textcolor{g r e e n}{\text{Putting "x =1/12->g(x)" we have}}$

$g \left(\frac{1}{12}\right) = \frac{1}{12} f \left(\frac{1}{6}\right) - f \left(\frac{1}{12} - \frac{1}{3}\right) + \frac{2}{3} \times \frac{1}{12}$

$\implies g \left(\frac{1}{12}\right) = \frac{1}{12} f \left(\frac{1}{6}\right) - f \left(- \frac{1}{4}\right) + \frac{1}{18}$

Now

$\implies f \left(\frac{1}{6}\right) = \left(\frac{1}{6} + \frac{1}{3}\right) \left(\frac{1}{6} + 2\right)$

$= \frac{3}{6} \times \frac{13}{6} = \frac{13}{12}$

And

$\implies f \left(- \frac{1}{4}\right) = \left(- \frac{1}{4} + \frac{1}{3}\right) \left(- \frac{1}{4} + 2\right)$

$= \frac{1}{12} \times \frac{7}{4} = \frac{7}{48}$

$\therefore g \left(\frac{1}{12}\right) = \frac{1}{12} \times \frac{13}{12} - \frac{7}{48} + \frac{1}{18}$

$= \frac{13 - 21 + 8}{144} = \frac{0}{144} = 0$

color(red)("This means " (12x-1)" is a factor of " g(x)