Algebra Exercise asked in the math group and edited by me in different way

Exercise:

Let $g\left( x \right)=$ $4{{x}^{4}}-4{{x}^{3}}-71{{x}^{2}}+36x+224$

Show that , $g(x)$  has 4 zeros to be determined

Solution: we have $g\left( x \right)=4{{x}^{4}}-4{{x}^{3}}-71{{x}^{2}}+36x+224=0$

it’s clear that $224=315-91$

$\Rightarrow 4{{x}^{4}}-4{{x}^{3}}-71{{x}^{2}}+36x+315-91=0$

$\Rightarrow 4{{x}^{4}}-4{{x}^{3}}-71{{x}^{2}}+36x+315=91$

To factorize $p\left( x \right)=4{{x}^{4}}-4{{x}^{3}}-71{{x}^{2}}+36x+315\in \mathbb{R}\left[ x \right]$

So we need to find all divisors for $315$ ${{D}_{315}}=\left\{ 1,3,5,7,9,15,21,35,45,63,105,315 \right\}$

$p\left( 1 \right)=4-4-71+36+315\ne 0$ , $p\left( 3 \right)=0$ hence $\left( x-3 \right)$ is a factor of $p\left( x \right)$

So $p\left( x \right)=\left( x-3 \right)q\left( x \right)$ where $\deg \left( q\left( x \right) \right)=3$ and $q\left( x \right)\in \mathbb{R}\left[ x \right]$

Thus $p\left( x \right)=\left( x-3 \right)\left( a{{x}^{3}}+b{{x}^{2}}+cx+d \right)$

Let’s expand $p\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx-3a{{x}^{3}}-3b{{x}^{2}}-3cx-3d$

$\Rightarrow p\left( x \right)=a{{x}^{4}}+{{x}^{3}}\left( b-3a \right)+{{x}^{2}}\left( c-3b \right)+x\left( d-3c \right)-3d$

Thus $a=4\,\,,\,\,\,\,-3d=315\,\,,\,\,\,\,b-3a=-4\,\,\,,\,\,c-3b=-71\,\,,\,\,d-3c=36$

So $\left\{ a,b,c,d \right\}=\left\{ 4,8,-47,-105 \right\}$

Thus $p\left( x \right)=\left( x-3 \right)\left( 4{{x}^{3}}+8{{x}^{2}}-47x-105 \right)$

Let’s find new factor for $q\left( x \right)=4{{x}^{3}}+8{{x}^{2}}-47x-105$

By the same miner we obtain after finding the divisors for 105 that $\left( x+3 \right)$ is factor of $q\left( x \right)$

So $q\left( x \right)=\left( x+3 \right)t\left( x \right)$ where $t\left( x \right)\in \mathbb{R}\left[ x \right]\,\,\,\And \,\,\deg \left( t\left( x \right) \right)=2$

So expanding $q\left( x \right)=\left( x+3 \right)\left( a{{x}^{2}}+bx+c \right)=a{{x}^{3}}+b{{x}^{2}}+cx+3a{{x}^{2}}+3bx+3c$

$\Rightarrow q\left( x \right)=a{{x}^{3}}+{{x}^{2}}\left( b+3a \right)+x\left( c+3b \right)+3c$

So $a=4\,\,,\,\,\,b+3a=8\,\,,\,\,\,c+3b=-47\,\,\,\And \,\,3c=-105$ i.e $\left\{ a,b,c \right\}=\left\{ 4,-4,-35 \right\}$

Hence $p\left( x \right)=\left( x-3 \right)\left( x+3 \right)\left( 4{{x}^{2}}-4x-35 \right)$

But $t\left( x \right)=4{{x}^{2}}-4x-35={{\left( 2x \right)}^{2}}-2\left( 2x \right)\left( 1 \right)+1-1=35={{\left( 2x-1 \right)}^{2}}=36$

So $t\left( x \right)={{\left( 2x-1 \right)}^{2}}-36=\left( 2x-1-6 \right)\left( 2x-1+6 \right)=\left( 2x-7 \right)\left( 2x+5 \right)$

Thus $p\left( x \right)=\left( x-3 \right)\left( x+3 \right)\left( 2x-7 \right)\left( 2x+5 \right)$

Therefore, $\left( 2x-7 \right)\left( x+3 \right)\left( x-3 \right)\left( 2x+5 \right)=91$

$\Rightarrow \left( 2{{x}^{2}}-x-21 \right)\left( 2{{x}^{2}}-x-15 \right)=91$ but $-21=-15-6$

$\Rightarrow \left( 2{{x}^{2}}-x-15-6 \right)\left( 2{{x}^{2}}-x-15 \right)=91$ put $w=2{{x}^{2}}-x-15$

$\Rightarrow \left( w-6 \right)w=91\Rightarrow {{w}^{2}}-6w-91=0\Rightarrow {{\left( w-3 \right)}^{2}}-9-91={{\left( w-3 \right)}^{2}}-100=0$

$\Rightarrow \left( w-3-10 \right)\left( w-3+10 \right)=0\Rightarrow \left( w-13 \right)\left( w+7 \right)=0$

Hence $w=13\,\,\,or\,\,w=-7$

$if\,\,w=13\Leftrightarrow 2{{x}^{2}}-x-15=13\Leftrightarrow 2{{x}^{2}}-x-28=0$

$if\,\,\,w=-7\Leftrightarrow 2{{x}^{2}}-x-15=-7\Leftrightarrow 2{{x}^{2}}-x-8=0$

$2{{x}^{2}}-x-28={{x}^{2}}-\frac{1}{2}x-14={{x}^{2}}-\frac{1}{2}x+\frac{1}{16}-\frac{1}{16}-14={{\left( x-\frac{1}{4} \right)}^{2}}-\frac{255}{16}=0$

$\Rightarrow {{\left( x-\frac{1}{4} \right)}^{2}}=\frac{255}{16}\Rightarrow x-\frac{1}{4}=\pm \frac{\sqrt{255}}{4}\Rightarrow x=\frac{1\pm \sqrt{255}}{4}$  (2 zeros )

Also $2{{x}^{2}}-x-8={{x}^{2}}-\frac{1}{2}x-4={{\left( x-\frac{1}{4} \right)}^{2}}-\frac{1}{16}-4={{\left( x-\frac{1}{4} \right)}^{2}}-\frac{65}{16}=0$

So $x-\frac{1}{4}=\pm \frac{\sqrt{65}}{4}\Leftrightarrow x=\frac{1\pm \sqrt{65}}{4}$ ( 2 zeros )