How do you factor the polynomial function \[5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8\]?

Hint: This type of problem is based on the concept of factoring a polynomial. First, we have to consider the polynomial with degree 4. First, we have to split the middle term of the polynomial in such a way that we get common terms from the first two terms and last two terms. Here, the middle term is \[-22{{x}^{2}}\]. We can split the middle as addition of \[-20{{x}^{2}}\] and \[-2{{x}^{2}}\]. Then, we need to take \[{{x}^{2}}\] common from the first three terms and -4 common from the last two terms. We find that \[\left( 5{{x}^{2}}+x-2 \right)\] is common in the two terms. On taking the common terms, we convert the polynomial into a product of two polynomials with degree two. Consider the two polynomials separately and find the factors.

Complete step by step solution:
According to the question, we are asked to find the factors of \[5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8\].
We have been given the polynomial is \[5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8\]. ---------(1)
The given polynomial is of degree 4 and variable x.
To find the factors, we have to consider the middle term and split the middle term in such a way that we get common terms from the first and last two terms.
We know that -22=-20-2.
Therefore, we get
\[5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8=5{{x}^{4}}+{{x}^{3}}+\left( -20-2 \right){{x}^{2}}-4x+8\]
Using the distributive property in the middle term, that is \[a\left( b+c \right)=ab+ac\].
Here, a=\[{{x}^{2}}\], b=-20 and c=-2.
\[\Rightarrow 5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8=5{{x}^{4}}+{{x}^{3}}+\left( -20 \right){{x}^{2}}+\left( -2 \right){{x}^{2}}-4x+8\]
\[\Rightarrow 5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8=5{{x}^{4}}+{{x}^{3}}-20{{x}^{2}}-2{{x}^{2}}-4x+8\]
We can rearrange the polynomial as
\[5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8=5{{x}^{4}}+{{x}^{3}}-2{{x}^{2}}-20{{x}^{2}}-4x+8\]
We can express the polynomial as
\[5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8=5{{x}^{4}}+{{x}^{3}}-2{{x}^{2}}-4\times 5{{x}^{2}}-4x+4\times 2\]
We find that \[{{x}^{2}}\] is common in the first three terms of the simplified polynomial and -4 are common in the last three terms of the simplified polynomial.
Let us take \[{{x}^{2}}\] and -4 common out of the bracket respectively.
\[\Rightarrow 5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8={{x}^{2}}\left( 5{{x}^{2}}+x-2 \right)-4\left( 5{{x}^{2}}+x-2 \right)\]
We find that \[5{{x}^{2}}+x-2\] is common in both the terms of the equation. On taking \[5{{x}^{2}}+x-2\] common, we get
\[5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8=\left( 5{{x}^{2}}+x-2 \right)\left( {{x}^{2}}-4 \right)\] -----------(2)
Now, let us consider \[\left( {{x}^{2}}-4 \right)\].
We know that 4 is the square of 2.
\[\Rightarrow {{x}^{2}}-4={{x}^{2}}-{{2}^{2}}\]
We know that \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]. Using this identity, we get
\[{{x}^{2}}-4=\left( x+2 \right)\left( x-2 \right)\] --------------(3)
Now, let us consider \[5{{x}^{2}}+x-2\].
We know that for a quadratic polynomial \[a{{x}^{2}}+bx+c\], the factors are \[\left( x-\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right)\].
On comparing with the quadratic polynomial, we get
a=5, b=1 and c=-2.
On substituting these values, we get
\[5{{x}^{2}}+x-2=\left( x-\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 5\times -2}}{2\times 5} \right)\]
On further simplification, we get
\[5{{x}^{2}}+x-2=\left( x-\dfrac{-1\pm \sqrt{1-4\times 5\times -2}}{10} \right)\]
\[\Rightarrow 5{{x}^{2}}+x-2=\left( x-\dfrac{-1\pm \sqrt{1-20\times -2}}{10} \right)\]
\[\Rightarrow 5{{x}^{2}}+x-2=\left( x-\dfrac{-1\pm \sqrt{1+40}}{10} \right)\]
\[\Rightarrow 5{{x}^{2}}+x-2=\left( x-\dfrac{-1\pm \sqrt{41}}{10} \right)\]
We know that 41 do not have a perfect square.
Therefore, we get
\[5{{x}^{2}}+x-2=\left( x-\dfrac{-1+\sqrt{41}}{10} \right)\left( x-\dfrac{-1-\sqrt{41}}{10} \right)\]
Let us take -1 common from the second term of the polynomial.
\[\Rightarrow 5{{x}^{2}}+x-2=\left( x-\dfrac{-1+\sqrt{41}}{10} \right)\left( x-\dfrac{-\left( 1+\sqrt{41} \right)}{10} \right)\]
On further simplification, we get
\[5{{x}^{2}}+x-2=\left( x-\dfrac{-1+\sqrt{41}}{10} \right)\left( x+\dfrac{1+\sqrt{41}}{10} \right)\] --------------(4)
Therefore, on substituting equation (3) and (4) in (2), we get
\[5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8=\left( x-\dfrac{-1+\sqrt{41}}{10} \right)\left( x+\dfrac{1+\sqrt{41}}{10} \right)\left( x-2 \right)\left( x+2 \right)\]
Therefore, the factors of \[5{{x}^{4}}+{{x}^{3}}-22{{x}^{2}}-4x+8\] are \[\left( x-\dfrac{-1+\sqrt{41}}{10} \right)\], \[\left( x+\dfrac{1+\sqrt{41}}{10} \right)\], x+2 and x-2.

Note: Whenever we get such a type of problem, we have to split the middle terms to find the factors. Since the given polynomial is of degree 4, we get 4 factors. Avoid calculation mistakes based on sign convention.