Question

How do you factor \[2{{x}^{4}}-19{{x}^{3}}+57{{x}^{2}}-64x+20\]?

Answer 1

Hint: For the given question we are given to factor the equation \[2{{x}^{4}}-19{{x}^{3}}+57{{x}^{2}}-64x+20\]. Let us assume the equation as y and then we have to use trial and error methods to find the first root after that we have to factor it out and then we have to find roots for the remaining equation.

Complete step-by-step solution:
To find the factor for the equation \[2{{x}^{4}}-19{{x}^{3}}+57{{x}^{2}}-64x+20\]. Let us assume the equation as y and consider it as equation (1).
By considering the given equation as equation (1), we get
\[y=2{{x}^{4}}-19{{x}^{3}}+57{{x}^{2}}-64x+20.................\left( 1 \right)\]
By the rational root theorem any rational zeros of y must be expressible in the form p/q for integer’s p and q with p as a divisor of the constant term 20 and q is a divisor of the coefficient 2 of the leading term.
So, the only possible roots are:
\[\pm \dfrac{1}{2},\text{ }\pm 1,\text{ }\pm 2,\text{ }\pm \dfrac{5}{2},\text{ }\pm 4,\text{ }\pm 5,\text{ }\pm 10,\text{ }\pm 20\]
Let us try these by substituting \[x=\dfrac{1}{2}\], we get
\[y=\dfrac{1}{8}-\dfrac{19}{8}+\dfrac{57}{4}-32+20=0\]
So, therefore we can say that \[x=\dfrac{1}{2}\] is a zero which means that \[\left( 2x-1 \right)\] is a factor.
\[\Rightarrow y=2{{x}^{4}}-19{{x}^{3}}+57{{x}^{2}}-64x+20=\left( 2x-1 \right)\left( {{x}^{3}}-9{{x}^{3}}+24x-20 \right)\]
Let us consider it as equation (2).
\[y=\left( 2x-1 \right)\left( {{x}^{3}}-9{{x}^{3}}+24x-20 \right)...........\left( 2 \right)\]
Let us check with other possible rational roots.
We find:
\[\begin{align}
  & g(1)=1-9+24-20=-4 \\
 & g\left( -1 \right)=-1-9-24-20=-54 \\
 & g\left( 2 \right)=8-36+48-20=0 \\
\end{align}\]
So, \[x=2\] is a zero and \[\left( x-2 \right)\] is a factor, we get
\[\Rightarrow y=\left( 2x-1 \right)\left( x-2 \right)\left( {{x}^{2}}-7x+10 \right)\]
Let us consider it as equation (3), we get
\[y=\left( 2x-1 \right)\left( x-2 \right)\left( {{x}^{2}}-7x+10 \right).................\left( 3 \right)\]
By solving quadratic equation by factorization method, we get
\[y=\left( 2x-1 \right)\left( x-2 \right)\left( x-2 \right)\left( x-5 \right)\]
Let us consider the above equation as equation (4), we get
\[y=\left( 2x-1 \right)\left( x-2 \right)\left( x-2 \right)\left( x-5 \right)..............\left( 4 \right)\]
Hence by equation (4) factors of \[2{{x}^{4}}-19{{x}^{3}}+57{{x}^{2}}-64x+20\] are \[\left( 2x-1 \right)\],\[\left( x-2 \right)\] , \[\left( x-2 \right)\] and \[\left( x-5 \right)\].

Note: We can also do this problem in a different way i.e. first we have to find the first root by trial and error method and after that we have to factor the equation by direct division method. In this way we have to do until the x power in the quotient term is 2.