Answer
\[2{{x}^{2}}-9x+10=\left( x-2 \right)\left( 2x-5 \right)\]
Work Step by Step
Let the factor of the provided expression is \[\left( ax+b \right)\left( cx+d \right)\]. Then,
\[\begin{align}
& 2{{x}^{2}}-9x+10=\left( ax+b \right)\left( cx+d \right) \\
& 2{{x}^{2}}-9x+10=ac{{x}^{2}}+\left( ad+bc \right)x+bd
\end{align}\]
Compare the coefficient,
\[\begin{align}
& ac=2 \\
& ad+bc=-9 \\
& bd=10
\end{align}\]
The positive factor of 2 is\[\left( 1,2 \right)\].
And the factors of 10 are \[\left( 1,10 \right)\], \[\left( -1,-10 \right)\], \[\left( 2,5 \right)\], and \[\left( -2,-5 \right)\].
Then, the possible factorization of the expression \[2{{x}^{2}}-9x+10\]can be:
\[\begin{align}
& \left( x+1 \right)\left( 2x+10 \right)=2{{x}^{2}}+12x+10 \\
& \left( x-1 \right)\left( 2x-10 \right)=2{{x}^{2}}+12x+10 \\
& \left( x+2 \right)\left( 2x+5 \right)=2{{x}^{2}}+9x+10 \\
& \left( x-2 \right)\left( 2x-5 \right)=2{{x}^{2}}-9x+10
\end{align}\]
\[2{{x}^{2}}-9x+10\]factors as \[\left( x-2 \right)\left( 2x-5 \right)\].
Therefore, the resultant factorization is:
\[2{{x}^{2}}-9x+10=\left( x-2 \right)\left( 2x-5 \right)\]
