Answer
\[{{x}^{2}}-8x+15=\left( x-3 \right)\left( x-5 \right)\]
Work Step by Step
Note that this trinomial has leading coefficient is 1, therefore, to factor it, follow the steps below:
Step1:
Find first two terms whose product is\[{{x}^{2}}\]:
\[{{x}^{2}}-8x+15=\left( x-3 \right)\left( x-5 \right)\]
Step2:
Find the last two terms that when multiplied give \[15\]:
\[\text{Factors of }15:\text{ }-1,-15\text{ }1,15\text{ }-3,-5\text{ 3},5\]
Step3:
Now try various combinations of these factors and look for the pair of factors whose sum is the coefficient of the middle term. That is, \[-8\]in this case.
\[\begin{align}
& \text{Factors of }15\text{ }:-1,-15\text{ }1,15\text{ }-3,-5\text{ 3},5 \\
& \text{Sum of factors : }-16\text{ 16 }-8\text{ 8 } \\
\end{align}\]
Thus, from above,\[-3,-5\] is the required factor of \[15\]that gives the desired sum.
So that,
\[{{x}^{2}}-8x+15=\left( x-3 \right)\left( x-5 \right)\].
