How do you factor by grouping \[3{{x}^{2}}-17x+10\] ?

Hint:
When we have a polynomial of the form \[a{{x}^{2}}+\text{ }bx\text{ }+\text{ }c\], we can factor this quadratic with splitting up the b term into two terms based on the signs of a and c terms. If we have the same sign for both a and c terms, then we split the b term into two parts. These parts are formed such that the sum of them is b term itself and their product is the same as that of the product of a and c terms.

Complete step by step answer:
In the given quadratic polynomial \[3{{x}^{2}}-17x+10\], the coefficient of \[{{x}^{2}}\] and constant terms are of the same sign and their product is 30. That is we have a and c coefficients with same sign in the polynomial of form \[a{{x}^{2}}+\text{ }bx\text{ }+\text{ }c\].
Hence, we would split -17, which is the coefficient of x, in two parts, whose sum is 17 and product is 30. These are 2 and 15.
So, we write it as
\[\Rightarrow 3{{x}^{2}}-17x+10\]
We now split -17x into -15x and -2x
\[\Rightarrow 3{{x}^{2}}-15x-2x+10\]
We take the terms common in first 2 terms and last 2 terms
\[\Rightarrow 3x(x-5)-2(x-5)\]
Here, we have (x – 5) in common then
\[\Rightarrow \left( 3x\text{ }\text{ }2 \right)\text{ }\left( x\text{ }\text{ }5 \right)\]
\[\therefore \left( 3x\text{ }\text{ }2 \right)\left( x\text{ }\text{ }5 \right)\] is the required answer.

Note:
If the sign of the coefficient of \[{{x}^{2}}\] and constant terms are different, then we factor the polynomial by splitting up the b term into two terms. Here also we have to split b terms such that the sum of those parts is b term and the product is the same as that of the product of a and c terms. Factoring by grouping will not always work. In such cases, we better go with the quadratic formula.