How do you factor by grouping \[12{{x}^{2}}+29x+15\]?

Hint: we solve this question using factorization method. We will use the middle term to factorize. We will split the middle term in a way that their sum is equal to the middle term and product is equal to product of first and last terms. Finally, we have taken common terms and we have to represent the equation as a product of \[\left( x-a \right)\left( x-b \right)\].

Complete step-by-step answer:
 Here we have been asked to factorize the quadratic polynomial \[12{{x}^{2}}+29x+15\] by grouping.
First we have to split the middle term into two in such a way that the sum is equal to \[29\] and the product is equal to the product of constant term \[15\]and \[12{{x}^{2}}\]that is \[180\].
To do this, first we need to find all the primes factors of \[180\]
So let us find prime factors for \[180\] which will satisfy the condition.
We can write \[180\] as \[180=2\times 2\times 5\times 3\times 3\] as a product of primes. Now we can group two \[2's\] and one \[5\] as one group and two \[3's\] as one group . so we can satisfy our conditions.
\[180=20\times 9\]
\[29=20+9\]
So these values satisfy the condition now we will split the middle term as \[20x+9x\]
Then the polynomial will look like
\[\Rightarrow 12{{x}^{2}}+20x+9x+15\]
Now we can take common terms out to make them as factors.
Here we can see from the first two terms we can take \[4x\] as common and from the next two terms we can \[3\] as common . After taking the common terms out the equation will look like
\[\Rightarrow 4x\left( 3x+5 \right)+3\left( 3x+5 \right)\]
Now in the above polynomial we can see \[x+5\] as common and we can take it out then the polynomial will look like
\[\Rightarrow \left( 3x+5 \right)\left( 4x+3 \right)\]
Now the polynomial is represented as a product of factors.
 So the factors of the polynomial \[12{{x}^{2}}+29x+15\] are \[\left( 3x+5 \right)\left( 4x+3 \right)\].

Note: we can also solve the above question using discriminant method also. In which we can solve the x for values a and b after that making them factors as \[\left( x-a \right)\left( x-b \right)\]. But in the question it is clearly mentioned to use factor method by grouping we used this method to solve the problem.