Factor using the ‘ac’ method:

$4y^2+8y+3$

We have the equation$4y^2+8y+3$.

We will first take out any common factor, if present, from the equation to reduce it to the simplest form.

Then we will find any two numbers whose product comes out to be the product of first and last term and the sum results in the middle term.

Then by splitting the middle term and writing it into the an equation consisting of those two numbers, we will get the factors by taking the common factors put from that particular equation.

We can see that there is no common factor in the equation. So, now we will look for two numbers whose product comes out to be the product of first and last term, i.e. ,$12$and sum comes out to be the middle term, i.e., $8$. The factors of$12$and their sum are-

Through this we can say that the two numbers will be$2and6$.

Two numbers which we got are $2,6$.

Splitting the middle term gives us,

$$\begin{gathered} =4y^2+8y+3 \\ =4y^2+2y+6y+3 \\ =2y(2y+1)+3(2y+1) \\ =(2y+1)(2y+3) \end{gathered}$$

This shows that factors are$(2y+1)(2y+3)$.

We will check the solution by multiplying. If we get the same given equation, our calculations are right.

So,

$$\begin{gathered} =(2y+1)(2y+3) \\ =4y^2+6y+2y+3 \\ =4y^2+8y+3 \end{gathered}$$