How do you factor \[5{y^2} + 12y - 32\] ?

Hint:In the given question, we have to find the factor of the given quadratic equation. For that we will put the given equation equal to zero and then we will convert the given equation in the standard equation form that is $a{x^2} + bx + c = 0$ . And then we compare the given equation and the standard equation and get the values of a, b and c.
Then we will try to solve the equation by factorization; in factorization, we write b as a sum of two numbers such that their product is equal to the product of a and c, that is, ${b_1} \times {b_2} = a \times c$ . We find the value of ${b_1}$ and ${b_2}$ by hit and trial, if we are not able to solve an equation by factorization then we move to the quadratic formula. Using this information, we can factor the given equation.

Complete step by step answer:
The given equation is \[5{y^2} + 12y - 32\]
It can be solved by factorization as follows –
\[
5{y^2} + 12y - 32 = 0 \\
\Rightarrow 5{y^2} + 20y - 8y - 32 = 0 \\
\Rightarrow 5y(y + 4) - 8(y + 4) = 0 \\
\Rightarrow (5y - 8)(y + 4) = 0 \\
\Rightarrow 5y - 8 = 0,\,y + 4 = 0 \\
\Rightarrow y = \dfrac{8}{5},\,y = - 4 \\
\]
Hence, the factors of the equation are $y - \dfrac{8}{5} = 0$ and $y + 4 = 0$ .

Note:
The equation given in the question has a degree equal to 2 so it is quadratic; a quadratic equation is a kind of polynomial equation. The values of the unknown variable at which the function comes out to be zero are known as the factors/roots/solutions/zeros of the equation. They are simply the x- intercepts as the value of y is zero at the x-axis. The factors of a quadratic equation can be found easily with the help of several methods like factorization, completing the square, quadratic formula, etc.