How do you solve $4{x^2} + 28x - 32 = 0$ by factoring?

Hint: In the method of factoring, taking the reference of the equation $a{x^2} + bx + c = 0$, we have to find any two numbers whose multiplication results in a product of term $a$ and $c$ . Also, the sum of two numbers, if the term $c$ is positive and the difference between two numbers, if the term $c$ is negative should result in the term $b$.

Complete step-by-step answer:
First of all, we compare the given equation $4{x^2} + 28x - 32 = 0$ with the standard equation $a{x^2} + bx + c = 0$ .
From the comparison, we get
$\Rightarrow$$a = 4$
$\Rightarrow$$b = 28$
$\Rightarrow$$c = - 32$
Here, the term $c$ is negative.
Hence, we need to find two numbers whose difference will lead to $b$ and their product will lead to a product of term $a$ and $c$ .
As per the problem, we need two numbers that follow the conditions
Product = $a \times c = 4 \times 32 = 128$
Difference = $b = 28$ .
For the product $\;128$ , the pairs can be $\;1$ and $\;128$ , $\;2$ and $\;64$ , $\;4$ and $\;32$ , $\;8$ and $\;16$
Out of these all pairs, the pair $\;4$ and $\;32$ seems to be valid because
Product = $4 \times 32 = 128 = a \times c$
Difference = $32 - 4 = 28 = b$
Now, we have two numbers for factoring.
Now, consider the equation
$\Rightarrow$$4{x^2} + 28x - 32 = 0$
Now, we can replace the term $\;28$ with the difference
$ \Rightarrow 4{x^2} + (32 - 4)x - 32 = 0$
Opening the brackets,
$ \Rightarrow 4{x^2} + 32x - 4x - 32 = 0$
Shifting the terms,
$ \Rightarrow 4{x^2} - 4x + 32x - 32 = 0$
Now, considering the first two terms and expanding them
$ \Rightarrow 4 \times x \times x - 4 \times x$
Here, in both terms $\;4$ and $x$ are common
So taking $\;4$ and $x$ common and putting the remaining in the brackets
$ \Rightarrow 4x(x - 1)$
Similarly, for the last two terms
$ \Rightarrow 32 \times x - 32$
Here, in both terms $\;32$ is common
So taking $\;32$ common and putting the remaining in the brackets
$ \Rightarrow 32(x - 1)$
Therefore, replacing the equations we get,
$ \Rightarrow 4x(x - 1) + 32(x - 1) = 0$
From this equation, taking the bracket $(x - 1)$ common, we get
$ \Rightarrow (x - 1)(4x + 32) = 0$
Here either of the two brackets is equal to zero.
$ \Rightarrow (x - 1) = 0$
$ \Rightarrow x = 1$
Also considering the other bracket
$ \Rightarrow (4x + 32) = 0$
Shifting $\;32$ on the right-hand side
$ \Rightarrow 4x = - 32$
$ \Rightarrow x = - \dfrac{{32}}{4}$
Hence the final answer can be written as
$ \Rightarrow x = - 8$
Hence, we have two values for $x$

$x = 1$ or $x = - 8$

Note:
Here, the conditions to remember for the two numbers depends on the sign of $c$. If $c$ is positive, we need to take the sum of two numbers to give the value $b$. If $c$ is negative, we need to take the difference of two numbers to give the value $b$ . The alternate formula to find the value of $x$ is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.