Solve x² + 3 = -4x: Finding the Value of X

Let's notice that this is a quadratic equation, therefore we'll first arrange it and move all terms to one side and have 0 on the other side:

$x^2+3=-4x \\ x^2+4x+3=0$

remembering that when moving a term to the other side its sign changes,

Now we want to solve this equation using factoring,

First we'll check if we can factor out a common factor, but this isn't possible, since there is no factor common to all three terms on the left side of the equation, so we'll look for trinomial factoring:

Note that the coefficient of the quadratic term (the term with power of 2) is 1, therefore we can try to use quick trinomial method: (this factoring is also called "automatic trinomial"),

But before we do this in our problem - let's recall the rule for quick trinomial method:

The rule states that for the quadratic algebraic expression in the general form:

$x^2+bx+c$

we can find a factored form if we can find two numbers $m,\hspace{4pt}n$such that the following conditions (quick trinomial method conditions) are met:

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$

If we can find such two numbers $m,\hspace{4pt}n$then we can factor the general expression above into a product form and present it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

which is its factored form (multiplication factors) of the expression,

Let's return now to the equation in our problem that we got in the last step after arranging it:

$x^2+4x+3=0$

Note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$ are:$\begin{cases} c=3\\ b=4 \end{cases}$where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side using quick trinomial method, above, so we'll look for a pair of numbers $m,\hspace{4pt}n$

that satisfy:

$\begin{cases} m\cdot n=3 \\ m+n=4 \end{cases}$

We'll try to find these numbers through logical thinking and using our knowledge of multiplication tables, starting with the multiplication of the two required numbers $m,\hspace{4pt}n$meaning - from the first row of requirements we mentioned in the last step:

$m\cdot n=3$

We can identify that their product needs to give a positive result, therefore we can conclude that their signs must be the same,

Next we'll consider the factors (whole numbers) of 3, and from our knowledge of multiplication tables we can know that there's only one possibility for such factors: 3 and 1, where we previously concluded that their signs must be identical, a quick check of this possibility regarding the second condition (and there's always an obligation to check the condition, even if there's only one possible pair of whole number factors as mentioned above):

$m+n=4$

will lead to the quick conclusion that the only possibility for both conditions above to be met together is:

$3,\hspace{4pt}1$meaning - for:

$m=3,\hspace{4pt}n=1$

(it doesn't matter which we call m and which we call n)

It holds that:

$\begin{cases} \underline{3}\cdot \underline{1}=3\\ \underline{3}+\underline{1}=4 \end{cases}$

From here - we understood what numbers we're looking for and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2+4x+3\\ \downarrow\\ (x+3)(x+1)$

Meaning we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

Therefore we factored the quadratic expression on the left side of the equation using quick trinomial method, and the equation is:

$x^2+4x+3=0 \\ \downarrow\\ (x+3)(x+1)=0\\$

Now that the expression on the left side is factored we'll continue to its quick solution,

Let's note a simple fact, on the left side there's a multiplication between two terms, and on the right side is 0,

Therefore we can conclude that the only two possibilities for which this equation will hold are if:

$x+3=0$

or if:

$x+1=0$

since only multiplying a number by 0 will give the result 0,

From here we'll solve the two new equations we got:

$x+3=0 \rightarrow x=-3\\ x+1=0\rightarrow x=-1$

where we solved each equation separately,

To summarize: We therefore got the solutions to the quadratic equation and used quick trinomial method to factor the quadratic expression on its left side:

$x^2+4x+3=0 \\ \downarrow\\ (x+3)(x+1)=0\\$which are:

$x_1=-3, \hspace{4pt}x_2=-1$

where substituting either of these solutions, the first or the second, in the equation - will yield a true statement,

Therefore the correct answer is answer B.