Solve 3x²+10x+17 = 2x²-3+x: 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:

$3x^2+10x+17=2x^2-3+x \\ 3x^2-2x^2+10x-x+17+3=0\\ x^2+9x+20=0$

where in the first stage we moved all terms to one side while remembering that when moving a term its sign changes and in the second stage we combined like terms,

We now 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, therefore, we'll look for trinomial factoring:

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

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

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

$x^2+bx+c$

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

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

If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression above into a product form and write 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 we got in the last stage after arranging it:

$x^2+9x+20=0$

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

$x^2+bx+c$ are:$\begin{cases}c=20 \\ b=9\end{cases}$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 factoring mentioned above, so we'll look for two numbers $m,\hspace{4pt}n$that satisfy:

$\begin{cases}m\cdot n=20 \\ m+n=9\end{cases}$

Let's try to find these numbers through logical thinking and using our multiplication table knowledge, 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 stage:

$m\cdot n=20$

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

Next we'll look at the factors (whole numbers) of 20, and from our multiplication table knowledge we know there are three possibilities for such factors: 2 and 10, 4 and 5, 20 and 1, where we previously concluded their signs must be identical, a quick check of these possibilities 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):

$m+n=9$

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

$4,\hspace{4pt}5$

meaning - for:

$m=4,\hspace{4pt}n=5$

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

It satisfies that:

$\begin{cases} \underline{4}\cdot\underline{5}=20 \\ \underline{4}+\underline{5}=9 \end{cases}$

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:

$x^2+9x+20\\ \downarrow\\ (x+4)(x+5)$

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 factoring, and the equation is:

$x^2+9x+20=0 \\ \downarrow\\ (x+4)(x+5)=0\\$

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 this equation to be satisfied are if:

$x+4=0$

or if:

$x+5=0$

since only multiplying a number by 0 gives the result 0,

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

$x+4=0 \rightarrow x=-4\\ x+5=0\rightarrow x=-5$

where we solved each equation separately,

Let's summarize: We therefore got the solutions to the quadratic equation and used quick trinomial factoring to factor the quadratic expression on its left side:

$x^2+9x+20=0 \\ \downarrow\\ (x+4)(x+5)=0\\$which are:

$x_1=-4,\hspace{4pt}x_2=-5$

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

Therefore the correct answer is answer B.