Factoring Trinomials: Solving the equation

Let's solve the given equation:

$x(x^2-7x-8)=0$

Let's try to factor the expression inside the parentheses:

$x^2-7x-8$

Note that in this expression the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-8\\ m+n=-7\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to be negative, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 8 are 2 and 4, or 8 and 1, fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have different signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-8\\ n=1 \end{cases}$

therefore we'll factor the expression on the left side of the equation to:

$x^2-7x-8\\ \downarrow\\ (x-8)(x+1)$

Now let's return to the original equation and apply the factoring we detailed earlier:

$x(x^2-7x-8)=0 \\ \downarrow\\ x(x-8)(x+1)=0$

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get three simple equations and solve them by isolating the variable on one side:

$\boxed{x=0}$

or:

$x-8=0\\ \boxed{x=8}$

or:

$x+1=0\\ \boxed{x=-1}$

Let's summarize then the solution of the equation:

$x(x^2-7x-8)=0 \\ \downarrow\\ x(x-8)(x-3)=0 \\ \downarrow\\ \boxed{x=0}\\ x-8=0\rightarrow\boxed{x=8}\\ x+1=0\rightarrow\boxed{x=-1}\\ \downarrow\\ \boxed{x=0,8,-1}$

Therefore the correct answer is answer C.

Let's solve the given equation:

$x^2-8x+15=3(x-3)$

$$First, let's organize the equation by opening parentheses (using the extended distribution law) and combining like terms:

$x^2-8x+15=3(x-3) \\ x^2-8x+15=3x-9 \\ x^2-8x+15-3x+9=0 \\ x^2-11x+24=0$

Now we notice that in the resulting equation the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=24\\ m+n=-11\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a positive result, therefore we can conclude that both numbers must have the same signs, according to multiplication rules, and now we'll remember that the possible factors of 24 are 6 and 4, 12 and 2, 8 and 3, or 24 and 1. Meeting the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are equal to each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-8\\ n=-3 \end{cases}$

Therefore we can factor the expression on the left side of the equation to:

$x^2-11x+24=0 \\ \downarrow\\ (x-8)(x-3)=0$

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we get two simple equations and we'll solve them by isolating the variable in each:

$x-8=0\\ \boxed{x=8}$

or:

$x-3=0\\ \boxed{x=3}$

Let's summarize then the solution of the equation:

$x^2-8x+15=3(x-3) \\ x^2-8x+15=3x-9 \\ x^2-11x+24=0 \\ \downarrow\\ (x-8)(x-3)=0 \\ \downarrow\\ x-8=0\rightarrow\boxed{x=8}\\ x-3=0\rightarrow\boxed{x=3}\\ \downarrow\\ \boxed{x=8,3}$

Therefore the correct answer is answer B.

Let's solve the following equation:

$-7(x+1)(x-3)(x+7)=0$

First, let's divide both sides of the equation by the number outside of the parentheses:

$-7(x+1)(x-3)(x+7)=0 \hspace{6pt}\text{/}:(-7)\\ (x+1)(x-3)(x+7)=0$

Remember that the product of an expression equals 0 only if at least one of the multiplying expressions equals zero,

Therefore we should obtain three simple equations and solve them by isolating the variable in each one:

$x+1=0\\ \boxed{x=-1}$or:

$x-3=0\\ \boxed{x=3}$or:

$x+7=0\\ \boxed{x=-7}$ Hence the solution to the equation is:

$\boxed{x=-1,3-7}$Therefore the correct answer is answer D.