Factorization Practice Problems - Master Algebraic Techniques

To solve this problem, we will factor the given polynomial expression:

Step 1: Identify the greatest common factor (GCF) in the equation $9x^3 - 12x^2 = 0$. The GCF of the terms $9x^3$ and $12x^2$ is $3x^2$.

Step 2: Factor out the GCF from the polynomial:

$9x^3 - 12x^2 = 3x^2(3x - 4) = 0$.

Step 3: Apply the zero-product property. Set each factor equal to zero:

• $3x^2 = 0$
• $3x - 4 = 0$

Step 4: Solve each equation for $x$:

For $3x^2 = 0$, divide by 3:

$x^2 = 0$ → $x = 0$.

For $3x - 4 = 0$, add 4 to both sides and then divide by 3:

$3x = 4$
$x = \frac{4}{3}$.

Thus, the solutions to the equation $9x^3 - 12x^2 = 0$ are $x = 0$ and $x = \frac{4}{3}$.

Therefore, the correct answer is:

$x=0, x=\frac{4}{3}$

To solve the equation $-9x + 3x^2 = 0$, follow these steps:

Step 1: Factor the equation.

Observe that both terms in the equation share a common factor, $3x$. We can factor this out:

$-9x + 3x^2 = 3x(-3 + x)$.

The factored equation is $3x(-3 + x) = 0$.

Step 2: Apply the zero product property.

According to the zero product property, if the product of two factors is zero, then at least one of the factors must be zero. This gives us two equations to solve:

• $3x = 0$
• $-3 + x = 0$

Step 3: Solve each equation.

• For $3x = 0$, divide both sides by 3 to solve for $x$:
• $x = 0$
• For $-3 + x = 0$, add 3 to both sides to solve for $x$:
• $x = 3$

Therefore, the solutions to the equation $-9x + 3x^2 = 0$ are $x = 0$ and $x = 3$.

Matching these solutions to the given choices, the correct answer is choice 3: $x = 0, x = 3$.

Thus, the values of $x$ that satisfy the equation are $x = 0$ and $x = 3$.

To solve this quadratic equation by factoring, follow these steps:

• Step 1: Write the equation: $x^2 - 6x + 8 = 0$.
• Step 2: Find two numbers that multiply to $+8$ (the constant term) and add up to $-6$ (the coefficient of $x$).

These numbers are $-2$ and $-4$, since $(-2) \times (-4) = 8$ and $(-2) + (-4) = -6$.

• Step 3: Rewrite the quadratic expression as $(x - 2)(x - 4) = 0$.
• Step 4: Solve each factor separately:
  • $x - 2 = 0 \Rightarrow x = 2$
  • $x - 4 = 0 \Rightarrow x = 4$

Therefore, the solutions to the quadratic equation $x^2 - 6x + 8 = 0$ are $x = 2$ and $x = 4$.

The correct choice for the solution is:

$x=2,x=4$ which corresponds to choice 4.

To solve this problem, we'll follow these steps:

• Step 1: Factor the equation $x^2 + x = 0$.
• Step 2: Use the Zero-Product Property to solve for $x$.

Now, let's work through each step:

Step 1: Start by factoring the left-hand side of the equation:
$x^2 + x = x(x + 1)$

Step 2: Apply the Zero-Product Property:
Since $x(x + 1) = 0$, we have two possible equations:
1) $x = 0$
2) $x + 1 = 0$

For the second equation, solve for $x$:
$x + 1 = 0$ implies $x = -1$

Therefore, the solutions to the equation are $x = 0$ and $x = -1$.

Hence, the value of the parameter $x$ is $x = 0, x = -1$.

Our goal is to factor the expression on the left side of the given equation:

$x^2+x-2=0$

Note that the coefficient of the quadratic term in the expression on the left side is 1, therefore, we can (try to) factor the expression by 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 the given values:

$m\cdot n=-2\\ m+n=1\\$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 the multiplication rules. Note that the possible factors of 2 are 2 and 1, fulfilling the second requirement mentioned. Furthermore the fact that the signs of the numbers are different from each other leads us to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-1\\ n=2 \end{cases}$

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

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

The correct answer is answer A.