Factorization Practice Problems - Master Algebraic Techniques

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 $x = 0, x = -1$.

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 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:

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

Let's observe that the given equation:

$x^2-5x-50=0$is a quadratic equation that can be solved using quick factoring:

$x^2-5x-50=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-50\\ \underline{?}+\underline{?}=-5\end{cases}\\ \downarrow\\ (x-10)(x+5)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-10)(x+5)=0 \\ \downarrow\\ x-10=0\rightarrow\boxed{x=10}\\ x+5=0\rightarrow\boxed{x=-5}\\ \boxed{x=10,-5}$Therefore, the correct answer is answer C.

Let's observe that the given equation:

$x^2-19x+60=0$is a quadratic equation that can be solved using quick factoring:

$x^2-19x+60=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=60\\ \underline{?}+\underline{?}=-19\end{cases}\\ \downarrow\\ (x-4)(x-15)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-4)(x-15)=0 \\ \downarrow\\ x-4=0\rightarrow\boxed{x=4}\\ x-15=0\rightarrow\boxed{x=15}\\ \boxed{x=4,15}$Therefore, the correct answer is answer A.