Factorization - Common Factor: Third power

First, we use the power law to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It is necessary to keep in mind that:

$x^4=x^3\cdot x$Next, we return to the problem and extract the greatest common factor for the numbers separately and for the letters separately,

For the numbers, the greatest common factor is

$4$and for the letters it is:

$x^3$and therefore for the extraction

$4x^3$outside the parenthesis

We obtain the expression:

$4x^3+8x^4=4x^3(1+2x)$To determine what the expression inside the parentheses is, we use the power law, our knowledge of the multiplication table, and the answer to the question: "How many times do we multiply the common factor that we took out of the parenthesis to obtain each of the terms of the original expression that we factored?

Therefore, the correct answer is: a.

It is always recommended to review again and check that you get each and every one of the terms of the expression that is factored when opening the parentheses (through the distributive property), this can be done in the margin, on a piece of scrap paper, or by marking the factor we removed and each and every one of the terms inside the parenthesis, etc.

First, we take out the smallest power

$6x^6-9x^4=$

$6x^4\left(x^2-1.5\right)=0$

If possible, we reduce the numbers by a common factor

Finally, we will compare the two sections with: $0$

$6x^4=0$

We divide by: $6x^3$

$x=0$

$x^2-1.5=0$

$x^2=1.5$

$x=\pm\sqrt{\frac{3}{2}}$

To solve the equation $x^4 - 8x = 0$, we'll follow these steps:

• Step 1: Factor out the greatest common factor.
• Step 2: Set each factor equal to zero and solve for $x$.

Now, let's work through each step:
Step 1: The equation given is $x^4 - 8x = 0$. Both terms on the left contain $x$ as a factor. We can factor out $x$ to rewrite the equation as:

$x(x^3 - 8) = 0$

Step 2: To find the solutions, set each factor to zero.

If $x = 0$, then one solution is:

$x = 0$

Next, solve for $x$ in the equation $x^3 - 8 = 0$:
Add 8 to both sides:

$x^3 = 8$

Take the cube root of both sides:

$x = \sqrt[3]{8} = 2$

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

Thus, the correct answer is: $x = 0, 2$.

To solve the equation $x^3 + x^2 - 12x = 0$, follow these steps:

• Step 1: Factor out the greatest common factor. The common factor here is $x$.
• Step 2: The equation becomes $x(x^2 + x - 12) = 0$.
• Step 3: Apply the zero-product property. This gives us two equations to solve: $x = 0$ and $x^2 + x - 12 = 0$.
• Step 4: Solve $x = 0$. This is a straightforward solution: $x = 0$.
• Step 5: Solve the quadratic equation $x^2 + x - 12 = 0$. We will factor it:
• Factor as $(x - 3)(x + 4) = 0$.
• Set each factor equal to zero: $x - 3 = 0$ or $x + 4 = 0$.
• Solving these, we obtain $x = 3$ and $x = -4$.

Therefore, the solutions to the equation are $x = 0$, $x = 3$, and $x = -4$.

Thus, the complete solution set for $x$ is $x = 0, 3, -4$.

To solve the given cubic equation $x^3 - 7x^2 + 6x = 0$, follow these steps:

• Step 1: Identify that the equation can be factored by its Greatest Common Factor (GCF).

There is an $x$ common in all terms: $x(x^2 - 7x + 6) = 0$

• Step 2: Factor the quadratic expression $x^2 - 7x + 6$.

Look for two numbers that multiply to $6$ (the constant term) and add up to $-7$ (the coefficient of the linear term). The numbers are $-1$ and $-6$. Thus:

$x^2 - 7x + 6 = (x - 1)(x - 6)$

• Step 3: Set each factor equal to zero to solve for $x$.

Now that the equation is fully factored as $x(x - 1)(x - 6) = 0$, apply the zero product property:

$x = 0$, $x - 1 = 0$ (so $x = 1$), $x - 6 = 0$ (so $x = 6$)

Thus, the solutions to the equation $x^3 - 7x^2 + 6x = 0$ are $x = 0$, $x = 1$, and $x = 6$.

To solve this problem, we need to factor the cubic polynomial equation $x^3 - x^2 - 4x + 4 = 0$. We'll begin by applying the Rational Root Theorem, which suggests that possible rational roots are factors of the constant term (4) divided by factors of the leading coefficient (1). This gives us potential roots: $\pm 1, \pm 2, \pm 4$.

Let's test these possible roots by substituting them into the polynomial:

• For $x = 1$, the polynomial evaluates to $1^3 - 1^2 - 4 \times 1 + 4 = 1 - 1 - 4 + 4 = 0$. Thus, $x = 1$ is a root.
• For $x = -1$, it evaluates to $(-1)^3 - (-1)^2 - 4(-1) + 4 = -1 - 1 + 4 + 4 = 6$. Thus, $x = -1$ is not a root.
• For $x = 2$, it evaluates to $2^3 - 2^2 - 4 \times 2 + 4 = 8 - 4 - 8 + 4 = 0$. Thus, $x = 2$ is a root.
• For $x = -2$, it evaluates to $(-2)^3 - (-2)^2 - 4(-2) + 4 = -8 - 4 + 8 + 4 = 0$. Thus, $x = -2$ is a root.

From these calculations, we identified $x = 1$, $x = 2$, and $x = -2$ as roots of the polynomial.

The polynomial can be factored as $(x - 1)(x - 2)(x + 2) = 0$. Solving each factor for zero, we obtain the roots $x = 1$, $x = 2$, and $x = -2$.

Therefore, the correct answer from the given choices is Answers a and c, which correspond to the roots $x = \pm 2$ and $x = 1$.