Extracting Square Roots: Factoring Out the Greatest Common Factor (GCF)

The equation in the problem is:

$x^2-x=0$

First let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $x$and this is because the first power is the lowest power in the equation and therefore is included both in the term with the second power and in the term with the first power, any power higher than this is not included in the term with the first power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

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

Let's continue and address the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$\boxed{x=0}$

or:

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

Let's summarize then the solution to the equation:

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

Therefore the correct answer is answer B.

To solve the equation $x^4 + 2x^2 = 0$, we will use the technique of factoring. Let's proceed step-by-step:

First, notice that both terms $x^4$ and $2x^2$ have a common factor of $x^2$. We can factor $x^2$ out from the equation:

$x^2(x^2 + 2) = 0$

Now, to solve for $x$, we apply the Zero Product Property, which gives us that at least one of the factors must be zero:

• $x^2 = 0$ or
• $x^2 + 2 = 0$

Solving the first case, $x^2 = 0$:

$x = 0$

For the second case, $x^2 + 2 = 0$, we reach:

$x^2 = -2$

Since $x^2 = -2$ has no real solutions (squares of real numbers are non-negative), we can conclude that this equation doesn't provide additional real solutions.

Therefore, the only real solution to the given equation is $x = 0$.

The correct choice from the provided options is:

$x=0$

The equation in the problem is:

$3x^2+9x=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $3x$because the first power is the lowest power in the equation and therefore is included both in the term with the second power and in the term with the first power. Any power higher than this is not included in the term with the first power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms for the letters,

For the numbers, note that 9 is a multiple of 3, therefore it is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

$3x^2+9x=0 \\ \downarrow\\ 3x(x+3)=0$

Let's continue and address the fact that on the left side of the equation we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$3x=0 \hspace{8pt}\text{/}:3\\ \boxed{x=0}$

In solving the above equation, we divided both sides of the equation by the term with the variable,

Or:

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

Let's summarize the solution of the equation:

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

Therefore the correct answer is answer C.

The equation in the problem is:

$x^5-4x^4=0$

First note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $x^4$since the fourth power is the lowest power in the equation and therefore is included both in the term with the fifth power and in the term with the fourth power, any power higher than this is not included in the term with the fourth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$x^5-4x^4=0 \\ \downarrow\\ x^4(x-4)=0$

Let's continue and address the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^4=0 \hspace{8pt}\text{/}\sqrt[4]{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$(In this case taking the even root of the right side of the equation will yield two possibilities - positive and negative, however since we're dealing with zero, we get only one solution)

Or:

$x-4=0\\ \downarrow\\ \boxed{x=4}$

Let's summarize the solution of the equation:

$x^5-4x^4=0 \\ \downarrow\\ x^4(x-4)=0 \\ \downarrow\\ x^4=0 \rightarrow\boxed{ x=0}\\ x-4=0 \rightarrow \boxed{x=4}\\ \downarrow\\ \boxed{x=0,4}$

Therefore the correct answer is answer C.

The equation in the problem is:

$x^6+x^5=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $x^5$because the fifth power is the lowest power in the equation and therefore is included both in the term with the sixth power and in the term with the fifth power. Any power higher than this is not included in the term with the fifth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression. We will continue and perform the factoring:

$x^6+x^5=0 \\ \downarrow\\ x^5(x+1)=0$

Let's continue and address the fact that on the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^5=0 \hspace{8pt}\text{/}\sqrt[5]{\hspace{6pt}}\\ \boxed{x=0}$(in this case taking the odd root of the right side of the equation will yield one possibility)

or:

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

Let's summarize the solution of the equation:

$x^6+x^5=0 \\ \downarrow\\ x^5(x+1)=0 \\ \downarrow\\ x^5=0 \rightarrow\boxed{ x=0}\\ x+1=0 \rightarrow \boxed{x=-1}\\ \downarrow\\ \boxed{x=0,-1}$

Therefore the correct answer is answer A.

The equation in the problem is:

$x^7-x^6=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $x^6$since the sixth power is the lowest power in the equation and therefore is included both in the term with the seventh power and in the term with the sixth power. Any power higher than this is not included in the term with the sixth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$x^7-x^6=0 \\ \downarrow\\ x^6(x-1)=0$

Let's continue and address the fact that in the left side of the equation we got from the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^6=0 \hspace{8pt}\text{/}\sqrt[6]{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$

(in this case taking the even root of the right side of the equation will yield two possibilities - positive and negative but since we're dealing with zero, we get only one answer)

or:

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

Let's summarize the solution of the equation:

$x^7-x^6=0 \\ \downarrow\\ x^6(x-1)=0 \\ \downarrow\\ x^6=0 \rightarrow\boxed{ x=0}\\ x-1=0 \rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}$

Therefore the correct answer is answer C.

The equation in the problem is:

$4x^4-12x^3=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $4x^3$because the third power is the lowest power in the equation and therefore is included in both the term with the fourth power and the term with the third power. Any power higher than this is not included in the term with the third power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor for the variables,

For the numbers, note that 12 is a multiple of 4, therefore 4 is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

$4x^4-12x^3=0 \\ \downarrow\\ 4x^3(x-3)=0$

Let's continue and address the fact that in the left side of the equation we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$4x^3=0 \hspace{8pt}\text{/}:4\\ x^3=0 \hspace{8pt}\text{/}\sqrt[3]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we first divided both sides of the equation by the term with the unknown and then extracted a cube root for both sides of the equation.

(In this case, extracting an odd root for the right side of the equation yielded one possibility)

Or:

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

Let's summarize the solution of the equation:

$4x^4-12x^3=0 \\ \downarrow\\ 4x^3(x-3)=0 \\ \downarrow\\ 4x^3=0 \rightarrow\boxed{ x=0}\\ x-3=0\rightarrow \boxed{x=3}\\ \downarrow\\ \boxed{x=0,3}$

Therefore the correct answer is answer A.

The equation in the problem is:

$7x^{10}-14x^9=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $7x^9$since the ninth power is the lowest power in the equation and therefore is included in both the term with the tenth power and the term with the ninth power. Any power higher than this is not included in the term with the ninth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms for the variables,

For the numbers, note that 14 is a multiple of 7, therefore 7 is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

$7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0$

Let's continue and address the fact that on the left side of the equation we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get a result of 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$7x^9=0 \hspace{8pt}\text{/}:7\\ x^9=0 \hspace{8pt}\text{/}\sqrt[9]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we first divided both sides of the equation by the term with the variable, and then we extracted a ninth root from both sides of the equation.

(In this case, extracting an odd root from the right side of the equation yielded one possibility)

Or:

$x-2=0 \\ \boxed{x=2}$

Let's summarize the solution of the equation:

$7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0\\ \downarrow\\ 7x^9=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}$

Therefore, the correct answer is answer A.

The equation in the problem is:

$7x^3-x^2=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $x^2$since the second power is the lowest power in the equation and therefore is included both in the term with the third power and in the term with the second power. Any power higher than this is not included in the term with the second power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$7x^3-x^2=0 \\ \downarrow\\ x^2(7x-1)=0$

Let's continue and address the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^2=0 \hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$(in this case taking the even root of the right side of the equation will indeed yield two possibilities, positive and negative, but since we're dealing with zero, we'll get only one possibility)

or:

$7x-1=0$Let's solve this equation to get the additional solutions (if they exist) to the given equation:

We got a simple first-degree equation which we'll solve by isolating the unknown on one side, we'll do this by moving terms and then dividing both sides of the equation by the coefficient of the unknown:

$7x-1=0 \\ 7x=1\hspace{8pt}\text{/}:7\\ \boxed{x=\frac{1}{7}}$

Let's summarize the solution of the equation:

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

Therefore the correct answer is answer C.

The problem at hand is to solve the equation $x^4 + x^2 = 0$.

Let's begin by factoring the expression:

The given equation is: $x^4 + x^2 = 0$

We can factor out the common factor of $x^2$ from both terms:

$x^2(x^2 + 1) = 0$

To solve for $x$, we set each factor equal to zero:

• $x^2 = 0$

Solving for $x$, we have:

$x = 0$

Next, consider the second factor:

• $x^2 + 1 = 0$

Solving for $x$, we have:

$x^2 = -1$

Since $x^2 = -1$ has no real solutions, we ignore these solutions in the real number system.

Thus, the only real solution to the equation $x^4 + x^2 = 0$ is:

$x = 0$

To solve this problem, we start by factoring the given equation:

The equation is $x^6 - 4x^4 = 0$. Notice that both terms contain a power of $x$. We can factor out the greatest common factor, which is $x^4$.

This yields $x^4(x^2 - 4) = 0$.

Next, we apply the zero-product property, which states that if a product of factors equals zero, at least one of the factors must be zero:

• First factor: $x^4 = 0$. This implies that $x = 0$.
• Second factor: $x^2 - 4 = 0$. We solve this quadratic equation by factoring further:

The quadratic equation $x^2 - 4 = 0$ can be factored using the difference of squares:

$(x - 2)(x + 2) = 0$.

Again applying the zero-product property, we set each factor equal to zero:

• For $x - 2 = 0$, $x = 2$.
• For $x + 2 = 0$, $x = -2$.

Thus, the complete set of solutions to the equation is $x = 0, x = 2, x = -2$.

Therefore, the solution to the problem is $x = 0, x = \pm 2$.

The equation in the problem is:

$x^{100}-9x^{99}=0$

First, note that in the left side we can factor out a common factor from the terms, the largest common factor for the numbers and letters in this case is $x^{99}$because the power of 99 is the lowest power in the equation and therefore is included both in the term with power of 100 and in the term with power of 99. Any power higher than this is not included in the term with the lowest power of 99, and therefore this is the term with the highest power that can be factored out as a common factor from all letter terms,

Let's continue and perform the factoring:

$x^{100}-9x^{99}=0 \\ \downarrow\\ x^{99}(x-9)=0$

Let's continue and address the fact that on the left side of the equation we received in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get a result of 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$x^{99}=0 \hspace{8pt}\text{/}\sqrt[99]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we extracted a 99th root from both sides of the equation.

(In this case, extracting an odd root from the right side of the equation yielded one possibility)

Or:

$x-9=0 \\ \boxed{x=9}$

Let's summarize the solution of the equation:

$x^{100}-9x^{99}=0 \\ \downarrow\\ x^{99}(x-9)=0 \\ \downarrow\\ x^{99}=0 \rightarrow\boxed{ x=0}\\ x-9=0\rightarrow \boxed{x=9}\\ \downarrow\\ \boxed{x=0,9}$

Therefore, the correct answer is answer C.

The equation in the problem is:

$x^7-5x^6=0$

First let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $x^6$since the sixth power is the lowest power in the equation and therefore is included both in the term with the seventh power and in the term with the sixth power, any power higher than this is not included in the term with the sixth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$x^7-5x^6=0 \\ \downarrow\\ x^6(x-5)=0$

Let's continue and address the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^6=0 \hspace{8pt}\text{/}\sqrt[6]{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$

(in this case taking the even root of the right side of the equation will yield two possibilities - positive and negative, however since we're dealing with zero, we get only one solution)

or:

$x-5=0\\ \downarrow\\ \boxed{x=5}$

Let's summarize the solution of the equation:

$x^7-5x^6=0 \\ \downarrow\\ x^6(x-5)=0 \\ \downarrow\\ x^6=0 \rightarrow\boxed{ x=0}\\ x-5=0 \rightarrow \boxed{x=5}\\ \downarrow\\ \boxed{x=0,5}$

Therefore the correct answer is answer A.

The equation in the problem is:

$15x^4-30x^3=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $15x^3$since the third power is the lowest power in the equation and therefore is included in both the term with the fourth power and the term with the third power, any power higher than this is not included in the term with the third power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms for the variables,

For the numbers, note that 30 is a multiple of 15, therefore 15 is the largest common factor for the numbers for both terms in the expression,

Let's continue and perform the factoring:

$15x^4-30x^3=0 \\ \downarrow\\ 15x^3(x-2)=0$

Let's continue and address the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 as a result of multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$15x^3=0 \hspace{8pt}\text{/}:15\\ x^3=0 \hspace{8pt}\text{/}\sqrt[3]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we first divided both sides of the equation by the term with the unknown and then extracted a cube root for both sides of the equation.

(In this case extracting an odd root for the right side of the equation yielded one possibility)

Or:

$x-2=0\\ \boxed{x=2}$

Let's summarize the solution of the equation:

$15x^4-30x^3=0 \\ \downarrow\\ 15x^3(x-2)=0 \\ \downarrow\\ 15x^3=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}$

Therefore the correct answer is answer B.

To solve this polynomial equation, we'll follow these steps:

• Step 1: Identify and factor out the greatest common factor (GCF).
• Step 2: Solve the resulting simpler equations for $x$.
• Step 3: Verify the solutions.

Now, let's work through each step:

Step 1: Factor out the Greatest Common Factor (GCF)

The given equation is $12x^4 - 3x^3 = 0$.
Both terms share a common factor of $3x^3$. Factoring out this common factor, we get:

$3x^3(4x - 1) = 0$

Step 2: Solve the factored equation

We now have two factors: $3x^3$ and $(4x - 1)$. Set each factor to zero to find possible solutions:

• For the factor $3x^3 = 0$:
  Solving this gives $x^3 = 0$, which implies $x = 0$.
• For the factor $4x - 1 = 0$:
  Solving this gives $4x = 1$, leading to $x = \frac{1}{4}$.

Step 3: Verification

Substitute $x = 0$ and $x = \frac{1}{4}$ back into the original equation to verify:

• Substituting $x = 0$:
  $12(0)^4 - 3(0)^3 = 0$, which is true.
• Substituting $x = \frac{1}{4}$:
  Calculations show $12\left(\frac{1}{4}\right)^4 - 3\left(\frac{1}{4}\right)^3 = 0$, which also holds true.

The correct choice from the given options is $x = 0, \frac{1}{4}$.

Therefore, the solution to the problem is $x=0,\frac{1}{4}$.

The equation in the problem is:

$7x^5-14x^4=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $7x^4$since the fourth power is the lowest power in the equation and therefore is included both in the term with the fifth power and the term with the fourth power, any power higher than this is not included in the term with the fourth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms for the variables,

For the numbers, note that 14 is a multiple of 7, therefore 7 is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

$7x^5-14x^4=0 \\ \downarrow\\ 7x^4(x-2)=0$

Let's continue and consider the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$7x^4=0 \hspace{8pt}\text{/}:7\\ x^4=0 \hspace{8pt}\text{/}\sqrt[4]{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$

In solving the equation above, we first divided both sides of the equation by the term with the variable, and then we took the fourth root of both sides of the equation.

(In this case, taking an even root of the right side of the equation will yield two possibilities - positive and negative but since we're dealing with zero, we get only one answer)

Or:

$x-2=0\\ \boxed{x=2}$

Let's summarize the solution of the equation:

$7x^5-14x^4=0 \\ \downarrow\\ 7x^4(x-2)=0 \\ \downarrow\\ 7x^4=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}$

Therefore the correct answer is answer A.

The equation in the problem is:

$7x^8-21x^7=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $7x^7$since the seventh power is the lowest power in the equation and therefore is included in both the term with the eighth power and the term with the seventh power. Any power higher than this is not included in the term with the seventh power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor for variables,

For the numbers, we notice that 21 is a multiple of 7, therefore 7 is the largest common factor for numbers in both terms of the expression,

Let's continue and perform the factoring:

$7x^8-21x^7=0 \\ \downarrow\\ 7x^7(x-3)=0$

Let's continue and address the fact that in the left side of the equation we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$7x^7=0 \hspace{8pt}\text{/}:7\\ x^7=0 \hspace{8pt}\text{/}\sqrt[7]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we first divided both sides of the equation by the term with the variable and then extracted a seventh root from both sides of the equation.

(In this case, extracting an odd root from the right side of the equation yielded one possibility)

Or:

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

Let's summarize the solution of the equation:

$7x^8-21x^7=0 \\ \downarrow\\ 7x^7(x-3)=0 \\ \downarrow\\ 7x^7=0 \rightarrow\boxed{ x=0}\\ x-3=0\rightarrow \boxed{x=3}\\ \downarrow\\ \boxed{x=0,3}$

Therefore, the correct answer is answer B.

The equation in the problem is:

$8x-x^4=0$

First note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $x$because the first power is the lowest power in the equation and therefore is included both in the term with the fourth power and in the term with the first power. Any power higher than this is not included in the term with the first power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$8x-x^4=0 \\ \downarrow\\ x(8-x^3)=0$

Let's continue and address the fact that on the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$\boxed{x=0}$

or:

$8-x^3=0\\ 8=x^3\hspace{8pt}\text{/}\sqrt[3]{\hspace{6pt}}\\ \downarrow\\ \boxed{x=2}$(in this case taking the odd root of the left side of the equation will yield only one possibility)

Let's summarize the solution of the equation:

$8x-x^4=0 \\ \downarrow\\ x(8-x^3)=0 \\ \downarrow\\ \boxed{ x=0}\\ 8-x^3=0 \rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}$

Therefore the correct answer is answer A.

To solve this problem, we need to apply the following steps:

• Step 1: Identify the highest factor common to all terms and factor it out.
• Step 2: Set each factor equal to zero and solve for $x$.
• Step 3: Validate solutions within the context of the problem statement.

Now, following these steps:

Step 1: Identify and factor out the greatest common factor:

The given equation is $28x^8 - 7x^7 = 0$.

The greatest common factor (GCF) of the terms $28x^8$ and $7x^7$ is $7x^7$.

We can factor the equation as:

$7x^7(4x - 1) = 0$.

Step 2: Set each factor equal to zero:

For $7x^7 = 0$, dividing both sides by 7 yields $x^7 = 0$, which implies $x = 0$.

For $4x - 1 = 0$, solve for $x$:

$4x = 1$

$x = \frac{1}{4}$

Step 3: Verify solutions:

The values $x = 0$ and $x = \frac{1}{4}$ both satisfy the original equation, as substituting them back results in $0$.

Thus, the solutions to the equation are $x = 0$ and $x = \frac{1}{4}$.

The answer, based on the choices provided, is: Answers a and b are correct.

To solve the equation $x^8 - 25x^6 = 0$, we start by noticing that both terms share a common factor of $x^6$. We can factor out $x^6$ from the expression:

$x^6(x^2 - 25) = 0$

According to the zero-product property, a product is zero if and only if at least one of the factors is zero. Therefore, we have two separate equations to solve:

• $x^6 = 0$
• $x^2 - 25 = 0$

For $x^6 = 0$:

$x = 0$

For $x^2 - 25 = 0$, this can be seen as a difference of squares, which factors as:

$(x - 5)(x + 5) = 0$

Again, using the zero-product property, we solve the factors:

• $x - 5 = 0$ gives $x = 5$
• $x + 5 = 0$ gives $x = -5$

The solutions to the equation are therefore $x = 0, x = 5,$ and $x = -5$.

The correct answer choice is "Answers a + b", where $\pm 5$ and $0$ are included as solutions.