Solving Equations by Factoring: Solving the problem

We will factor using the shortened multiplication formulas:

$a^2-b^2=(a-b)(a+b)$$(a-b)^2=a^2-2ab+b^2$

$(a+b)^2=a^2+2ab+b^2$

Let's remember that there might be more than one solution for the value of x.

According to one solution, we'll take the square root:

$(x-5)^2=0$

$x-5=0$

$x=5$

According to the second solution, we'll use the shortened multiplication formula:

$(x-5)^2=x^2-10x+25=0$

We'll use the trinomial:

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

$x-5=0$

$x=5$

or

$x-5=0$

$x=5$

Therefore, according to all calculations, $x=5$

We will factor using the shortened multiplication formulas:

$a^2-b^2=(a-b)(a+b)$$(a-b)^2=a^2-2ab+b^2$

$(a+b)^2=a^2+2ab+b^2$

Remember that there might be more than one solution for the value of x.

According to the first formula:

$x^2=a^2$

We'll take the square root:

$x=a$

$25=b^2$

We'll take the square root:

$b=5$

We'll use the first shortened multiplication formula:

$a^2-b^2=(a-b)(a+b)$

$x^2-25=(x-5)(x+5)=0$

Therefore:

$x+5=0$

$x=-5$

Or:

$x-5=0$

$x=5$

We will factor using trinomials, remembering that there is more than one solution for the value of X:

$2x^2-7x+5=0$

We will factor -7X into two numbers whose product is 10:

$2x^2-5x-2x+5=0$

We will factor out a common factor:

$2x(x-1)-5(x-1)=0$

$(2x-5)(x-1)=0$

Therefore:

$x-1=0$$x=1$

Or:

$2x-5=0$

$2x=5$

$x=2.5$

First, factor using trinomials and remember that there might be more than one solution for the value of $x$:

$-x^2-7x-12=0$

Divide by -1:

$x^2+7x+12=0$

$(x+3)(x+4)=0$

Therefore:

$x+4=0$

$x=-4$

Or:

$x+3=0$

$x=-3$

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

• Step 1: Expand each term:
  $-2x(3-x) = -6x + 2x^2$ and $(x-3)^2 = x^2 - 6x + 9$.
• Step 2: Substitute the expanded terms into the equation:
  $-6x + 2x^2 + x^2 - 6x + 9 = 9$.
• Step 3: Combine like terms:
  $(2x^2 + x^2) + (-6x - 6x) + 9 = 9$.
• Step 4: Simplify further:
  $3x^2 - 12x + 9 = 9$.
• Step 5: Move all terms to one side to form a quadratic equation:
  $3x^2 - 12x + 9 - 9 = 0$.
• Step 6: Simplify the expression:
  $3x^2 - 12x = 0$.
• Step 7: Factor out the common term:
  $3x(x - 4) = 0$.
• Step 8: Solve for $x$:
  Since $3x = 0$ or $x - 4 = 0$, we find $x = 0$ or $x = 4$.

Therefore, the values of $x$ that satisfy the equation are $\mathbf{x = 0}$ and $\mathbf{x = 4}$.

Let's open the parentheses, remembering that there might be more than one solution for the value of X:

$(x-4)^2+x(x-12)=16$

$x^2-8x+16+x^2-12x=16$

$2x^2-20x=0$

$2x(x-10)=0$

Therefore:

$x-10=0$

$x=10$

Or:

$2x=0$

$x=0$

To solve the problem, we follow these steps:

• Step 1: Identify the given cubic equation, $12x^3 - 9x^2 - 3x = 0$.
• Step 2: Look for common factors in the terms of the equation.
• Step 3: Apply the zero-product property to factor and solve the equation.

Let's work through the solution:

Step 1: Observe that each term in the equation $12x^3 - 9x^2 - 3x$ has a common factor of $3x$. So, we can factor $3x$ out of the equation, giving us:

$3x (4x^2 - 3x - 1) = 0$

Step 2: Having factored out $3x$, we now have a product of terms equaling zero. According to the zero-product property, at least one of the factors must be zero:

$3x = 0 \quad \text{or} \quad 4x^2 - 3x - 1 = 0$

This gives us one solution directly:

$x = 0$

Step 3: Solve the quadratic equation $4x^2 - 3x - 1 = 0$ using the quadratic formula, where $a = 4$, $b = -3$, and $c = -1$:

The quadratic formula is:

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$

Applying it to our equation:

$x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4 \cdot 4 \cdot (-1)}}}}{2 \cdot 4}$

$x = \frac{{3 \pm \sqrt{{9 + 16}}}}{8}$

$x = \frac{{3 \pm \sqrt{25}}}{8}$

$x = \frac{{3 \pm 5}}{8}$

This gives us two solutions:

When $\sqrt{25} = 5$, $x = \frac{{3 + 5}}{8} = 1$.

When $\sqrt{25} = -5$, $x = \frac{{3 - 5}}{8} = -\frac{1}{4}$.

Therefore, the solutions to the equation $12x^3 - 9x^2 - 3x = 0$ are $x = 0$, $x = 1$, and $x = -\frac{1}{4}$.

Verifying against the provided choices, the correct choice is choice 2: $x=0,x=1,x=-\frac{1}{4}$.

To solve the equation $(x+5)^2 = 0$, we will use the fact that a perfect square is zero only when the quantity being squared is zero itself.

• Step 1: Set $x+5 = 0$ since it is the term being squared.
• Step 2: Solve for $x$ by subtracting 5 from both sides: $x+5 = 0$ $x = -5$

Therefore, the solution to the equation is $x = -5$.