Solving Equations by Factoring - Examples, Exercises and Solutions

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:

$x+3=0$

$x=-3$

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:

$2x-5=0$

$2x=5$

$x=2.5$

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 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 $\mathbf{x = 4}$.

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 $x = -5$.

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 find the lengths of the sides of the right triangle, we will apply the Pythagorean theorem, which states $a^2 + b^2 = c^2$ for a right triangle, where $c$ is the hypotenuse.

Given the side lengths are $x + 9$, $x + 2$, and $x + 10$, we assume $x + 10$ is the hypotenuse because it is the largest value and confirm it by checking with the theorem.

Substitute into the Pythagorean theorem:
$(x + 2)^2 + (x + 9)^2 = (x + 10)^2$

Let's expand each side and solve for $x$:
$(x + 2)^2 = x^2 + 4x + 4$
$(x + 9)^2 = x^2 + 18x + 81$
$(x + 10)^2 = x^2 + 20x + 100$

Combine these into a single equation:
$x^2 + 4x + 4 + x^2 + 18x + 81 = x^2 + 20x + 100$

Simplify and combine like terms:
$2x^2 + 22x + 85 = x^2 + 20x + 100$

Rearrange to form a quadratic equation:
$2x^2 + 22x + 85 - x^2 - 20x - 100 = 0$
$x^2 + 2x - 15 = 0$

Factor the quadratic equation:
$(x + 5)(x - 3) = 0$

Solve for $x$:
$x + 5 = 0 \Rightarrow x = -5$ (Not valid as $x > 1$)
$x - 3 = 0 \Rightarrow x = 3$

Therefore, substituting $x = 3$ will provide the side lengths:
$<begin>$ - Short side: $x + 2 = 3 + 2 = 5$
$<same time>$ - Other side: $x + 9 = 3 + 9 = 12$
$<sell>$ - Hypotenuse: $x + 10 = 3 + 10 = 13$

These side lengths $5$, $12$, and $13$ form a well-known Pythagorean triple. Therefore, the solution to the problem is $5, 12, 13$.

To solve this problem, we begin by using the Pythagorean theorem, as the triangle is right-angled. Let's identify the hypotenuse:

• The side lengths are given as $x$, $x + 2$, and $x + 4$.
• Since $x + 4$ is the largest, it will serve as the hypotenuse, so we'll denote the sides as follows: $a = x$, $b = x + 2$, and $c = x + 4$.

Using the Pythagorean theorem, we write:

$x^2 + (x + 2)^2 = (x + 4)^2$

Let's expand and simplify the equation:

$x^2 + (x^2 + 4x + 4) = x^2 + 8x + 16$

Simplifying further:

$2x^2 + 4x + 4 = x^2 + 8x + 16$

Rearrange all terms to one side:

$2x^2 + 4x + 4 - x^2 - 8x - 16 = 0$

Simplifying gives:

$x^2 - 4x - 12 = 0$

This is a standard quadratic equation, which we can solve using factoring. By factoring, we find:

$(x - 6)(x + 2) = 0$

Setting each factor equal to zero gives solutions $x = 6$ and $x = -2$. Since $x > 1$, we discard $x = -2$.

The valid solution is $x = 6$.

Now, substitute $x = 6$ back into the expressions for the side lengths:

• The first side: $x = 6$
• The second side: $x + 2 = 6 + 2 = 8$
• The hypotenuse: $x + 4 = 6 + 4 = 10$

Therefore, the lengths of the sides of the triangle are $6$, $8$, and $10$, which matches choice 4.

Therefore, the correct answer is choice 4: $6, 8, 10$.

To solve for the side length of the square, we follow these steps:

• Step 1: Given each side of the square as $x+4$ cm, use the area formula for a square: $\text{Area} = (\text{side length})^2$.
• Step 2: Since the area is 36 cm$^2$, set up the equation:

$(x+4)^2 = 36$.

• Step 3: Solve for $x+4$:

Taking the square root of both sides,

$x+4 = 6$ or $x+4 = -6$.

• Step 4: Solve each equation for $x$:
• From $x+4 = 6$, we get $x = 2$.
• From $x+4 = -6$, we get $x = -10$.
• Step 5: Apply the condition $x > -4$:

Only $x = 2$ satisfies the condition $x > -4$.

• Step 6: Calculate the side length using $x = 2$:

Side length = $x+4 = 2+4 = 6$ cm.

Therefore, the length of the sides of the square is $6$ cm.

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

• Step 1: Set up the equation using the area of the square.
• Step 2: Solve for $x$ using algebraic methods.
• Step 3: Validate the solution against the given condition.

Now, let's work through each step:
Step 1: Given that the area of the square is 16, we use the area formula: $(x+2)^2 = 16$.
Step 2: Solving the equation, take the square root of both sides:
$x + 2 = \pm 4$ This produces two solutions: $x + 2 = 4 \quad \text{or} \quad x + 2 = -4$ Step 3: Solve each equation for $x$:
For $x + 2 = 4$, we have: $x = 4 - 2 = 2$ For $x + 2 = -4$, we have: $x = -4 - 2 = -6$ Since $x > -2$, we discard the solution $x = -6$ because it does not satisfy the condition.
Thus, the acceptable value is $x = 2$.

The length of the sides of the square is $x + 2 = 2 + 2 = 4$ cm.

Therefore, the solution to the problem is 4, and it matches with choice 1.

The problem involves finding the side lengths of an isosceles right triangle given its area. Let's proceed with the solution.

Since the triangle is an isosceles right triangle, the two legs are equal, and the area $A$ is provided by the formula: $A = \frac{1}{2} \cdot \text{base} \cdot \text{height}$ For this triangle, both the base and the height are $x + 5$.

The area is given as 12.5, so we set up the equation: $12.5 = \frac{1}{2} \cdot (x + 5) \cdot (x + 5)$ $12.5 = \frac{1}{2} \cdot (x + 5)^2$

Multiply both sides by 2 to solve for $(x + 5)^2$: $25 = (x + 5)^2$

Take the square root of both sides: $x + 5 = \sqrt{25}$ $x + 5 = 5$

Solve for $x$: $x = 5 - 5$ $x = 0$

Therefore, the length of each leg of the triangle is $x + 5 = 5$ cm.

For the hypotenuse $c$, use the Pythagorean theorem $c^2 = a^2 + a^2$, where $a = x + 5 = 5$: $c^2 = 5^2 + 5^2 = 25 + 25 = 50$ $c = \sqrt{50} = 5\sqrt{2}$

Thus, the lengths of the sides of the triangle are $5$, $5$, and $5\sqrt{2}$.

Therefore, the correct solution is $5, 5, 5\sqrt{2}$.

To solve this problem, we'll utilize the properties of an isosceles right triangle and the area formula:

• Step 1: Identify the expression for the legs of the triangle as $x + 8$.
• Step 2: Use the area formula for a right triangle, $\frac{1}{2} \times (\text{leg})^2 = 32$.
• Step 3: Given that the triangle is isosceles, solve for $x$ by substituting and expanding the terms.
• Step 4: Derive the hypotenuse using the known leg value and calculate $\text{leg} \times \sqrt{2}$.

Now, let's work through each step:

Step 1: Recognize both legs of the isosceles triangle are $x + 8$.

Step 2: Apply the area formula for right triangles:
We know $\frac{1}{2} \times (\text{leg}) \times (\text{leg}) = 32$. Therefore, the equation is:

Step 3: Simplify and solve for $x$.

Given the constraint $x > -8$, we discard $x + 8 = -8$ since it violates the condition. Therefore,

Recalculate $x + 8$, which states the leg is 8.

Step 4: Determine the hypotenuse.

Therefore, the side lengths of the triangle are $8, 8, 8\sqrt{2}$. Match this to the choices given, which is option 3.

The lengths of the sides of the triangle are $8, 8, 8\sqrt{2}$.