Problem 65 - Algebra 1: Concepts and Skills

Question

Solve the equation by factoring. $$ x^{2}+2 x-15=0 $$

Step-by-Step Solution

Answer

The solutions to the equation \( x^2 +2x -15 = 0 \) are \( x = 3 \) and \( x = -5 \)

Step 1: Identify coefficients

Rearrange the equation to match the general form of a quadratic equation \( ax^2 + bx + c = 0 \). From this, identify the coefficients \( a \), \( b \), and \( c \) as follows: \( a = 1 \), \( b = 2 \), and \( c = -15 \).

Step 2: Express middle term

Look for two numbers that multiply to give the product \( ac = (1)*(-15) = -15 \) and add up to \( b = 2 \). The numbers that satisfy these conditions are 5 and -3, since \( 5*(-3)=-15 \) and \( 5 + (-3) = 2 \). Therefore, we express the middle term as the sum of 5x and -3x: \( x^2 + 5x - 3x - 15 = 0 \).

Step 3: Factor by grouping

Next, factor by grouping. Combine the first two terms, getting a common factor of \( x \), and the last two terms, getting a common factor of -3: \( x(x + 5) - 3(x + 5) = 0 \). Notice that the expressions in the parenthesis are the same. We can again factor out \( (x + 5) \) as a common factor to give the final factored form: \( (x - 3)(x + 5) = 0 \).

Step 4: Solve for \( x \)

Finally, solve for \( x \) by setting each factor equal to zero and solving the resulting equations: \( x -3 = 0 \) gives \( x = 3 \) and \( x + 5 = 0 \) gives \( x = -5 \).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that typically take the form \( ax^2 + bx + c = 0 \). This equation involves the unknown variable \( x \) raised to the second power, making it a "quadratic." The coefficients \( a \), \( b \), and \( c \) represent real numbers with \( a eq 0 \). The term \( ax^2 \) is the quadratic term, \( bx \) is the linear term, and \( c \) is the constant term.

Quadratic equations can arise in various real-world scenarios, such as calculating projectile motion or finding optimal solutions in business problems. Solving quadratic equations involves finding the values of \( x \) that make the equation true, also called the "roots" or "solutions" of the equation.

One of the effective ways to solve these equations is by factoring. Factoring allows us to break down the quadratic into simpler expressions that we can easily solve.

Factoring by Grouping

Factoring by grouping is a method used to simplify and solve certain types of polynomial equations, including quadratics. This technique involves grouping terms with common factors and then factoring these groups step by step.

For the equation \( x^2 + 2x - 15 = 0 \), the goal first is to split the middle term \( 2x \) into two terms whose coefficients add to 2 and multiply to \( ac = -15 \). Therefore, the numbers 5 and -3 work because \( 5 + (-3) = 2 \) and \( 5 \times (-3) = -15 \).

Once the middle term is expressed as \( 5x - 3x \), the equation becomes \( x^2 + 5x - 3x - 15 = 0 \). We then group the terms: \((x^2 + 5x)\) and \((-3x - 15)\).

By factoring out the common terms from the groups, \((x \cdot (x + 5)) - 3(x + 5)\), both groups contain \((x + 5)\). Factoring out this common binomial gives us \((x - 3)(x + 5)\), resulting in a fully factored quadratic.

Solving Equations

Once a quadratic equation is factored, solving it becomes straightforward. The factored form \((x - 3)(x + 5) = 0\) allows us to apply the "zero product property." This property states that if a product of two factors equals zero, at least one of the factors must be zero.

Thus, setting \( (x - 3) = 0 \) gives \( x = 3 \), and setting \( (x + 5) = 0 \) results in \( x = -5 \). Both \( x = 3 \) and \( x = -5 \) are the solutions of the given quadratic equation.

Factoring is a powerful tool because it breaks down complex equations, making them more manageable. It also provides a clear path to finding the roots, helping verify the equation's original correctness. Always verify the solutions by substituting them back into the original equation to ensure they satisfy the equation.