Problem 12 - Beginning and Intermediate Algebra

Question

Complete the factorization. $$t^{2}-5 t+4=(t-4)(\quad)$$

Step-by-Step Solution

Answer

The complete factorization is \(t^2 - 5t + 4 = (t - 4)(t - 1)\).

Step 1: Locate the provided factor

The exercise already provides one factor of the expression, which is (t-4).

Step 2: Apply the distributive property

In order to find the second factor, we need to multiply (t-4) by an unknown term to obtain the expression \(t^2 - 5t + 4 \). Let's call the unknown term (t - x), where x is a constant we need to find.

Step 3: Multiply the factors (Distributive Property)

Now, we will multiply \((t-4)\) by \((t-x)\) to reproduce the original expression \(t^2 - 5t + 4\). \((t-4)(t-x) = t^2 - (4+x)t + 4x\)

Step 4: Compare the coefficients and find the value of x

We need to compare the coefficients of the like terms in both expressions. \(t^2 - 5t + 4 = t^2 - (4+x)t + 4x\) Here, we see that the coefficients of the linear term (-5) and the constant term (4) must be equal between both sides of the equation.

Step 5: Solve for x

Since the linear term is -5t, we can set up an equation to solve for x: 4 + x = 5 (because -5t = - (4+x)t) x = 1

Step 6: Write the complete factorization

Now that we have the value of x, we can replace it in the second factor: (t - x) = (t - 1) So, our complete factorization is: \(t^2 - 5t + 4 = (t - 4)(t - 1)\)

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest degree of the variable is two. These equations typically take the standard form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. These equations form a parabola when graphed on a coordinate plane. What makes quadratic equations interesting is that they can appear in many real-world scenarios, such as projectile motion or calculating areas.

Factorization of quadratics often involves breaking down an equation into simpler, multiplied expressions, known as factors. These factorizations allow you to see the root or roots of the equation, which are the values of \( x \) that make the equation equal to zero. In our example, the quadratic \( t^2 - 5t + 4 \) is about finding two expressions that multiply to give the original quadratic. This approach simplifies solving or analyzing the equation further.

Distributive Property

The distributive property is a fundamental algebraic property used to simplify expressions and perform multiplication across terms inside parentheses. This property states that \( a(b + c) = ab + ac \). It's a powerful tool that helps in expanding or factoring polynomials, particularly in quadratic equations.

In our exercise, the distributive property is used to multiply the given factor \((t-4)\) by an unknown factor \((t-x)\). By applying the distributive property, the expression \((t-4)(t-x)\) is expanded into \( t^2 - (4+x)t + 4x \).

Utilizing the distributive property correctly is essential when attempting to match the expanded equation back to its original quadratic form through identifying common terms.

Coefficient Comparison

Coefficient comparison is a useful technique employed to find unknown values when dealing with polynomial equations. It involves equating corresponding coefficients of like terms from two polynomial expressions.

In solving our exercise, once the expression is expanded using the distributive property, it generates \( t^2 - (4+x)t + 4x \). By comparing this to the original quadratic equation \( t^2 - 5t + 4 \), you can directly match the coefficients of each term.

• For the \( t \) terms: Compare \(- (4+x)\) with \(-5\) to solve for \( x \).
• For the constant term: Ensure that \(4x\) equals the constant from the original equation.

Solving these comparisons gives you the necessary values to complete the factorization. This method provides a clear and straightforward way to handle polynomial equations when other methods might seem too complex.