Problem 26
Question
Factor completely, if possible. Check your answer. $$z^{2}-10 z+24$$
Step-by-Step Solution
Verified Answer
The completely factored expression is \( (z - 4)(z - 6) \).
1Step 1: Identify the coefficients
In the given quadratic expression \(z^2 - 10z + 24\), the coefficient of the quadratic term is 1, the coefficient of the linear term is -10, and the constant term is 24.
2Step 2: Find two numbers that multiply to give 24 and add to -10
We need to find two numbers that multiply to give 24 and add to give -10. Although there are many ways to find these numbers, one simple method is trial-and-error:
- \(1 \times 24 = 24\), but \(1 + 24 ≠ -10\)
- \((-1) \times (-24) = 24\), and \((-1) + (-24) = -25 ≠ -10\)
- \(2 \times 12 = 24\), but \(2 + 12 ≠ -10\)
- \((-2) \times (-12) = 24\), and \((-2) + (-12) = -14 ≠ -10\)
- \(3 \times 8 = 24\), but \(3 + 8 ≠ -10\)
- \((-3) \times (-8) = 24\), and \((-3) + (-8) = -11 ≠ -10\)
- \(4 \times 6 = 24\), but \(4 + 6 ≠ -10\)
- \((-4) \times (-6) = 24\), and \((-4) + (-6) = -10\)
Hence, the two numbers are -4 and -6.
3Step 3: Rewrite the expression as the product of two binomials
Since -4 and -6 are the numbers we're looking for, we can rewrite the given expression as the product of two binomials: \((z - 4)(z - 6)\)
4Step 4: Check the answer
To check if our solution is correct, we can use the FOIL method to expand the product of binomials:
\((z - 4)(z - 6) = z^2 - 6z - 4z + 24 = z^2 - 10z + 24\)
As the expanded expression is the same as the given expression, our solution is correct.
#Result#: The completely factored expression is \( (z - 4)(z - 6) \).
Key Concepts
Quadratic EquationsFactoring MethodsPolynomial Expressions
Quadratic Equations
Quadratic equations are a powerful way to describe a variety of real-world problems. They are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.
The variable \( x \) represents an unknown that we need to solve.What makes quadratic equations unique is their highest exponent, which is 2. This gives the equation its parabolic shape when graphed.
The solutions to quadratic equations are the values of \( x \) where the equation equals zero.Typically, these solutions can be found via:
Understanding and being able to identify elements like the coefficients and constants play essential roles in solving these equations.
The variable \( x \) represents an unknown that we need to solve.What makes quadratic equations unique is their highest exponent, which is 2. This gives the equation its parabolic shape when graphed.
The solutions to quadratic equations are the values of \( x \) where the equation equals zero.Typically, these solutions can be found via:
- Factoring
- Completing the square
- Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Understanding and being able to identify elements like the coefficients and constants play essential roles in solving these equations.
Factoring Methods
Factoring quadratic expressions involves rewriting them as a product of simpler terms.
This makes it easier to solve for the variable. Let's delve into the common factoring methods often used.Firstly, identifying numbers that add and multiply to specific values is crucial.
Let's consider the quadratic expression \( z^2 - 10z + 24 \). Here, the key task is choosing two numbers that multiply to give the constant term 24, while also adding up to the linear coefficient -10.
These numbers will help rewrite the quadratic expression as a product of two binomials.Trial-and-Error:
This makes it easier to solve for the variable. Let's delve into the common factoring methods often used.Firstly, identifying numbers that add and multiply to specific values is crucial.
Let's consider the quadratic expression \( z^2 - 10z + 24 \). Here, the key task is choosing two numbers that multiply to give the constant term 24, while also adding up to the linear coefficient -10.
These numbers will help rewrite the quadratic expression as a product of two binomials.Trial-and-Error:
- We try combinations like \((-4)\) and \((-6)\) since \((-4) \times (-6) = 24\) and \((-4) + (-6) = -10\).
- Once these numbers are identified, rewrite the quadratic expression as \((z - 4)(z - 6)\).
- To verify, expand the binomials back to the original quadratic expression to ensure correctness.
Polynomial Expressions
Polynomial expressions are mathematical expressions involving sums of powers in one or more variables. The term "polynomial" comes from "poly-" for many and "-nomial" meaning terms.
They can have any number of terms with non-negative integer exponents.In the context of quadratic expressions, they're a special kind of polynomial expression where the highest power of the variable is 2.
Polynomials like \( z^2 - 10z + 24 \) are standard forms in algebra and are crucial for understanding more complicated mathematical concepts.Properties of polynomials include being closed under addition, subtraction, and multiplication.
This means you can add, subtract, or multiply them without changing the polynomial's essential nature.
This enables analyzing and solving equations like quadratics by recognizing patterns and employing appropriate factoring methods.
They can have any number of terms with non-negative integer exponents.In the context of quadratic expressions, they're a special kind of polynomial expression where the highest power of the variable is 2.
Polynomials like \( z^2 - 10z + 24 \) are standard forms in algebra and are crucial for understanding more complicated mathematical concepts.Properties of polynomials include being closed under addition, subtraction, and multiplication.
This means you can add, subtract, or multiply them without changing the polynomial's essential nature.
- Each individual term in a polynomial has a coefficient, which is the number in front of the variable.
- The highest degree term degree dictates the polynomial's degree. Here, \(z^2\) means it's a second-degree polynomial or quadratic.
This enables analyzing and solving equations like quadratics by recognizing patterns and employing appropriate factoring methods.
Other exercises in this chapter
Problem 26
Solve each equation. $$10 a^{2}+20 a=0$$
View solution Problem 26
Factor out the greatest common factor. Be sure to check your answer. $$\frac{1}{8} k^{2}+\frac{7}{8} k$$
View solution Problem 27
Factor completely. $$2 p^{4}-24 p^{3}+72 p^{2}$$
View solution Problem 27
Solve each equation. $$d^{2}-15 d=-54$$
View solution