Question
d. \( (x+5)(x+8) \) e. \( (x-7)(x+2) \) f. \( (-5 x+1)(x+6) \)
Ask by Curry Perkins. in the United States
Feb 04,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
Here are the expanded forms of the given binomials:
d. \( (x + 5)(x + 8) = x^2 + 13x + 40 \)
e. \( (x - 7)(x + 2) = x^2 - 5x - 14 \)
f. \( (-5x + 1)(x + 6) = -5x^2 - 29x + 6 \)
Solution
Certainly! Let's expand each of the given binomial products:
---
**d. \( (x + 5)(x + 8) \)**
To expand, use the distributive property (also known as the FOIL method for binomials):
\[
\begin{align*}
(x + 5)(x + 8) &= x \cdot x + x \cdot 8 + 5 \cdot x + 5 \cdot 8 \\
&= x^2 + 8x + 5x + 40 \\
&= x^2 + 13x + 40
\end{align*}
\]
**Expanded Form:** \( x^2 + 13x + 40 \)
---
**e. \( (x - 7)(x + 2) \)**
Apply the distributive property:
\[
\begin{align*}
(x - 7)(x + 2) &= x \cdot x + x \cdot 2 - 7 \cdot x - 7 \cdot 2 \\
&= x^2 + 2x - 7x - 14 \\
&= x^2 - 5x - 14
\end{align*}
\]
**Expanded Form:** \( x^2 - 5x - 14 \)
---
**f. \( (-5x + 1)(x + 6) \)**
Again, use the distributive property:
\[
\begin{align*}
(-5x + 1)(x + 6) &= (-5x) \cdot x + (-5x) \cdot 6 + 1 \cdot x + 1 \cdot 6 \\
&= -5x^2 - 30x + x + 6 \\
&= -5x^2 - 29x + 6
\end{align*}
\]
**Expanded Form:** \( -5x^2 - 29x + 6 \)
---
If you have any further questions or need additional assistance, feel free to ask!
Reviewed and approved by the UpStudy tutoring team
Like
error msg
Explain 
Simplify this solution Mind Expander
Did you know that factoring expressions like \( (x+5)(x+8) \) can actually help you understand quadratic equations better? When you expand it, you end up with a standard form \( x^2 + 13x + 40 \), which shows the relationship between the roots and the coefficients. This insight is crucial for anyone diving into algebra! Now, let’s talk real-world applications! Factoring expressions like \( (x-7)(x+2) \) is not just an academic exercise; it can be applied in areas such as engineering and economics where you need to optimize functions. Whether you're maximizing profit or minimizing material costs, being able to quickly factor polynomials is a handy tool in the toolbox!
