Question
2) \( 7 b^{2}+15 b+2 \)
Ask by Chang Norton. in the United States
Mar 12,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( 7b^2 + 15b + 2 = (7b + 1)(b + 2) \)
Solution
1. Identify the quadratic expression:
\[
7b^2 + 15b + 2
\]
2. Compute the product of the coefficient of \( b^2 \) and the constant term:
\[
7 \times 2 = 14
\]
3. Find two numbers that multiply to \( 14 \) and add to \( 15 \); these numbers are \( 14 \) and \( 1 \) since:
\[
14 \times 1 = 14 \quad \text{and} \quad 14 + 1 = 15
\]
4. Rewrite the middle term \( 15b \) as \( 14b + b \):
\[
7b^2 + 14b + b + 2
\]
5. Factor by grouping:
- Group the first two terms and the last two terms:
\[
(7b^2 + 14b) + (b + 2)
\]
- Factor out the greatest common factor from each group:
\[
7b(b+2) + 1(b+2)
\]
6. Factor out the common binomial factor \( (b+2) \):
\[
(b+2)(7b+1)
\]
Thus, the factored form of \( 7b^2 + 15b + 2 \) is:
\[
(7b+1)(b+2)
\]
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Simplify this solution Extra Insights
To factor the quadratic expression \( 7b^{2} + 15b + 2 \), you can look for two numbers that multiply to \( 7 \times 2 = 14 \) and add up to \( 15 \). Those numbers are \( 14 \) and \( 1 \). This allows us to split the middle term: \( 7b^{2} + 14b + 1b + 2 \). Now, group the terms: \( (7b^{2} + 14b) + (1b + 2) \). Factoring out the common factors from each group gives: \( 7b(b + 2) + 1(b + 2) \). You can now factor out \( (b + 2) \): \( (b + 2)(7b + 1) \). So the fully factored form is: \( (b + 2)(7b + 1) \). Isn't it satisfying when a quadratic nicely breaks down like this?
