Question
Factor each completely. \( \begin{array}{ll}\text { 1) } b^{2}+8 b+7 & \text { 2) } n^{2}-11 n+10 \\ \text { 3) } m^{2}+m-90 & \text { 4) } n^{2}+4 n-12\end{array} \)
Ask by Harrington Chang. in the United States
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
1. \( (b + 7)(b + 1) \)
2. \( (n - 10)(n - 1) \)
3. \( (m + 10)(m - 9) \)
4. \( (n + 6)(n - 2) \)
Solution
Sure, let's factor each quadratic expression completely:
1) **\( b^{2} + 8b + 7 \)**
We need two numbers that multiply to **+7** and add to **+8**.
The numbers are **7** and **1**.
**Factored form:** \( (b + 7)(b + 1) \)
2) **\( n^{2} - 11n + 10 \)**
We need two numbers that multiply to **+10** and add to **-11**.
The numbers are **-10** and **-1**.
**Factored form:** \( (n - 10)(n - 1) \)
3) **\( m^{2} + m - 90 \)**
We need two numbers that multiply to **-90** and add to **+1**.
The numbers are **10** and **-9**.
**Factored form:** \( (m + 10)(m - 9) \)
4) **\( n^{2} + 4n - 12 \)**
We need two numbers that multiply to **-12** and add to **+4**.
The numbers are **6** and **-2**.
**Factored form:** \( (n + 6)(n - 2) \)
**Summary of Factored Forms:**
1. \( (b + 7)(b + 1) \)
2. \( (n - 10)(n - 1) \)
3. \( (m + 10)(m - 9) \)
4. \( (n + 6)(n - 2) \)
Reviewed and approved by the UpStudy tutoring team
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Explain 
Simplify this solution Mind Expander
To factor these expressions completely, let's take a closer look at each one: 1) \( b^{2}+8b+7 \) can be factored as \( (b+1)(b+7) \). This is because 1 and 7 are the two numbers that multiply to 7 and add up to 8. 2) \( n^{2}-11n+10 \) factors into \( (n-1)(n-10) \), with -1 and -10 being the numbers that multiply to 10 and add to -11. 3) For \( m^{2}+m-90 \), it factors to \( (m-9)(m+10) \) because -9 and +10 multiply to -90 and combine to give +1. 4) Lastly, \( n^{2}+4n-12 \) factors into \( (n+6)(n-2) \). Here, 6 and -2 multiply to -12 and add up to 4. So the complete factorizations are: 1) \( (b+1)(b+7) \) 2) \( (n-1)(n-10) \) 3) \( (m-9)(m+10) \) 4) \( (n+6)(n-2) \)
