13. $x^2 - 5x - 6 = (x - 6)(quad)$
14. $x^2 + 18x + 81 = (x + 9)(quad)$
15. $x^2 - 16x + 63 = (x - 9)(quad)$
16. $6x^2 - 41x + 30 = (6x - 5)(quad)$
17. $2x^2 - 3x - 27 = (2x - 9)(quad)$
18. $6x^2 - 53x - 70 = (6x + 7)(quad)$

Question

Accepted Answer

### Problem 13:
# Explanation:
## Step1: Recall factoring quadratic.
We know that $(x - 6)(x + a)=x^{2}+(a - 6)x-6a$. Compare with $x^{2}-5x - 6$. So, $-6a=-6$ gives $a = 1$, and $(a - 6)=-5$ (since $1-6=-5$). So the missing factor is $x + 1$.
## Step2: Verify by multiplying $(x - 6)(x + 1)$.
$(x - 6)(x + 1)=x^{2}+x-6x - 6=x^{2}-5x - 6$, which matches the left - hand side.
# Answer: $x + 1$

### Problem 14:
# Explanation:
## Step1: Recall perfect square trinomial or factoring.
We know that $(x + 9)(x + b)=x^{2}+(9 + b)x+9b$. Compare with $x^{2}+18x + 81$. So, $9b = 81$ gives $b = 9$, and $9 + b=18$ (since $9 + 9 = 18$). So the missing factor is $x + 9$.
## Step2: Verify by multiplying $(x + 9)(x + 9)$.
$(x + 9)(x + 9)=x^{2}+9x+9x + 81=x^{2}+18x + 81$, which matches the left - hand side.
# Answer: $x + 9$

### Problem 15:
# Explanation:
## Step1: Recall factoring quadratic.
We know that $(x - 9)(x - c)=x^{2}-(9 + c)x+9c$. Compare with $x^{2}-16x + 63$. So, $9c = 63$ gives $c = 7$, and $-(9 + c)=-16$ (since $-(9 + 7)=-16$). So the missing factor is $x - 7$.
## Step2: Verify by multiplying $(x - 9)(x - 7)$.
$(x - 9)(x - 7)=x^{2}-7x-9x + 63=x^{2}-16x + 63$, which matches the left - hand side.
# Answer: $x - 7$

### Problem 16:
# Explanation:
## Step1: Recall factoring quadratic with leading coefficient.
We know that $(6x - 5)(x - d)=6x^{2}-(6d + 5)x+5d$. Compare with $6x^{2}-41x + 30$. So, $5d = 30$ gives $d = 6$, and $-(6d + 5)=-41$ (since $-(6\times6 + 5)=-(36 + 5)=-41$). So the missing factor is $x - 6$.
## Step2: Verify by multiplying $(6x - 5)(x - 6)$.
$(6x - 5)(x - 6)=6x^{2}-36x-5x + 30=6x^{2}-41x + 30$, which matches the left - hand side.
# Answer: $x - 6$

### Problem 17:
# Explanation:
## Step1: Recall factoring quadratic with leading coefficient.
We know that $(2x - 9)(x + e)=2x^{2}+(2e - 9)x-9e$. Compare with $2x^{2}-3x - 27$. So, $-9e=-27$ gives $e = 3$, and $2e - 9=-3$ (since $2\times3-9 = 6 - 9=-3$). So the missing factor is $x + 3$.
## Step2: Verify by multiplying $(2x - 9)(x + 3)$.
$(2x - 9)(x + 3)=2x^{2}+6x-9x - 27=2x^{2}-3x - 27$, which matches the left - hand side.
# Answer: $x + 3$

### Problem 18:
# Explanation:
## Step1: Recall factoring quadratic with leading coefficient.
We know that $(6x + 7)(x - f)=6x^{2}-(6f - 7)x-7f$. Compare with $6x^{2}-53x - 70$. So, $-7f=-70$ gives $f = 10$, and $-(6f - 7)=-53$ (since $-(6\times10 - 7)=-(60 - 7)=-53$). So the missing factor is $x - 10$.
## Step2: Verify by multiplying $(6x + 7)(x - 10)$.
$(6x + 7)(x - 10)=6x^{2}-60x+7x - 70=6x^{2}-53x - 70$, which matches the left - hand side.
# Answer: $x - 10$

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QUESTION IMAGE

Question

Explanation:

Problem 13:

Step1: Recall factoring quadratic.

Step2: Verify by multiplying \((x - 6)(x + 1)\).

Step1: Recall perfect square trinomial or factoring.

Step2: Verify by multiplying \((x + 9)(x + 9)\).

Step1: Recall factoring quadratic.

Step2: Verify by multiplying \((x - 9)(x - 7)\).

Answer:

Problem 14: