Question

(n+1)(n-1)

Answer 1

The answers are the boxed ones in the photo

Answer 2

For this question, you have to use one of the common products:

•  (u + a) · (u + b)

Expand it by applying the distrubutive property:

(u + a) · (u + b) = (u + a) · u + (u + a) · b

(u + a) · (u + b) = u · u + a · u + u · b + a · b

Take out the common factor  u  from those terms in between, and you finally have

(u + a) · (u + b) = u² + (a + b) · u + ab    <———    use this formula.

The key is just plugging in the appropriate values for  u,  a  and  b  into the formula above. Observe:

————

a)  (n + 1) · (n – 1)    ———>    u = n,  a = 1,  b = –1.  So,

(n + 1) · (n – 1) = n² + (1 – 1) · n + 1 · (– 1)

(n + 1) · (n – 1) = n² + 0n – 1

(n + 1) · (n – 1) = n² – 1          ✔

—————

b)  (x + 6) · (x + 8)    ———>    u = x,  a = 6,  b = 8. So,

(x + 6) · (x + 8) = x² + (6 + 8) · x + 6 · 8

(x + 6) · (x + 8) = x² + 14x + 48          ✔

—————

c)  (k + 1) · (k + 6)    ———>  u = k,  a = 1,  b = 6.

(k + 1) · (k + 6) = k² + (1 + 6) · k + 1 · 6

(k + 1) · (k + 6) = k² + 7k + 6          ✔

—————

d)  (x + 7) · (x + 5)    ———>    u = x,  a = 7,  b = 5.

(x + 7) · (x + 5) = x² + (7 + 5) · x + 7 · 5

(x + 7) · (x + 5) = x² + 12x + 35          ✔

—————

e)  (p – 8) · (p – 1)    ———>    u = p,  a = – 8,  b = – 1.

(p – 8) · (p – 1) = p² + (– 8 – 1) · p + (– 8) · (– 1)

(p – 8) · (p – 1) = p² – 9p + 8          ✔


I hope this helps. =)