explain how to factorise i,j and n

Answer:

Explanation:

I).
(x+5)^2 - 16 = 0

It’s in a form of: a^2 - b^2 = (a-b)(a+b)

(x + 5)^2 - 4^2 = 0
(x + 5 - 4)(x + 5 + 4) = 0
(x + 1)(x + 9) = 0

J.
(x - 4)^2 - 9 = 0

Same form: a^2 - b^2 = (a-b)(a+b)

(x - 4)^2 - 3^2 = 0
(x - 4 - 3)(x - 4 + 3) = 0
(x - 7)(x - 1) = 0

n).
(2y + 7)^2 - y^2 = 0
(2y + 7 - y)(2y + 7 + y) = 0
(y + 7)(3y + 7) = 0

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jimrgrant1 jimrgrant1 · Answer 2 · 2020-09-03T10:22:16+02:00

Answer:

see explanation

Step-by-step explanation:

A difference of squares is factored in general as

a² - b² = (a - b)(a + b)

(i)

(x + 5)² - 16 ← is a difference of squares, that is

(x + 5)² - 4²

= (x + 5 - 4)(x + 5 + 4)

= (x + 1)(x + 9)

(j)

(x - 4)² - 9 ← is a difference of squares, that is

(x - 4)² - 3²

= (x - 4 - 3)(x - 4 + 3)

= (x - 7)(x - 1)

(n)

(2y + 7)² - y² ← is a difference of squares

= (2y + 7 - y)(2y + 7 + y)

=(y + 7)(3y + 7)