Notice that 25r²+20r+4 is a perfect square trinomial because it can be written in the form a²+2ab+b2, where a is 5r and b is 2.

a²+2ab+b²

(5r)²+25r2+2²

25r²+20r+4

Now use the formula for factorizing perfect square trinomials.

a²+2ab+b² = (a+b)²

(5r)²+25r2+2² = (5r+2)²

25r²+20r+4 = (5r+2)²

The factorised form of 25r²+20r+4 is (5r+2)².

Finally, check your work.

(5r+2)²

(5r+2)(5r+2) Expand

25r²+10r+10r+4 Apply the distributive property (FOIL)

25r²+20r+4

Yes, 25r²+20r+4=(5r+2)².

Notice that 9x²–1 is a difference of squares, because it can be written in the form a²–b², where a is 3x and b is 1.

a²–b²

(3x)²–1²

9x²–1

Now use the formula for factorising a difference of squares.

a²–b² = (a+b)(a–b)

(3x)²–1² = (3x+1)(3x–1)

9x²–1 = (3x+1)(3x–1)

The factorised form of 9x²–1 is (3x+1)(3x–1).

Finally, check your work.

(3x+1)(3x–1)

9x²+3x–3x–1 Apply the distributive property (FOIL)

9x²–1

Yes, 9x²–1=(3x+1)(3x–1).

Notice that 9g²–25 is a difference of squares, because it can be written in the form a²–b², where a is 3g and b is 5.

a²–b²

(3g)²–5²

9g²–25

Now use the formula for factorising a difference of squares.

a²–b² = (a+b)(a–b)

(3g)²–5² = (3g+5)(3g–5)

9g²–25 = (3g+5)(3g–5)

The factorised form of 9g²–25 is (3g+5)(3g–5).

Finally, check your work.

(3g+5)(3g–5)

9g²+15g–15g–25 Apply the distributive property (FOIL)

9g²–25

Yes, 9g²–25=(3g+5)(3g–5).