Question

factorise : (5r+2/3)²-(2r-1/3)² using the identities (a+b)² and (a-b)²​

Answer 1

Step-by-step explanation:

(a+b)²=a²+b²+2ab

(a-b) ²=a²+b²-2ab

(5r+2/3)²=25r²+4/9+20r/3

(2r-1/3)²=4r²+1/9-4r/3

25r²-4r²+4/9-1/9+20r/3+4r/3=0

21r²+3/9+24r/3=0

21r²+1/3+8r=0

63r²+1+24r=0×3

63r²+24r+1=0

63r²+21r+3r+1=0

21r(3r+1)+1(3r+1)

(21r+1)(3r+1)

21r+1=0 21r=-1 r=-1/21

3r+1=0 3r=-1 r=-1/3

Answer 2

Answer:

$(5r+ \frac{2}{3} )^2 - ((2r - \frac{1}{3} )^2 )= 21r^2 + \frac{24r}{3} + \frac{1}{3}$

Step-by-step explanation:

Identities:

(a + b)^2 = a^2 + 2ab + b^2

(a - b)^2 = a^2 - 2ab + b^2

$(5r+ \frac{2}{3} )^2 = (5r)^2 + (2 \times 5r \times \frac{2}{3} ) + (\frac{2}{3} )^2\\= 25r^2 + 10 r\times \frac{2}{3} + \frac{4}{9} \\= 25r^2 + \frac{20r}{3} + \frac{4}{9}$

$(2r - \frac{1}{3} )^2 = (2r)^2 - (2 \times 2 r \times \frac{1}{3} ) + (\frac{1}{3} )^2\\= 4r^2 - 4 r\times \frac{1}{3} + \frac{1}{9} \\= 4r^2 - \frac{4r}{3} + \frac{1}{9}$

$(5r+ \frac{2}{3} )^2 - ((2r - \frac{1}{3} )^2 )\\= (25r^2 + \frac{20r}{3} + \frac{4}{9}) - (4r^2 - \frac{4r}{3} + \frac{1}{9} )\\= 25r^2 + \frac{20r}{3} + \frac{4}{9} - 4r^2 + \frac{4r}{3} - \frac{1}{9} \\= 25r^2 - 4r^2 + \frac{20r}{3} + \frac{4r}{3}+ \frac{4}{9} - \frac{1}{9}\\= (25 - 4)r^2 + \frac{20r + 4r}{3} + \frac{4 - 1}{9} \\= 21r^2 + \frac{24r}{3} + \frac{3}{9} \\= 21r^2 + \frac{24r}{3} + \frac{1}{3}$