19) 67x²+115x+48
20) (w+8)²
21) (d+11)²
22) (r+7)(r-7)
Explanation:
19) The area of the ceiling is given by (10x+9)(7x+7). Multiplying we have:
10x*7x+7*10x+9*7x+9*7
=70x²+70x+63x+63
=70x²+133x+63
The area of the skylight is given by (x+5)(3x+3). Multiplying we have:
x*3x+3*x+5*3x+5*3
=3x²+3x+15x+15
=3x²+18x+15
To find the remaining area of the ceiling we subtract:
(70x²+133x+63)-(3x²+18x+15)
=70x²+133x+63-3x²-18x-15
=67x²+115x+48
20) To factor, we want to find factors of c, 64, that sum to b, 16. 8*8=64 and 8+8=16; therefore we have:
(w+8)(w+8)
Since this is the same binomial twice, we write it as (w+8)²
21) Again we look for factors of c, 121, that sum to b, 22. 11*11=121 and 11+11=22, so:
(d+11)(d+11)
Since this is the same binomial twice, we have (d+11)²
22) Factors of -49 that sum to 0 (there is no r term, just r²): -7*7=-49 and -7+7=0, so:
(r-7)(r+7)