1. 8 + 45r – 18r2
a. (4 – 2r)(2 + 9r)
b. (4 + 18r)(2 – r)
c. (8 – 3r)(1 + 6r)
d. prime
2. 35y2 + 2y – 24
a. (7y + 6)(5y – 4)
b. (7y + 12)(5y – 2)
c. (35y – 24)(y + 1)
d. prime
2007-12-11 05:05:02 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
0 = 8 + 45r - 18r²
The middle term is + 45r
Find the sum of the missle term
Multiply the first term 8 times the last term 18 equals 144 and factor
Factors of 144
1 x 144
2 x 72
3 x 48. . .←. .Use these factors
4 x 36
6 x 24
8 x 18
9 x 16
12 x 12
+ 48 and - 3 satisfy the sum of the middle term
Insert + 48r and - 3y into the equation
0 = 8 + 45r - 18r²
0 = 8 + 48r - 3r - 18r²
Group factor
0 = (8 + 48r) - (3r - 18r²)
0 = 8(1 + 6) - 3r(1 + 6r)
0 = (8 - 3r((1 + 6r)
- - - - - - - -s-
2007-12-11 05:28:42 · answer #1 · answered by SAMUEL D 7 · 0⤊ 0⤋
i think that they are both prime???
just kidding!!!!!!!
Number one is C and number two is D
2007-12-11 05:14:10 · answer #2 · answered by TM 2 · 0⤊ 0⤋