5t^2+28t+32
2007-06-19 15:12:22 · 5 answers · asked by Leo 1 in Science & Mathematics ➔ Mathematics
5t^2 + 28t + 32
= 5t^2 + 20t + 8t + 32
= 5t(t+4) + 8(t+4)
= (t+4)(5t+8)
2007-06-19 15:15:55 · answer #1 · answered by gudspeling 7 · 1⤊ 0⤋
Find 2 numbers when times together = 32 x 5 = 160, and when added is 28. The two numbers are 20 and 8.
Now we can break up the middle term:
5t^2 + 20t + 8t + 32
(5t^2 + 20t) + (8t + 32)
5t (t + 4) + 8 (t + 4)
(t + 4)(5t + 8)
So t = -4, or t = -8/5
2007-06-19 16:24:04 · answer #2 · answered by Eleckid 2 · 0⤊ 1⤋
5t^2+28t+32=0
(5t+8)(t+4)=0
t= -8/5
= -1&3/5
t= -4
2007-06-19 16:47:14 · answer #3 · answered by curiosity2414 1 · 0⤊ 1⤋
5t^2+28t+32
(5y+8)(t+4)
2007-06-19 15:26:13 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
Factors of 5t² are fixed as 5t and t:-
(5t-----).(t------)
Must now find factors of 32 that combine with 5t and t to give 28t.
Try 8 and 4:-
(5t + 8).(t + 4) is correct.
2007-06-19 22:56:52 · answer #5 · answered by Como 7 · 0⤊ 0⤋