How to factor 20r(r)+27r-8
2007-03-09 03:57:09 · 8 answers · asked by Buddriver 2 in Science & Mathematics ➔ Mathematics
20r(r)+27r-8
=20r^2+27r-8
=20r^2+32t-5r-8
=4r(5r+8)-1(5r+8)
=(5r+8)(4r-1)
2007-03-09 04:06:44 · answer #1 · answered by alpha 7 · 1⤊ 1⤋
20r(r)+27r-8
=20r^2 +27r -8
=(4r-1)(5r+8)
2007-03-09 04:12:45 · answer #2 · answered by ironduke8159 7 · 1⤊ 0⤋
20 r^2 + 27r - 8
= (4r - 1) (5r + 8)
2007-03-09 04:06:45 · answer #3 · answered by schatz101 3 · 1⤊ 0⤋
20r(r)+27r-8= (4r-1)(5r+8)
2007-03-09 04:03:29 · answer #4 · answered by Masry_c777 2 · 1⤊ 0⤋
(5r+8)(4r-1)
Basically u want something that when multiplied gives you -8:
Combinations possible: (+/-) 8x1, 4x2
Then you want two numbers when multiplied gives you 20:
Combinations possible: 20x1, 10x2, 5x4
Then you want the combination of numbers which when you add them up using the numbers above will give you 27:
This gives us:
(+/-) 20x8 - 1x1 =158 x wrong x
(+/-) 20x1 - 8x1 = 12 x wrong x
(+/-) 20x4 - 1x2 = 78 x wrong x
(+/-) 20x2 - 1x4 = 96 x wrong x
(+/-) 10x8 - 2x1 = 78 x wrong x
(+/-) 10x1 - 2x8 = 2 x wrong x
(+/-) 10x4 -1x2 = 38 x wrong x
(+/-) 10x2 -1x4 = 16 x wrong x
(+/-) 5x8 - 4x1 = 36 x wrong x
(+/-) 5x1 - 4x8 = -27/27 x RIGHTx
(+/-) 5x4 - 4x2 = 12 x wrong x
(+/-) 5x2 - 4x4 = -6/6 x wrong x
So basically... Your combination has to have:
5,4,1 and 8... with 1 and 8 being the last two numbers
(5r+8) (4r-1)
The more you practice.. the easier it gets
2007-03-09 04:29:25 · answer #5 · answered by Yahya d 3 · 1⤊ 0⤋
easy!!
20r(squared)+27r-8
47r(cubed)-8
that's your answer
(i think that's factoring, or maybe its the distributive property.)
2007-03-09 04:04:55 · answer #6 · answered by Jackie 2 · 0⤊ 3⤋
(4r-1)(5r+8)
2007-03-09 04:19:04 · answer #7 · answered by Anonymous · 0⤊ 0⤋
= (5r + 8).(4r - 1)
2007-03-09 04:05:17 · answer #8 · answered by Como 7 · 1⤊ 1⤋