It is (x-1)(x-8) when factored
x2 - 9x + 8 = (x-1)(x-8)
(x+1)(x+8)
x2-9x+8=0 has two solutions:x = 8x = 1
This statement is already nearly factored. The best we can do is pull out the 9 from each term:9x + 9y + 72 = 9(x + y + 8)
(9x + 8)(9x - 8)
Integral : 9x + 8 = 9(x squared)/2 + 8x Differential : 9x +8 = 9
x2 - 9x + 8 = (x-1)(x-8)
(x+1)(x+8)
x2-9x+8=0 has two solutions:x = 8x = 1
This statement is already nearly factored. The best we can do is pull out the 9 from each term:9x + 9y + 72 = 9(x + y + 8)
(x - 8)(x - 6)
the answer is (3x-2)(9x squared+6x+4)
It is: (c-4)(c-8) when factored
(9x + 8)(9x - 8)
n3 + 8n2 + 12n = n(n+2)(n+6)
(3x + 8)(3x - 8)
b^(2) + 8x + 7 Does NOT factor . Reason; you have two unknowns, viz. 'b' & 'x'. However, if you means b^(2) + 8b + 7 . Herewe have one unknown viz. 'b'. This factors to ( b + 7)(b + 1)