Hi. To see if you can factorise this equation, which I'm guessing is meant to be t squared, you should use the quadratic formula which is b2-4ac.
in your question
ax2+bx+c (x is the same as t)
a = 1 as it is the co-efficient of t2 (the coefficient is the number in front of the t2)
b = -18 (this is because it is in front of the t)
c = 81
so (-18)2- 4*1*81 = 0
as this equals 0 this means that there is only one solution so when your factorise both of the brackets will contain the same thing.
so if you say what two numbers multiply together to get 81.
say 9*9
consider these numbers for adding together to the b value. if you add these you get 18. this is no the b value. therefore we can try -9 and -9. adding these together gives you -18 which is the b value and it works for the multiply part aswell as the minus cancel.
so from here we can say that
t2-18t+81 = (t-9)(t-9) or could be written as (t-9)2
hope this helps you.
19 is not a factor of 81. Factors of 81 are 1, 3, 9, 27, and 81
9x9=81. Proving that one of the factor trees for 81 is 9xp.
The greatest common factor of 52 and 81 is 1
81
The factors of 81 are: 1, 3, 9, 27, 81.
Type your answer here... 81
(m + 9)(m + 9)
4x2-36x+81 (2x-9)(2x-9)
(3X + 9) (X + 9)
It is 8(80b+81) when factored
(X-9)2
3(x + 3)(x2 - 3x + 9)
It can be. 81 is a factor of 162.
(2x - 9)(2x - 9)
19 is not a factor of 81. Factors of 81 are 1, 3, 9, 27, and 81
78 + 79 + 80 + 80 + 81 + 81 = 479
1,2,3,4,6,8,9,12,18,24,36,72 factor of 72 1,3,9,27,and 81 is factor of 81