Please help, I'm really confused.
2007-01-14 07:54:10 · 16 answers · asked by coffeekeepsmeawake88 2 in Science & Mathematics ➔ Mathematics
5t²+5t+10=5(t²+t+2)
2007-01-14 07:58:19 · answer #1 · answered by A 150 Days Of Flood 4 · 2⤊ 0⤋
5 is a common factor of each term.
5 ( t^2 + t + 2)
If you multiply '5' into each term you will come back to the original statement:-
5 x t^2 = 5t^2
5 x t = 5t
5 x 2 = 10
The term remaining in the brackets CANNOT be factorised further.
So :- 5(t^2 + t + 2) is fully factorised.
2007-01-14 08:16:26 · answer #2 · answered by lenpol7 7 · 0⤊ 0⤋
5t^2 + 5t +10
this is in the form ax + by + c
where 5 is a, 5 is b and 10 is c.
when you factorise (putting this qudratic into 2 brackets etc), b is always the sum of the numbers that make c as their product. if a is a prime number you can just put it in the front bit of either bracket and then allocate numbers that make up b and c appropriately.
far as i can see, there is no way this quadratic can be put into 2 brackets and factorised, as there are no factors of 10 that add up to 5.
2007-01-17 07:06:41 · answer #3 · answered by ChristopheraX 4 · 0⤊ 0⤋
I suspect you've written the question down wrong - should it be:
5t² + 5t - 10?
In that case, first you can take out the common factor of 5:
5(t² + t - 2)
and then factorise out in your favourite way to give:
5(t-1)(t+2)
2007-01-14 21:56:38 · answer #4 · answered by robcraine 4 · 0⤊ 0⤋
5xtxt + 5t + 10=5(txt + t + 2) there is no more factors, but check your signs once again to make sure what sign 10 really has, if eventually you find out later that 10 has minus sign, then the solution would be thus 5(txt + t - 2) the factors of 2, which when added algebraically will give the coefficient of the middle number,t (which is 1) are -1 &+2 thus: 5(t -1)(t + 2)
2007-01-15 07:03:40 · answer #5 · answered by Reginald N 1 · 0⤊ 0⤋
5t² + 5t + 10
= 5(t² + t + 2)
Note that t² + t + 2 cannot be factorised further.
2007-01-14 14:40:25 · answer #6 · answered by Kemmy 6 · 0⤊ 0⤋
Hello Coffeekee!
Yes, it looks confusing, so don't worry! Watch!
With the two 5s and a 10, what is a common number divisible by all the numbers? 5 right? Now, let's take 5 into the whole deal!
5(t*2+t+2) okay? I hope this helps!
2007-01-14 08:06:12 · answer #7 · answered by bemeup 2 · 0⤊ 0⤋
5(t^2+t+2)
2007-01-14 08:46:09 · answer #8 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋
5t² + 5t + 10
t²+t+2
Take out the common factor which is 5
as you can divide each part by 5
2007-01-14 08:01:39 · answer #9 · answered by Anonymous · 0⤊ 0⤋
in the expression a=5; b=5 and c=10
b²-4ac= -175 [this is <0; or negative]
if this happens.. then the expression ''can not be factorised'' if you are in high school.
this same expression has imaginary roots.. which you can find if you are A level.. after you have learnt about complex numbers..
2007-01-14 08:31:13 · answer #10 · answered by joe 1 · 0⤊ 0⤋