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i need to know how to do t²+8t+12=0 and x²+24x+144=0 i have to factor them and its been so long i forgot how.

2007-01-25 09:23:37 · 14 answers · asked by Anonymous in Science & Mathematics Mathematics

14 answers

I'll solve the 1st equation - you solve the sec :-)
Solve t² + 8t + 12 = 0...

First: multiply the 1st and 3rd coefficient to get "12." Find two numbers that give you "12" when multiplied and "8" when added/subtracted. The numbers are: 6 and 2

Sec: rewrite the equation and include the new middle coefficients...

t² + 6t + 2t + 12 = 0

*When you have 4 terms - group "like" terms and factor...

(t² + 6t) + (2t + 12) = 0
t(t + 6) + 2(t + 6) = 0
(t + 6)(t+2) = 0

Third: solve the two t-variable by setting the parenthesis to "0"...

a. t + 6 = 0
t + 6 - 6 = 0 - 6
t = - 6

b. t + 2 = 0
t + 2 - 2 = 0 - 2
t = - 2

Solutions: -2 and -6

2007-01-25 09:33:51 · answer #1 · answered by ♪♥Annie♥♪ 6 · 0 0

t²+8t+12 factors into (t+6)(t+2)
x²+24x+144 factors into (t+12)^2

To factor, you look at the factors of the last number, the one without a variable, and find a set that adds up to the second number. So, in you're fist example 6*2=12 and 6+2=8.

2007-01-25 09:28:41 · answer #2 · answered by Grace J 1 · 0 0

t² + 8t + 12
(t +6) (t +2)

x² + 24x + 144
(x +12)²

2007-01-25 09:29:17 · answer #3 · answered by Anonymous · 0 0

t^2 + 8t + 12 = (t + 6) (t + 2)
x^2 + 24x + 144 = (x + 12)^2

2007-01-25 09:28:26 · answer #4 · answered by feanor 7 · 0 0

You just look at them sort of guessing based on the terms.
(t+2)(t+6) = 0 = t^2 + 8t + 12

(x+12)(x+12) = (x+12)^2 = 0 = x^2 + 24x + 144

factor the costan term into numbers that add up to the multiplier of the x or t term (the raised to the one term).

2007-01-25 09:30:02 · answer #5 · answered by KingGeorge 5 · 0 0

t^2 + 8t +12
(t+6)(t+2)

x^2 +24x + 144
(x+12)(x+12)

2007-01-25 09:28:16 · answer #6 · answered by blni0602 1 · 0 0

(t+6)(t+2) = 0
(t+12)²=0

2007-01-25 09:27:51 · answer #7 · answered by bequalming 5 · 0 0

What two numbers multiply to 12 and add up to 8? 6 and 2

(t+2)(t+6)

What two numbers multiply to 144 and add up to 24? 12 and 12
(x+12)(x+12)=(x+12)^2

2007-01-25 09:27:46 · answer #8 · answered by Professor Maddie 4 · 1 0

just split the two straight into brackets:
they are both positive and must add together to make 8 and multiply to make 12


(t+6 ) (t+2 ) = 0

so therefore t = -6 or -2

for the second one, i would use the formula:
t= -b +/- the sqare root of b² - 2ac all over 4a

so

a = 1 b = 24 and c = 144

which works out as t = -72 or -2.33 something

i hope thats right and helps ;) x

2007-01-25 09:42:53 · answer #9 · answered by Lauren v 2 · 0 0

take 12 and what multiples that add up because 8 is positive give you 12 when they multiply and when they add give you 8...
6*2=12 and 6+2=8
so (t+6) (t+2) tats how you factor

same for 144
12*12=144 and 12+12=24
so (x+12) (x+12)

2007-01-25 09:32:15 · answer #10 · answered by dream 2 · 0 0

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