how do I solve this problem Factor t2-3t-18?

Answer 1

FIRST: Find two numbers that when multiplied equal -18, and add up to -3 (that is, the coeficients of the second term* and the third term*).

Answer = 3 and -6. Because (3)(-6) = -18 and (3)+(-6)=-3

THEN: Multiply two binomials of the form (x + a) (x + b), where "x" will be the square root of the first term* of the original equation, and "a" and "b", will be 3 and -6 (respectively, or the other way around, it doesn't matter)


FACTORS: (t + 3) ( t - 6) or (t - 6) (t + 3)


PROOF:
(t + 3) (t - 6) = t² - 6t + 3t -18 = t² - 3t - 18....or
(t - 6) (t + 3) = t² + 3t - 6t -18 = t² - 3t - 18.



*TERM: Each part of the equation separated by a plus or minus sign.

Answer 2

Simple...

you can rewrite the equation as t^2 + 3t - 6t - 18
Now take out the common factors from each pair
=====> t (t+3) - 6(t+3)
=====> (t+3) (t-6)
=====> so t can either = -3 or +6

GoodLuck.

Answer 3

If the factors are not obvious, the easy way is to equate it to zero and use the quadratic formula; if this gives roots r1 and r2, the factors are (t - r1)(t - r2). In this case, we want two numbers whose product is 18 and difference is 3, and it is obvious by inspection that 3 and 6 qualify. It is only slightly more work to go from there to the factorization (t - 6)(t + 3).

Answer 4

the key here, if is factorable, the factor of 18 are as follows.

1*18
2*9
3*6
and the only ones that add up, algebraically to the middle coefficients are 3 and 6..
such as 3-6 =-3
so you can write
(x-6)(x+3)

as you can see,when you multiply them, you should get back your pruduct.. above..

hope that explanation helps..

Mr. C

Answer 5

do you mean t^2-3t-18?
t -6 =0
t +3=0
so t=6 and -3
its kinda hard to explain it here
you got to learn the technique somewhere

Answer 6

This can be written as
t2-3t-18
=t2-6t+3t-18
=t(t-6)+3(t-6)
=(t-6)(t+3)............answer

Answer 7

t^2-3t-18
(t-6)(t+3)
3*-6=-18
3+-6=-3

Answer 8

It's been forever, (years and years trusst me). But, here goes.
2t-3t-18=?
2t-3t=18
-t=18