t3 -2t2 - 15t
factorise it....
Answers
Answered by
15
t3 - 2t - 15t = 0
Taking t common
t (t2 - 2t - 15) = 0
By splitting the middle term
t {t2 - (- 3t + 5t) - 15} = 0
t (t2 + 3t - 5t - 15) = 0
t {t(t + 3) - 5(t + 3)} = 0
t (t + 3)(t - 5) = 0
⇒ t = - 3, 0, 5
Verification:
Sum of the zeroes = - (coefficient of x2) ÷ coefficient of x3
α + β + γ = - b/a
(0) + (- 3) + (5) = - (- 2)/1
= 2 = 2
Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x3
αβ + βγ + αγ = c/a
(0)(- 3) + (- 3) (5) + (0) (5) = - 15/1
= - 15 = - 15
Product of all the zeroes = - (constant term) ÷ coefficient of x3
αβγ = - d/a
(0)(- 3)(5) = 0
0 = 0
Hope it helps!! ❤✌
Answered by
2
1) take t as common I.e (t^2 -2t-15)t=0
2)t=0 or t^2-2t-15=0
3) t^2-5t+3t-15=0
4)t(t-5)+3(t-5)=0
5)(t+3),(t-5)=0
6)t=-3 & t=5
is the ans ....
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