find the zeroes of polynomial y3-7y+6.plz tell me..can we find it by applying sipliting method?
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Answer:1 , 2, -3... by spilliting the middle term
Step-by-step explanation:
y3-7y+6
splitting the middle term
y3-y-6y+6 as multiplication is 6 and 6+1 is 7
y(y2-1)-6(y-1)
y((y-1)(y+1)-6(y-1)
(y-1) common
(y-1)(y(y+1)-6)
(y-1)(y2+y-6)
(y-1)(y2+3y-2y-6) again splitting the middle term in second bracket expression
(y-1)(y(y+3)-2(y+3))
(y-1)((y-2)(y+3))
(y-1)(y-2)(y+3)
therefore by equating the three factors to 0,
zeroes will be 1,2,-3
hope you find it useful
akash12852:
sorry...it is too complicated method..i cannot understand
it can be solved by easier method..but you specifically asked with the method of splitting the middle term..so..i answered for that..do u need the another method?
Let p (y) = y3 — 7y + 6 putting y = 1, we get, p(l) = (1)t — 7 x 1 + 6 p(l) = 1 — 7 + 6 = 0 Since, p (1) = 0 then the zero of polynomial = 1
put y = 2 in p (y), we get p (2) = (2)t — 7 (2) — 6 = 8 — 14 + 6 = 0 So, y = 2 is a zero of p (y). put y = — 3 in p (y), we get p (-3) = (-3)3 — 7 (-3) + 6 = —27 + 21 + 6 =
So, y = — 3 is a zero of p (y). Hence the required zeroes of p (y) are : 1, 2 and —3
put y = 2 in p (y), we get p (2) = (2)t — 7 (2) — 6 = 8 — 14 + 6 = 0 So, y = 2 is a zero of p (y). put y = — 3 in p (y), we get p (-3) = (-3)3 — 7 (-3) + 6 = —27 + 21 + 6 =
So, y = — 3 is a zero of p (y). Hence the required zeroes of p (y) are : 1, 2 and —3
Let p (y) = y3 — 7y + 6
putting y = 1, we get,
p(l) = (1)t — 7 x 1 + 6 p(l) = 1 — 7 + 6 = 0
Since, p (1) = 0 then the zero of polynomial = 1
put y = 2 in p (y),
we get p (2) = (2)t — 7 (2) — 6
= 8 — 14 + 6 = 0
So, y = 2 is a zero of p (y).
put y = — 3 in p (y),
we get p (-3) = (-3)3 — 7 (-3) + 6
= —27 + 21 + 6 = 0
So, y = — 3 is a zero of p (y).
Hence the required zeroes of p (y) are : 1, 2 and —3
putting y = 1, we get,
p(l) = (1)t — 7 x 1 + 6 p(l) = 1 — 7 + 6 = 0
Since, p (1) = 0 then the zero of polynomial = 1
put y = 2 in p (y),
we get p (2) = (2)t — 7 (2) — 6
= 8 — 14 + 6 = 0
So, y = 2 is a zero of p (y).
put y = — 3 in p (y),
we get p (-3) = (-3)3 — 7 (-3) + 6
= —27 + 21 + 6 = 0
So, y = — 3 is a zero of p (y).
Hence the required zeroes of p (y) are : 1, 2 and —3
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