3x^3-4x^2-12x+16 , find zeros
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Let p(x) = 3x3-4x2-12x+16
Put x = 2 in p(x)
p(2) = 3(2)3 – 4(2)2 – 12(2) + 16
= 3(8) – 4(4) – 24 + 16
= 0
Hence (x – 2) is a factor of p(x)
Now on dividing p(x) with (x – 2) we get the quotient as (3x2 + 2x – 8)
⇒ p(x) = (x – 2)(3x2 + 2x – 8)
= (x – 2)(3x2 + 6x – 4x – 8)
= (x – 2)[3x(x+2) – 4(x + 2)]
= (x – 2)(x + 2)(3x – 4)
Zeroes :-
x - 2 = 0
x = 2
,
x + 2 = 0
x = -2
,
3x - 4 = 0
3x = 4
x = 4/3
Therefore zeroes are :-
2 , -2 , 4/3 ( Ans )
Thank You
Let p(x) = 3x3-4x2-12x+16
Put x = 2 in p(x)
p(2) = 3(2)3 – 4(2)2 – 12(2) + 16
= 3(8) – 4(4) – 24 + 16
= 0
Hence (x – 2) is a factor of p(x)
Now on dividing p(x) with (x – 2) we get the quotient as (3x2 + 2x – 8)
⇒ p(x) = (x – 2)(3x2 + 2x – 8)
= (x – 2)(3x2 + 6x – 4x – 8)
= (x – 2)[3x(x+2) – 4(x + 2)]
= (x – 2)(x + 2)(3x – 4)
Zeroes :-
x - 2 = 0
x = 2
,
x + 2 = 0
x = -2
,
3x - 4 = 0
3x = 4
x = 4/3
Therefore zeroes are :-
2 , -2 , 4/3 ( Ans )
Thank You
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