Question

3x^3-4x^2-12x+16 , find zeros

Answer 1

Hi there

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

Answer 2

$3x^{3} - 4x \: ^{2} - 12x + 16 = 0 \\ {x}^{2} (3x - 4) - 4(3x - 4) = 0 \\ ( {x}^{2} - 4)(3x - 4) = 0 \\ {x}^{2} - 4 = 0 \\ {x}^{2} = 4 \\ x = ±2 \\x = 2 \: , \: - 2 \\ and \ \\ 3x - 4 = 0 \\ 3x = 4 \\ x = \frac{4}{3}$