Question

find the zeroes of the polynomial 3x³ +10x² -9x - 4 if its one zero is 1 .

Answer 1

$3 {x }^{2} (x - 1) + 13x(x - 1) + 4(x - 1) = 0 \\ (x - 1)(3 {x}^{2} + 13x + 4) = 0 \\ (x - 1)(x + 4)(3x + 1) = 0 \\ x = 1 \\ x = - 4 \\ x = - \frac{1}{3}$

Answer 2

Answer:

All zeroes are 1 , -4 & -1/4

Step-by-step explanation:

let g(x) = 3x³ + 10x² - 9x - 4  and 1 is a zero

⇒ one factor of p(x) = ( x - 1 )

for other factors , divide g(x) by ( x - 1 )

On division , we get is 3x² + 13x + 4

⇒ g(x) = ( x - 1 ) ( 3x² + 13x + 4 )

          = ( x - 1 ) ( 3x² + 12x + x + 4 )

          = ( x - 1 ) [ 3x(x + 4) + (x + 4) ]

          = ( x - 1 ) ( x + 4 ) ( 3x + 1 )

For zeroes put g(x) = 0

⇒ ( x - 1 ) ( x + 4 ) ( 3x + 1 ) = 0

x + 4 = 0   & 3x + 1 = 0

x = -4   &  x = -1/4

Therefore, All zeroes of g(x) are 1 , -4 & -1/4