Question

obtain all the zeros of the polynomial 3x^3+10x^2-9x-4 if one of its zero is 1

Answer 1

Answer:

All the zeros are 1, -4 and -1/3

Step-by-step explanation:

The zeros of the given polynomial is found by factorization method.

since x=1 is a zero of the polynomial,

we can write

$3x^3+10x^2-9x-4=(x-1)(3x^2+px+4)$

Equating coefficient of x on both sides

-9 = 4 - p

-13 = -p

p = 13

other quadratic factor is

$3x^2+13x+4\\\\=3x^2+12x+x+4\\\\=3x(x+4)+1(x+4)\\\\=(x+4)(3x+1)\\\\3x^2+13x+4=(x+4)(3x+1)$

other zeros are -4 and -1/3

Answer 2

Answer:

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

Step-by-step explanation:

We are given with one Zero of polynomial 3x³ + 10x² - 9x - 4

let say, p(x) = 3x³ + 10x² - 9x - 4 & zero = 1

Thus, one factor of p(x) = ( x - 1 )

We get another factor of p(x) by dividing it with x - 1

On division, quotient we get is 3x² + 13x + 4

⇒ p(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 p(x) = 0

⇒ ( x - 1 ) ( x + 4 ) ( 3x + 1 ) = 0

x + 4 = 0   & 3x + 1 = 0

x = -4   &  x = -1/4

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