Find all zeroes of the polynomial  $3x^3\ +\ 10x^2\ -\ 9x\ –\ 4$, if one of its zeroes is 1.

Given:

Given polynomial is $3x^3\ +\ 10x^2\ -\ 9x\ –\ 4$ and one of its zeros is $1$.

To do:

We have to find all the zeros of the given polynomial.

Solution:

If $a$ is a zero of $f(x)$ then $(x-a)$ is a factor of $f(x)$.

Therefore,

$x-(1)=x-1$ is a factor of the given polynomial.

On applying the division algorithm,

Dividend$=3x^3 +10x^2 - 9x - 4$

Divisor$=x-1$

$x-1$)$3x^3+10x^2-9x-4$($3x^2+13x+4$

            $3x^3-3x^2$

           -----------------------------

                      $13x^2-9x-4$

                      $13x^2-13x$

                     -----------------------

                                    $4x-4$  

                                    $4x-4$

                                  --------------

                                          $0$  

Therefore,

Quotient$=3x^2+13x+4$

This implies,

$3x^3+10x^2-9x-4=(x-1)(3x^2+13x+4)$

To get the other zeros, put $3x^2+13x+4=0$.

$3x^2+12x+x+4=0$

$3x(x+4)+1(x+4)=0$

$(3x+1)(x+4)=0$

$3x+1=0$ or $x+4=0$

$3x=-1$ or $x=-4$

$x=-\frac{1}{3}$ or $x=-4$

All the zeros of the given polynomial are $-4$, $-\frac{1}{3}$ and $1$.

Tutorialspoint

Updated on: 10-Oct-2022

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