Question

find all the zeros of 2x^4 - 3x^3 - 3x^2 + 6x - 2 if it is given that two of its zeros are 1 and 1/2

Answer 1

GIVEN :

Find all the zeros of $2x^4 - 3x^3 - 3x^2 + 6x - 2$ if it is given that two of its zeros are 1 and $\frac{1}{2}$

TO FIND :

The zeros of $2x^4 - 3x^3 - 3x^2 + 6x - 2$ if it is given that two of its zeros are 1 and $\frac{1}{2}$

SOLUTION :

Given polynomial is $2x^4 - 3x^3 - 3x^2 + 6x - 2$

Also given that 1 and $\frac{1}{2}$ are the two zeros of the given polynomial.

By using the Synthetic Division we can solve it :

Since 1 and $\frac{1}{2}$ are the two zeros we have that

 1_| 2    -3       -3      6     -2

      0     2        -1     -4      2

     ____________________

$\frac{1}{2}$_| 2       -1       -4      2      0

    0        1        0     -2

    ___________________

    2       0       -4      0

Hence we have the quadratic equation $2x^2-4=0$

$2x^2=4$

$x^2=2$

$x=\pm \sqrt{2}$  

$x=\sqrt{2}$ and $x=-\sqrt{2}$

∴ $x=\sqrt{2}$ and $x=-\sqrt{2}$ are the other two zeros of the given polynomial.