Question

Find all zeros of polynomial f(x)2x4-3x3-9x2+15x-5, if two of its zeroes are root 5 and root-5

Answer 1

Answer:

zeros of f(x) are $\sqrt5,-\sqrt5,\frac{1}{2},1$

Step-by-step explanation:

$f(x)=2x^4-3x^3-9x^2+15x-5$

Given zeros are $\sqrt5\:and\:-\sqrt5$

sum of the zeros

$=\sqrt5+(-\sqrt5)$

$=0$

product of the zeros

$=(\sqrt5)(-\sqrt5)$

$=-5$

corresponding polynomial is

$x^2-5$

Now,

$2x^4-3x^3-9x^2+15x-5=(x^2-5)(2x^2+px+1)$

Equating coefficients of x on both sides we get

$15=-5p$

$p=-3$

other quadratic polynomial is

$2x^2-3x+1$

$=2x^2-2x-x+1$

$=2x(x-1)-1(x-1)$

$=(2x-1)(x-1)$

corresponding zeros are $\frac{1}{2}\:and\:1$

Therefore, zeros of f(x) are $\sqrt5,-\sqrt5,\frac{1}{2},1$

Answer 2

Answer:

Step-by-step explanation:

f(x)=2x^4-3x^3-9x^2+15x-5

Given zeros are \sqrt5\:and\:-\sqrt5

sum of the zeros

=\sqrt5+(-\sqrt5)

=0

product of the zeros

=(\sqrt5)(-\sqrt5)

=-5

corresponding polynomial is

x^2-5

Now,

2x^4-3x^3-9x^2+15x-5=(x^2-5)(2x^2+px+1)

Equating coefficients of x on both sides we get

15=-5p

p=-3

other quadratic polynomial is

2x^2-3x+1

=2x^2-2x-x+1

=2x(x-1)-1(x-1)

=(2x-1)(x-1)

corresponding zeros are \frac{1}{2}\:and\:1

Therefore, zeros of f(x) are \sqrt5,-\sqrt5,\frac{1}{2},1

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