Nataliya D. answered • 10/21/13

Patient and effective tutor for your most difficult subject.

^{4}- 7x

^{2}- 18 ......

*(1)*

To find the zeros of this polynomial function we need to set f(x) equal to 0 and solve it for "x" .

Assume that x

^{2}= y , then

y

^{2}- 7y - 18 = 0 .......

*(2)*

Let's factor the equation

*(2)*

- 18 = (- 9) * 2

- 7 = - 9 + 2

(y + 2)(y - 9) = 0

y

_{1}= 9 ----> x

^{2}= 9 -----> x

_{12}= ±√9 = ± 3

y

_{2}= - 2 -----> x

^{2}= - 2 < 0 this equation does not have real roots.

Thus, given polynomial function has two

*real zeros*

**±3**Nataliya D.

"To bound on the real zeros of the polynomial function" means to find an interval where graph of function intercepts the x-axis.

For this method, the leading coefficient must be "1" (that what we have).

Let's write down all the coefficients: 1, - 7, - 18.

Let's drop the leading coefficient and remove the minus signs: 7, 18.

BOUND 1 is the largest value plus 1 is (18 + 1) = 19

BOUND 2 is sum of all values, that is (7 + 18) = 25

My apology once again :)

10/21/13

Nataliya D.

-19, 19

10/21/13

Bella B.

10/21/13

Bella B.

10/21/13