Write the equation, in standard form, of a polynomial of the least degree with integral coefficients that has the given zeros: 6, - 7/2

f

(

x

)

=

3

x

3

−

5

x

2

−

47

x

−

15

Explanation:

If the zero is c, the factor is (x-c).

So for zeros of

−

3

,

−

1

3

,

5

, the factors are

(

x

+

3

)

(

x

+

1

3

)

(

x

−

5

)

Let's take a look at the factor

(

x

+

1

3

)

. Using the factor in this form will not result in integer coefficients because

1

3

is not an integer.

Move the

3

in front of the x and leave the

1

in place:

(

3

x

+

1

)

.

When set equal to zero and solved, both

(

x

+

1

3

)

=

0

and

(

3

x

+

1

)

=

0

result in

x

=

−

1

3

.

f

(

x

)

=

(

x

+

3

)

(

3

x

+

1

)

(

x

−

5

)

Multiply the first two factors.

f

(

x

)

=

(

3

x

2

+

10

x

+

3

)

(

x

−

5

)

Multiply/distribute again.

f

(

x

)

=

3

x

3

+

10

x

2

+

3

x

−

15

x

2

−

50

x

−

15

Combine like terms.

f

(

x

)

=

3

x

3

−

5

x

2

−

47

x

−

15

use formula for the circle a=3.14

step-by-step explanation:

is there any photos or whatever?