Polynomial: degree 3, zeroes -1/2, 2, 3, leading coeff 12

[thankyou77]

I don't know how to start this problem, any help would be appreciated!

Find a polynomial of degree 3 with constant coefficient 12 and zeros (-1/2), 2, and 3.

 

[Deleted member 4993]

Re: Finding Polynomials

I don't know how to start this problem, any help would be appreciated!

Find a polynomial of degree 3 with constant coefficient 12 and zeros (-1/2), 2, and 3.

To start -

The general form of a polynomial of degree 2 with zeroes at 'a' and 'b' is:

f(x) = A (x-a)(x-b)

The general form of a polynomial of degree 3 with zeroes at 'a' and 'b' and 'c' is:

f(x) = ???

Please show your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[fasteddie65]

"Find a polynomial of degree 3 with constant coefficient 12 and zeros (-1/2), 2, and 3."

Using the ideas of the previous submission, the factors of the polynomial are x + 1/2, x - 2, and x - 3.

a(x + 1/2)(x - 2)(x - 3) = a(x + 1/2)(x^2 - 5x + 6) = a(x^3 + x^2/2 - 5x^2 - 5x/2 + 6x + 3) = a(x^3 - 9x^2/2 + 7x/2 + 3)

The constant term is 12, so use a = 4.

4x^3 - 18x^2 + 14x + 12 is the polynomial.

 

[mmm4444bot]

... Using the ideas of the previous submission, ...

Hey Fast Eddie:

Thank you for showing your work. Here's a challenge question.

Is the following equation true or false?

thankyou77 = fasteddie65

Cheers,

~ Mark :twisted: