Find a polynomial function of degree 3 with the given numbers as zeros.

Question

-5, sqrt3, -sqrt3

Answer 1

The general cubic with given roots r1,r2,r3 is given by:
y=k(x-r1)(x-r2)(x-r3)
where k is a constant.
I will leave it to you to complete the answer. You are welcome to post your answer for verification.

Answer 2

f(-5)= 1^3 +5(1)^2-3(1)-15 = -12.

f(sqrt3) = 1^3 +5(1)^2-3(1)-15 = -12

f(sqrt-3) = 1^3 +5(1)^2-3(1)-15 = -12

Are these correct?

Answer 3

Not quite. However, it was good that you verify the value of f(x) at the zeroes.

Let the roots r1=-5, r2=√3, r3=-√3.
The function should be equal to zero when the roots are substituted for x, i.e. f(-5)=0, f(sqrt3)=0 and f(-sqrt3)=0.
The desired function is:
f(x)=k(x-r1)(x-r2)(x-r3)
=k(x-(-5))(x-√3)(x--√3)
Now check that f(r1), f(r2) and f(r3) all equal zero.

Answer 4

What is r2 and r3 stand for? Not too sure how to check this. Can you show me please I got 10 more problems like this. Thanks!

Answer 5

Let the roots r1=-5, r2=√3, r3=-√3.

Answer 6

Sorry, problem of encoding.
Let the roots r1=-5, r2=√3, r3=-√3.
These are the roots given in the question.

Answer 7

Ok so the function is

f(x)=k(x-r1)(x-r2)(x-r3)
=k(x-(-5))(x-�ã3)(x--�ã3)

Now to check this?

Answer 8

You will need to multiply the terms out to get the polynomial your teacher needs.
The last two terms should give (x²-3) and the whole function is thus:
f(x)= kx&sup3 + 5kx² -3kx -15k

Answer 9

In fact, since the question requires "Find a polynomial...", you could leave out the factor k to get
f(x)=x³+5x²-3x-15