Find all roots of the function

Question

f(x)=x³+5x²+9x+9

Answer 1

It's a cubic polynomial, so it probably has only 1 real root, or possibly 3 real roots

There are no positive roots, since any positive x will make the cubic positive

The real root(s) have to be negative

try -1, -2, -3, easy integers. x=-3 works, Divide the cubic by x+3 and get a quadratic polynomial: x^2+2x+3. It has a negative discriminate, so you know it has no real roots. Use the quadratic formula to solve for 2 imaginary roots, a conjugate pair.

Answer 2

The general form of a cubic expression is f(x)=ax³+bx²+cx+d. To factor this, we need to look at our d-value as one of its factors will always be a root of the expression. In this case, our d-value is 9. So, what are the factors of 9? (or what is 9 divisible by?) It should be 1,3,9 and their negative counterparts.

Given these factors, we can apply the factor theorem to plug these numbers into our original expression because at least one of them should be a root, so it should equal out to 0. Let's do them one by one, first positive, then negative:

f(1)=(1)³+5(1)²+9(1)+9 = 24, so not a root

f(-1)=(-1)³+5(-1)²+9(-1)+9 = 4, so not a root

f(3) = (3)³+5(3)²+9(3)+9 = 108, so not a root

f(-3)=(-3)³+5(-3)²+9(-3)+9 = 0, so this is our first root

We know now that x=-3 is one of our roots, so we need to divide the term (x+3) out of the original expression through synthetic division. (You may want to refresh yourself on that process, it is very important for higher degree polynomials) After dividing, our expression should now look like this: (x+3)(x²+2x+3).

To find the last two roots, we must use the quadratic we now have in the expression: x²+2x+3. We can plug this into the quadratic formula to find our last two roots. (Again, another process you should be familiar with when working with factoring). We should find that our last two roots are imaginary: -1±√(2)i.

This means that the roots for our function are -3, -1+√(2)i, -1-√(2)i. Let me know if you have any questions!