Using the value a=-14 and b=6, find the non-integer roots of the equation f(x)=0 in the form p+√q, where p and q are integers.

Question

where f(x)= x³+x² +ax+b

Answer 1

Graph the equation!

3 is a real root.

Divide it out and solve the resulting quadratic.

If this is insufficient help, please contact me.

Answer 2

I must confess I once knew an easier way to solve cubic functions, but I can't seem to find it. I will go over the method used in this video here: https://www.youtube.com/watch?v=vUNgcN6MbjA

The first step, the rational roots theorem, is actually used to solve binomials quickly. It's the same math that tells us, given ax² + bx + c, that c is the product of the two roots and b is the sum of the two roots. You can test this by solving x² + 6x + 5 using the theorem in the video.

For the problem, we have x³ + x² - 14x + 6. The highest degree term has a coefficient of 1 and the lowest has a coefficient of 6. Using the rational roots theorem, we can find any root that is a rational integer. This doesn't work if all of the roots are non-integer terms. Luckily, this one has one root that is an integer.

Factoring out 6, we get (1,2,3,6). For 1 we get (1). Easy. Now we divide the factors of the lowest degree coefficient (6) by the factors of the highest degree coefficient (1). Since we're dividing by 1, the new coefficients are still (1, 2, 3, 6). Not so bad.

Now, we plug each one of those factors for x into f(x) and see if any of them make the equation equal 0

If x = 1, we get:

f(1) = 1 + 1 - 14 + 6 = -10

This does not equal 0. Let's try x = 3

f(3) = (3)³ + (3)² - 14(3) + 6 = 27 + 9 - 42 + 6 = 0

Yay! This means we have a root at positive 3, so one of our cubic function's root terms is the monomial (x-3).

(x-3) equals 0 when x = 3, so we need our monomial to be subtracting.

Now, let's just put in a place-holder binomial to see what we have left. We can do this because a cubic polynomial is the product of a monomial and a binomial.

(x-3) * (ax² + bx + c)

Cool. Well, we know the product of our monomial and binomial needs to be f(x), right? That's our product that's given to us in the question and what we should get when we multiply all the terms out. So,

(x-3) * (ax² + bx + c) = x³ + x² - 14x + 6

We just made an equality we can use to solve this thing! If we multiply out our two terms on the left side of the equation, we can make a system of equations using the like terms.

ax³ + (-3ax² + bx²) + (-3bx + cx) - 3c = x³ + x² - 14x + 6

We can now make four equations out of this:

1) ax³ = x³

2) -3ax² + bx² = x²

3) -3bx + cx = -14x

4) -3c = 6

Cancelling out the x³ terms and solving Equation 1 gives us a = 1

Cancelling out the x² terms and plugging a = 1 into Equation 2 gives us b = 4

Cancelling out the x terms and plugging b = 4 into Equation 3 gives us c = -2

Plugging in c = -2 verifies Equation 4

Now we can take all that information and make ourselves a polynomial with all the coefficients:

(x-3)(x² + 4x - 2)

Gee whiz that was a lot of work, right? We're not done yet, though. One last thing. You can now factor out our binomial term using the quadratic equation to get the p√q terms that you need.

Quadratic Equation: (-b ± √(b² - 4ac)) ÷ 2a

Plugging in our a, b, and c, we get:

(-4 ± √(16- 8)) ÷ 2 = -2 ± √24

There are your non-integer roots, where p = -2 and q = 24

If anyone else has an easier way, please speak up