I like to be explicit in my divisions and explain what I'm dividing by, but yes it is good to say explicitly which factors could work, or you'll be trying them all!
In this example, the equation had 4 integer roots, but not every problem is like this. If there were imaginary roots present, the method of finding the roots through the rational zero theorem will work until we have a degree 2 polynomial. It is from there where we would use quadratic formula, because we cannot fact it any other way. If the number in the square root is negative, there cannot be real roots, seeing as you cannot square root a negative number; hence why it would be imaginary. In that case, you would then leave your factored form as from before your used the quadratic formula on the degree two polynomial, and if you were to take out the numbers from the brackets to find x, you would only include the factors that are not imaginary, as those are the only real roots. Good question
Nice simple explanation... thank yiu
I like to be explicit in my divisions and explain what I'm dividing by, but yes it is good to say explicitly which factors could work, or you'll be trying them all!
Try x = 1
(1) - (1) - 11 + 9 + 18 =/= 0
Try x = -1
(1) + (1) - 11 - 9 + 18 = 0, so x = -1 is a solution
Synthetic division:
- 1| 1 -1 -11 9 18
| -1 2 9 -18
---------
1 -2 -9 18
= (x + 1)(x^3 - 2x^2 - 9x + 18)
New cubic: x^3 - 2x^2 - 9x + 18
Grouping:
= x^2(x - 2) - 9(x - 2)
= (x^2 - 9)(x - 2)
Finally, the quadratic: x^2 - 9
Difference of squares:
(x + 3)(x - 3)
So the fully factored quartic is
x^4 - x^3 - 11x^2 + 9x + 18 =
(x + 1)(x - 2)(x + 3)(x - 3)
What if it has an imaginary solution?
In this example, the equation had 4 integer roots, but not every problem is like this. If there were imaginary roots present, the method of finding the roots through the rational zero theorem will work until we have a degree 2 polynomial. It is from there where we would use quadratic formula, because we cannot fact it any other way. If the number in the square root is negative, there cannot be real roots, seeing as you cannot square root a negative number; hence why it would be imaginary. In that case, you would then leave your factored form as from before your used the quadratic formula on the degree two polynomial, and if you were to take out the numbers from the brackets to find x, you would only include the factors that are not imaginary, as those are the only real roots.
Good question
@@professorbirman thanks for the explanation ❤️👌🏻
Thanks