how can i find the real and complex zeros of f(x)= x^5 + x^3 +2x^2 - 12x +8 ??

Question

thank you!

i dont understand how to use long division though...i mean i did but i kinda forgot. please tell me step by step how i am supposed to solve this.

Answer 1

By inspection, x=1 is a root

Divide f(x) by x-1 to get the remainder function of the order 4.

Try inspection again.

You find x=1 is a root again.

Divide again by x-1

The cubic polynomial can seen to have a root of x = -2 by using the Rational Root Theorem (very handy).

Divide by x+2

The quadratic that u end up with is x^2 + 4.

This has two roots +2i and - 2i.

Therefore there is four unique roots: 1, -2, 2i and -2i.

Answer 2

By inspection, I can see that x = 1 is a zero, so (x - 1) is a factor of the polynomial. Using polynomial long division to divide out this factor, I am left with x^4 + x^3 + 2x^2 + 4x - 8. By inspection of this polynomial, x = -2 is a zero, so (x + 2) is a factor. Using polynomial long division once again, I am left with x^3 - x^2 + 4x - 4, which I can factor on sight into (x -1)(x^2 + 4). So x = 1 is a double root and there is an additional real root of x = -2. The complex roots are 2i and -2i.

The method of polynomial long division is completely analagous to long division of numbers. The constant term is the "ones" digit. The x term is the "tens" digit. The x^2 term is the "hundreds" (or 10^2) digit. The x^3 term is the "thousands" (or 10^3) digit. Just remember to write in the terms with coefficients of zero. So the original polynomial becomes x^5 + 0x^4 + x^3 + 2x^2 - 12x + 8.

Answer 3

f(x) has only two sign changes so there are either 2 or 0 positive roots. f(-x) has only one sign change, so there is one negative root.

A quick check shows that x=1 is a root since. 1+1+2-12+8=0, so (x-1) is a factor. x=1 is one positive root and there must be one more positive root and it could also be 1 (a double root). If you take the derivative of f(x) and evaluate it at x =1, you find the derivative is 0 indicating that f(x) is tangent to the x=axis at x=1 so x=1 is indeeed a double root.

The negative root can also be found by inspection to be -2 since f(-2) = 0.

Thus we have found the roots x=1,1,-2. The remaining 2 roots must be imaginary. You can find them by dividing x^5+x^3+2x^2-12x+8 by (x-1)^2(x+2) which will give you a quadratic equation which you can solve using the quadratic formula. The quadratic equation you get will be x^2+4 = 0 so the other two roots are +2i and -2i which are pure imaginary numbers, not complex numbers.

So you can see how using DeCarte's law of signs helps you to quickly identify how many positive roots, negative roots and imaginary roots there are.

You should also use the fact that the possible roots are the factors of the constant term divided by the factors of the coefficient of x^5.
So possible factors are +/-(8/1, 4/1, 2/1, 1/1) which gives
the -2 and the +1 we found by inspection.

Answer 4

The easiest way is by Horner's rule: try divisors of last term 8, they are ±1, ±2, ±4, ±8 like this:
|1 0 1 2 -12 8
-----------------------
1|1 1 2 4 -8 0, so one root is 1;
1|1 2 4 8 0, so again 1 is a root; now seek only negative roots:
-2|1 0 4, so -2 is a root.
We reached the equation x^2 + 4 = 0 whose complex roots are ±2i, so the final answer is:
x1 = x2 = 1 (double root), x3 = -2, x4 = 2i, x5 = -2i.