5.
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Solving a 5th grade polynomial
We want to find the answer of the following polynomial:
We can see that the last term is -72
We want to find all the possible numbers that can divide it. Those are:
{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}
We want to factor this polynomial in order to find all the possible x-values. In order to factor it we will have to find some binomials that can divide it using the set of divisors of -72.
We know that if
(x - z) is a divisor of this polynomial then z might be a divisor of the last term -72.
We will verify which is a divisor using synthetic division. If it is a divisor then we can factor using it:
Let's begin with
(x-z) = (x - 1)
We want to divide
Using synthetic division we have that if the remainder is 0 it will be a factor
We can find the remainder by replacing x = z in the polynomial, when it is divided by (x - z). It is to say, that if we want to know if (x -1) is a factor of the polynomial we just need to replace x by 1, and see the result:
If the result is 0 it is a factor
If it is different to 0 it is not a factor
Replacing x = 1
If we replace x = 1, we will have that:
Then the remainder is not 0, then (x - 1) is not a factor.
Similarly we are going to apply this until we find factors:
(x - z) = (x + 1)
We replace x by -1:
Then, (x + 1) is a factor.
Using synthetic division we have that:
Then:
Now, we want to factor the 4th grade polynomial.
Let's remember our possibilities:
{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}
Since we verified ±1, let's try with ±2 as we did before.
(x - z) = (x - 2)
We want to divide:
We replace x by z = 2:
Then (x - 2) is a factor. Let's do the synthetic division:
Then,
Then, our original polynomial is:
Now, let's prove if (x +2) is a factor, using the new 3th grade polynomial.
(x - z) = (x + 2)
We replace x by z = -2:
Since the remainder is not 0, (x +2) is not a factor.
All the possible cases are:
{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}
let's prove with +4
(x - z) = (x + 4)
We want to divide:
Let's replace x by z = -4 in order to find the remainder:
Then (x + 4) is a factor. Let's do the synthetic division:
Then,
Since
x² + 9 cannot be factor, we have completed our factoring:
Now, we have the following expression:
Then, we have five posibilities:
(x - 1) = 0
or (x - 2) = 0
or (x + 4) = 0
or (x² + 9) = 0
Then, we have five solutions;
x - 1 = 0 → x₁ = 1
x - 2 = 0 → x₂ = 2
x + 4 = 0 → x₃ = -4
x² + 9 = 0 → x² = -9 → x = ±√-9 = ±√9√-1 = ±3i
→ x₄ = 3i
→ x₅ = -3i
<h2><em>Answer- the solutions of the polynomial are: x₁ = 1, x₂ = 2, x₃ = -4, x₄ = 3i and x₅ = -3i</em></h2>