Solving a 5th grade polynomial
We want to find the answer of the following polynomial:
![x^5+3x^4+3x^3+19x^2-54x-72=0](https://tex.z-dn.net/?f=x%5E5%2B3x%5E4%2B3x%5E3%2B19x%5E2-54x-72%3D0)
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
![\frac{(x^5+3x^4+3x^3+19x^2-54x-72)}{x-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%28x%5E5%2B3x%5E4%2B3x%5E3%2B19x%5E2-54x-72%29%7D%7Bx-1%7D)
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:
![\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ 1^5+3\cdot1^4+3\cdot1^3+19\cdot1^2-54\cdot1-72 \\ =1+3+3+19-54-72 \\ =-100 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E5%2B3x%5E4%2B3x%5E3%2B19x%5E2-54x-72%20%5C%5C%20%5Cdownarrow%20%5C%5C%201%5E5%2B3%5Ccdot1%5E4%2B3%5Ccdot1%5E3%2B19%5Ccdot1%5E2-54%5Ccdot1-72%20%5C%5C%20%3D1%2B3%2B3%2B19-54-72%20%5C%5C%20%3D-100%20%5Cend%7Bgathered%7D)
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:
![\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ (-1)^5+3\cdot(-1)^4+3\cdot(-1)^3+19\cdot(-1)^2-54\cdot(-1)-72 \\ =-1+3-3+19+54-72 \\ =0 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E5%2B3x%5E4%2B3x%5E3%2B19x%5E2-54x-72%20%5C%5C%20%5Cdownarrow%20%5C%5C%20%28-1%29%5E5%2B3%5Ccdot%28-1%29%5E4%2B3%5Ccdot%28-1%29%5E3%2B19%5Ccdot%28-1%29%5E2-54%5Ccdot%28-1%29-72%20%5C%5C%20%3D-1%2B3-3%2B19%2B54-72%20%5C%5C%20%3D0%20%5Cend%7Bgathered%7D)
Then, (x + 1) is a factor.
Using synthetic division we have that:
Then:
![x^5+3x^4+3x^3+19x^2-54x-72=(x+1)(x^4+2x^3+x^2+18x-72)](https://tex.z-dn.net/?f=x%5E5%2B3x%5E4%2B3x%5E3%2B19x%5E2-54x-72%3D%28x%2B1%29%28x%5E4%2B2x%5E3%2Bx%5E2%2B18x-72%29)
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:
![\frac{x^4+2x^3+x^2+18x-72}{x-2}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E4%2B2x%5E3%2Bx%5E2%2B18x-72%7D%7Bx-2%7D)
We replace x by z = 2:
![\begin{gathered} x^4+2x^3+x^2+18x-72 \\ \downarrow \\ 2^4+2\cdot2^3+2^2+18\cdot2-72 \\ =16+16+4+36-72 \\ =0 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E4%2B2x%5E3%2Bx%5E2%2B18x-72%20%5C%5C%20%5Cdownarrow%20%5C%5C%202%5E4%2B2%5Ccdot2%5E3%2B2%5E2%2B18%5Ccdot2-72%20%5C%5C%20%3D16%2B16%2B4%2B36-72%20%5C%5C%20%3D0%20%5Cend%7Bgathered%7D)
Then (x - 2) is a factor. Let's do the synthetic division:
Then,
![x^4+2x^3+x^2+18x-72=(x-2)(x^3+4x^2+9x+36)](https://tex.z-dn.net/?f=x%5E4%2B2x%5E3%2Bx%5E2%2B18x-72%3D%28x-2%29%28x%5E3%2B4x%5E2%2B9x%2B36%29)
Then, our original polynomial is:
![\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =\mleft(x+1\mright)\mleft(x^4+2x^3+x^2+18x-72\mright) \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E5%2B3x%5E4%2B3x%5E3%2B19x%5E2-54x-72%20%5C%5C%20%3D%5Cmleft%28x%2B1%5Cmright%29%5Cmleft%28x%5E4%2B2x%5E3%2Bx%5E2%2B18x-72%5Cmright%29%20%5C%5C%20%3D%28x-1%29%28x-2%29%28x%5E3%2B4x%5E2%2B9x%2B36%29%20%5Cend%7Bgathered%7D)
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:
![\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-2)^3+4(-2)^2+9(-2)+36 \\ =-8+16-18+36 \\ =26 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E3%2B4x%5E2%2B9x%2B36%20%5C%5C%20%5Cdownarrow%20%5C%5C%20%28-2%29%5E3%2B4%28-2%29%5E2%2B9%28-2%29%2B36%20%5C%5C%20%3D-8%2B16-18%2B36%20%5C%5C%20%3D26%20%5Cend%7Bgathered%7D)
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:
![\frac{x^3+4x^2+9x+36}{x+4}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E3%2B4x%5E2%2B9x%2B36%7D%7Bx%2B4%7D)
Let's replace x by z = -4 in order to find the remainder:
![\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-4)^3+4(-4)^2+9(-4)+36 \\ =-64+64-36+36 \\ =0 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E3%2B4x%5E2%2B9x%2B36%20%5C%5C%20%5Cdownarrow%20%5C%5C%20%28-4%29%5E3%2B4%28-4%29%5E2%2B9%28-4%29%2B36%20%5C%5C%20%3D-64%2B64-36%2B36%20%5C%5C%20%3D0%20%5Cend%7Bgathered%7D)
Then (x + 4) is a factor. Let's do the synthetic division:
Then,
![x^3+4x^2+9x+36=(x+4)(x^2+9)](https://tex.z-dn.net/?f=x%5E3%2B4x%5E2%2B9x%2B36%3D%28x%2B4%29%28x%5E2%2B9%29)
Since
x² + 9 cannot be factor, we have completed our factoring:
![\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \\ =(x-1)(x-2)(x+4)(x^2+9) \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E5%2B3x%5E4%2B3x%5E3%2B19x%5E2-54x-72%20%5C%5C%20%3D%28x-1%29%28x-2%29%28x%5E3%2B4x%5E2%2B9x%2B36%29%20%5C%5C%20%3D%28x-1%29%28x-2%29%28x%2B4%29%28x%5E2%2B9%29%20%5Cend%7Bgathered%7D)
Now, we have the following expression:
![(x-1)(x-2)(x+4)(x^2+9)=0](https://tex.z-dn.net/?f=%28x-1%29%28x-2%29%28x%2B4%29%28x%5E2%2B9%29%3D0)
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>