Question

Determine the equation of this degree 5 polynomial? Much appreciated

Answer 1

From the graph we can see

There are 3 exact zeroes

x = -4

x = -1

x = 1

Then there are 3 factors

(x + 4), (x + 1), (x - 1)

Then we will let the other 2 factors are (x - a) and (x - b)

Then we will use the exact vertices (2, 3) and (4, -3) to find a and b

$y=(x+4)(x+1)(x-1)(x-a)(x-b)$

We will start by multiplying (x - 1) by (x + 1)

$(x-1)(x+1)=x^2-1$

Now, multiply the answer by (x + 4)

$(x^2-1)(x+4)=x^3+4x^2-x-4$

Multiply this answer by (x - a)

$\begin{gathered} (x-a)(x^3+4x^2-x-4)=x^4+4x^3-x^2-4x-ax^3-4ax^2+ax+4a= \\ x^4+(4-a)x^3-(1+4a)x^2+(a-4)x+4a \end{gathered}$

Then the polynomial is

$y=[x^4+(4-a)x^3-(1+4a)x^2+(a-4)x+4a](x-b)$

Substitute y by 3 and x by 2 (one of the vertex (2, 3))

$\begin{gathered} 3=[16+(4-a)8-(1+4a)4+(a-4)2+4a](2-b) \\ 3=[16+32-8a-4-16a+2a-8+4a](2-b) \\ 3=[36-18a][2-b] \end{gathered}$

Multiply the 2 brackets

$\begin{gathered} 3=72-36b-36a+18ab \\ 36a+36b-18ab=69\rightarrow Divide\text{ by 3} \\ 12a+12b-6ab=23\rightarrow(1) \end{gathered}$

We will substitute y by -3 and x by 4 (the second vertex (4, -3)

$\begin{gathered} -3=[256+(4-a)64-(1+4a)16+(a-4)4+4a][4-b] \\ -3=[256+256-64a-16-64a+4a-16+4a][4-b] \\ -3=[480-120a][4-b] \end{gathered}$

Multiply the 2 brackets

$\begin{gathered} -3=1920-480b-480a+120ab \\ 480a+480b-120ab=1923 \\ 160a+160b-40ab=641\rightarrow(2) \end{gathered}$

Now we need to solve equations (1) and (2) to find a and b

$\begin{gathered} -\frac{40}{3}(12a)-\frac{40}{3}(12b)-\frac{40}{3}(-6ab)=-\frac{40}{3}(23) \\ -160a-160b+80ab=-\frac{920}{3}\rightarrow(3) \\ Add(2),(3) \\ 40ab=\frac{1003}{3} \\ \\ ab=\frac{1003}{120} \end{gathered}$

Then substitute the value of ab

$\begin{gathered} (x-a)(x-b)=x^2-bx-ax+ab \\ =(x^2-bx-ax+\frac{1003}{120}) \\ y=(x^3+4x^2-x-4)(x^2-ax-bx+\frac{1003}{120}) \end{gathered}$

Substitute x by -2 and y by 0 (y-intercept)

$\begin{gathered} 0=(-8+16+2-4)(4+(a+b)2+\frac{1003}{120}) \\ 0=6(4+2(a+b)+\frac{1003}{120}) \\ 0=24+12(a+b)+\frac{1003}{20} \\ a+b=-\frac{1483}{240} \end{gathered}$$y=(x^3+4x^2-x-4)(x^2+\frac{1483}{240}x+\frac{1003}{120})$

We can simplify it by multiplying the 2 brackets to get the final form of the equation

$y=x^5+\frac{2443}{240}x^4+\frac{1283}{40}x^3+\frac{5581}{240}x^2-\frac{1323}{40}x-\frac{1003}{30}$

This is the final answer of the polynomial of the 5th degree