first off, let's notice the graph touches the x-axis at -1 and 3, namely, those are the zeros/solutions/roots of the polynomial and therefore, the factors come from those points.
now, at -1, the graph doesn't cross the x-axis, instead it <u>simply bounces off</u> of it, that means the zero of x = -1, has an even multiplicity, could be 4 or 2 or 6, but let's go with 2.
at x = 3, the graph does cross the x-axis, meaning it has an odd multiplicity, could be 3 or 1, or 7 or 9, but let's use 1.
![\bf \begin{cases} x=-1\implies &x+1=0\\ x=3\implies &x-3=0 \end{cases}~\hspace{5em}\stackrel{\textit{even multiplicity}}{(x+1)^2}\qquad \stackrel{\textit{odd multiplicity}}{(x-3)^1}=\stackrel{y}{0} \\\\\\ (x^2+2x+1)(x-3)=y\implies x^3+2x^2+x-3x^2-6x-3=y \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill x^3-x^2-5x-3=y~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%3D-1%5Cimplies%20%26x%2B1%3D0%5C%5C%20x%3D3%5Cimplies%20%26x-3%3D0%20%5Cend%7Bcases%7D~%5Chspace%7B5em%7D%5Cstackrel%7B%5Ctextit%7Beven%20multiplicity%7D%7D%7B%28x%2B1%29%5E2%7D%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bodd%20multiplicity%7D%7D%7B%28x-3%29%5E1%7D%3D%5Cstackrel%7By%7D%7B0%7D%20%5C%5C%5C%5C%5C%5C%20%28x%5E2%2B2x%2B1%29%28x-3%29%3Dy%5Cimplies%20x%5E3%2B2x%5E2%2Bx-3x%5E2-6x-3%3Dy%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20x%5E3-x%5E2-5x-3%3Dy~%5Chfill)
Answer:
Step-by-step explanation:
x 6 a
8 48 c
-4 b 20
Let the unknown numbers of the multiplication grid are a, b and c.
1). 6 × 8 = 48
2). (-4)×6 = b
b = -24
3). (-4) × a = 20
a = -5
4). 8 × a = c
8 × (-5) = c
c = -40
Therefore, missing in the given multiplication grid are,
x 6 -5
8 48 -40
-4 -24 20
When angles and sides are all equal
Answer:
uh just answer a question
Step-by-step explanation: