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:
8 portraits are going to be used in the display.
Step-by-step explanation:
This problem can be solved as a rule of three problem.
Each yd has 36 inches.
The display is to span a total width of 10 yd. So it is a total width of 360 inches.
Each portrait has a width of 45 in. How many portraits can there be?
1 portrait - 45 in
x portraits - 360 in
8 portraits are going to be used in the display.
Answer:
Step-by-step explanation: d I think
Answer:
14x²-12y²
Step-by-step explanation:
Answer:
68°
Step-by-step explanation:
PRQ = SRQ gives:
3x-8 = 2x+6
x = 6+8 = 14
So PRQ = SRQ = 34
and
PRS = PRQ+SRQ = 68
